Why doesn't a const reference extend the life of a temporary object passed via a function? The 2019 Stack Overflow Developer Survey Results Are InWhy use “b < a ? a : b” instead of “a < b ? b : a” to implement max template?How come a non-const reference cannot bind to a temporary object?Does a const reference class member prolong the life of a temporary?How do I achieve the theoretical maximum of 4 FLOPs per cycle?Why not non-const reference to temporary objects?Temporary object and non-const referenceconst reference to temporary referenceC++11 scoping and lifetime of temporary bound to a (const) reference (GCC)Why am I using the deleted function 'void std::ref(const _Tp&&) [with _Tp = int]'Can a Second Const Reference Extend the Lifetime of a TemporaryExtending the life of a temporary object by getting a reference to a subobject

Flight paths in orbit around Ceres?

Why are there uneven bright areas in this photo of black hole?

How to display lines in a file like ls displays files in a directory?

Can you cast a spell on someone in the Ethereal Plane, if you are on the Material Plane and have the True Seeing spell active?

Falsification in Math vs Science

Inverse Relationship Between Precision and Recall

Ubuntu Server install with full GUI

Correct punctuation for showing a character's confusion

How do you keep chess fun when your opponent constantly beats you?

Relationship between Gromov-Witten and Taubes' Gromov invariant

Why couldn't they take pictures of a closer black hole?

How to notate time signature switching consistently every measure

How did passengers keep warm on sail ships?

Why does the nucleus not repel itself?

Getting crown tickets for Statue of Liberty

Button changing its text & action. Good or terrible?

Pokemon Turn Based battle (Python)

I am an eight letter word. What am I?

What is the most efficient way to store a numeric range?

Is it safe to harvest rainwater that fell on solar panels?

How much of the clove should I use when using big garlic heads?

Match Roman Numerals

Why isn't the circumferential light around the M87 black hole's event horizon symmetric?

What is preventing me from simply constructing a hash that's lower than the current target?



Why doesn't a const reference extend the life of a temporary object passed via a function?



The 2019 Stack Overflow Developer Survey Results Are InWhy use “b < a ? a : b” instead of “a < b ? b : a” to implement max template?How come a non-const reference cannot bind to a temporary object?Does a const reference class member prolong the life of a temporary?How do I achieve the theoretical maximum of 4 FLOPs per cycle?Why not non-const reference to temporary objects?Temporary object and non-const referenceconst reference to temporary referenceC++11 scoping and lifetime of temporary bound to a (const) reference (GCC)Why am I using the deleted function 'void std::ref(const _Tp&&) [with _Tp = int]'Can a Second Const Reference Extend the Lifetime of a TemporaryExtending the life of a temporary object by getting a reference to a subobject



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








38















In the following simple example, why can't ref2 be bound to the result of min(x,y+1)?



#include <cstdio>
template< typename T > const T& min(const T& a, const T& b) return a < b ? a : b ;

int main()
int x = 10, y = 2;
const int& ref = min(x,y); //OK
const int& ref2 = min(x,y+1); //NOT OK, WHY?
return ref2; // Compiles to return 0



live example - produces:



main:
xor eax, eax
ret









share|improve this question



















  • 9





    This program produces no output and exits with code 0. With -O3 optimization flag all the statements inside of main are discarded.

    – VTT
    Apr 8 at 7:10






  • 4





    What error does it give?

    – fredrik
    Apr 8 at 7:11






  • 6





    This is actually a very interesting question. I'll tag language lawyer and hope one of the chiefs picks this one up. I've neither the time nor the expertise. It's all to do with the fact that lifetime extension is not transitive, and where the original objects are located. Go on @StoryTeller.

    – Bathsheba
    Apr 8 at 7:16







  • 5





    @Bathsheba Could you quickly explain why b < a ? b : a that advantageous?

    – lubgr
    Apr 8 at 7:24






  • 5





    @Michiel The thing is, if the binding was direct (e.g. if min was returning by value), ref2 would extend the lifetime of the bound temporary to that of itself. Hence the question, I guess.

    – Angew
    Apr 8 at 7:26

















38















In the following simple example, why can't ref2 be bound to the result of min(x,y+1)?



#include <cstdio>
template< typename T > const T& min(const T& a, const T& b) return a < b ? a : b ;

int main()
int x = 10, y = 2;
const int& ref = min(x,y); //OK
const int& ref2 = min(x,y+1); //NOT OK, WHY?
return ref2; // Compiles to return 0



live example - produces:



main:
xor eax, eax
ret









share|improve this question



















  • 9





    This program produces no output and exits with code 0. With -O3 optimization flag all the statements inside of main are discarded.

    – VTT
    Apr 8 at 7:10






  • 4





    What error does it give?

    – fredrik
    Apr 8 at 7:11






  • 6





    This is actually a very interesting question. I'll tag language lawyer and hope one of the chiefs picks this one up. I've neither the time nor the expertise. It's all to do with the fact that lifetime extension is not transitive, and where the original objects are located. Go on @StoryTeller.

    – Bathsheba
    Apr 8 at 7:16







  • 5





    @Bathsheba Could you quickly explain why b < a ? b : a that advantageous?

    – lubgr
    Apr 8 at 7:24






  • 5





    @Michiel The thing is, if the binding was direct (e.g. if min was returning by value), ref2 would extend the lifetime of the bound temporary to that of itself. Hence the question, I guess.

    – Angew
    Apr 8 at 7:26













38












38








38


6






In the following simple example, why can't ref2 be bound to the result of min(x,y+1)?



#include <cstdio>
template< typename T > const T& min(const T& a, const T& b) return a < b ? a : b ;

int main()
int x = 10, y = 2;
const int& ref = min(x,y); //OK
const int& ref2 = min(x,y+1); //NOT OK, WHY?
return ref2; // Compiles to return 0



live example - produces:



main:
xor eax, eax
ret









share|improve this question
















In the following simple example, why can't ref2 be bound to the result of min(x,y+1)?



#include <cstdio>
template< typename T > const T& min(const T& a, const T& b) return a < b ? a : b ;

int main()
int x = 10, y = 2;
const int& ref = min(x,y); //OK
const int& ref2 = min(x,y+1); //NOT OK, WHY?
return ref2; // Compiles to return 0



live example - produces:



main:
xor eax, eax
ret






c++ language-lawyer temporary-objects






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 8 at 14:18









Boann

37.5k1290122




37.5k1290122










asked Apr 8 at 7:09









Khurshid NormuradovKhurshid Normuradov

492410




492410







  • 9





    This program produces no output and exits with code 0. With -O3 optimization flag all the statements inside of main are discarded.

    – VTT
    Apr 8 at 7:10






  • 4





    What error does it give?

    – fredrik
    Apr 8 at 7:11






  • 6





    This is actually a very interesting question. I'll tag language lawyer and hope one of the chiefs picks this one up. I've neither the time nor the expertise. It's all to do with the fact that lifetime extension is not transitive, and where the original objects are located. Go on @StoryTeller.

    – Bathsheba
    Apr 8 at 7:16







  • 5





    @Bathsheba Could you quickly explain why b < a ? b : a that advantageous?

    – lubgr
    Apr 8 at 7:24






  • 5





    @Michiel The thing is, if the binding was direct (e.g. if min was returning by value), ref2 would extend the lifetime of the bound temporary to that of itself. Hence the question, I guess.

    – Angew
    Apr 8 at 7:26












  • 9





    This program produces no output and exits with code 0. With -O3 optimization flag all the statements inside of main are discarded.

    – VTT
    Apr 8 at 7:10






  • 4





    What error does it give?

    – fredrik
    Apr 8 at 7:11






  • 6





    This is actually a very interesting question. I'll tag language lawyer and hope one of the chiefs picks this one up. I've neither the time nor the expertise. It's all to do with the fact that lifetime extension is not transitive, and where the original objects are located. Go on @StoryTeller.

    – Bathsheba
    Apr 8 at 7:16







  • 5





    @Bathsheba Could you quickly explain why b < a ? b : a that advantageous?

    – lubgr
    Apr 8 at 7:24






  • 5





    @Michiel The thing is, if the binding was direct (e.g. if min was returning by value), ref2 would extend the lifetime of the bound temporary to that of itself. Hence the question, I guess.

    – Angew
    Apr 8 at 7:26







9




9





This program produces no output and exits with code 0. With -O3 optimization flag all the statements inside of main are discarded.

– VTT
Apr 8 at 7:10





This program produces no output and exits with code 0. With -O3 optimization flag all the statements inside of main are discarded.

– VTT
Apr 8 at 7:10




4




4





What error does it give?

– fredrik
Apr 8 at 7:11





What error does it give?

– fredrik
Apr 8 at 7:11




6




6





This is actually a very interesting question. I'll tag language lawyer and hope one of the chiefs picks this one up. I've neither the time nor the expertise. It's all to do with the fact that lifetime extension is not transitive, and where the original objects are located. Go on @StoryTeller.

– Bathsheba
Apr 8 at 7:16






This is actually a very interesting question. I'll tag language lawyer and hope one of the chiefs picks this one up. I've neither the time nor the expertise. It's all to do with the fact that lifetime extension is not transitive, and where the original objects are located. Go on @StoryTeller.

– Bathsheba
Apr 8 at 7:16





5




5





@Bathsheba Could you quickly explain why b < a ? b : a that advantageous?

– lubgr
Apr 8 at 7:24





@Bathsheba Could you quickly explain why b < a ? b : a that advantageous?

– lubgr
Apr 8 at 7:24




5




5





@Michiel The thing is, if the binding was direct (e.g. if min was returning by value), ref2 would extend the lifetime of the bound temporary to that of itself. Hence the question, I guess.

– Angew
Apr 8 at 7:26





@Michiel The thing is, if the binding was direct (e.g. if min was returning by value), ref2 would extend the lifetime of the bound temporary to that of itself. Hence the question, I guess.

– Angew
Apr 8 at 7:26












3 Answers
3






active

oldest

votes


















36














It's by design. In a nutshell, only the named reference to which the temporary is bound directly will extend its lifetime.




[class.temporary]



5 There are three contexts in which temporaries are destroyed at a
different point than the end of the full-expression. [...]



6 The third context is when a reference is bound to a temporary.
The temporary to which the reference is bound or the temporary that is
the complete object of a subobject to which the reference is bound
persists for the lifetime of the reference except:



  • A temporary object bound to a reference parameter in a function call persists until the completion of the full-expression containing
    the call.

  • The lifetime of a temporary bound to the returned value in a function return statement is not extended; the temporary is destroyed
    at the end of the full-expression in the return statement.

  • [...]



You didn't bind directly to ref2, and you even pass it via a return statement. The standard explicitly says it won't extend the lifetime. In part to make certain optimizations possible. But ultimately, because keeping track of which temporary should be extended when a reference is passed in and out of functions is intractable in general.



Since compilers may optimize aggressively on the assumption that your program exhibits no undefined behavior, you see a possible manifestation of that. Accessing a value outside its lifetime is undefined, this is what return ref2; does, and since the behavior is undefined, simply returning zero is a valid behavior to exhibit. No contract is broken by the compiler.






share|improve this answer




















  • 1





    Thx a lot. I wrongly thought that, y+1 temporary object bind to b and it's live extends via return function. .

    – Khurshid Normuradov
    Apr 8 at 9:38


















15














This is intentional. A reference can only extend the lifetime of a temporary when it is bound to that temporary directly. In your code, you are binding ref2 to the result of min, which is a reference. It doesn't matter that that reference references a temporary. Only b extends the lifetime of the temporary; it doesn't matter that ref2 also refers to that same temporary.



Another way to look at it: You can't optionally have lifetime extension. It's a static property. If ref2 would do the Correct Thingtm, then depending on the runtime values of x and y+1 the lifetime is extended or not. Not something the compiler is able to do.






share|improve this answer






























    7














    I will answer the question first, and then provide some context for the answer. The current working draft contains the following wording:




    The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference if the glvalue to which the reference is bound was obtained through one of the following:



    • a temporary materialization conversion ([conv.rval]),


    • ( expression ), where expression is one of these expressions,

    • subscripting ([expr.sub]) of an array operand, where that operand is one of these expressions,

    • a class member access ([expr.ref]) using the . operator where the left operand is one of these expressions and the right operand designates a non-static data member of non-reference type,

    • a pointer-to-member operation ([expr.mptr.oper]) using the .* operator where the left operand is one of these expressions and the right operand is a pointer to data member of non-reference type,

    • a const_­cast ([expr.const.cast]), static_­cast ([expr.static.cast]), dynamic_­cast ([expr.dynamic.cast]), or reinterpret_­cast ([expr.reinterpret.cast])
      converting, without a user-defined conversion, a glvalue operand that is one of these expressions to a glvalue that refers to the object designated by the operand, or to its complete object or a subobject thereof,

    • a conditional expression ([expr.cond]) that is a glvalue where the second or third operand is one of these expressions, or

    • a comma expression ([expr.comma]) that is a glvalue where the right operand is one of these expressions.



    According to this, when a reference is bound to a glvalue returned from a function call, lifetime extension does not occur, because the glvalue was obtained from the function call, which is not one of the permitted expressions for lifetime extension.



    The lifetime of the y+1 temporary is extended once when bound to the reference parameter b. Here, the prvalue y+1 is materialized to yield an xvalue, and the reference is bound to the result of the temporary materialization conversion; lifetime extension thus occurs. When the min function returns, however, ref2 is bound to the result of the call, and lifetime extension does not occur here. Therefore, the y+1 temporary is destroyed at the end of the definition of ref2, and ref2 becomes a dangling reference.




    There has historically been some confusion on this topic. It is well-known that the OP's code and similar code result in a dangling reference, but the standard text, even as of C++17, did not provide an unambiguous explanation as to why.



    It is often claimed that lifetime extension only applies when the reference binds "directly" to the temporary, but the standard has never said anything to that effect. Indeed, the standard defines what it means for a reference to "bind directly", and that definition (e.g., const std::string& s = "foo"; is an indirect reference binding) is clearly not relevant here.



    Rakete1111 has said in a comment elsewhere on SO that lifetime extension only applies when the reference binds to a prvalue (rather than some glvalue that was obtained through a previous reference binding to that temporary object); they appear to be saying something similar here by "bound ... directly". However, there is no textual support for this theory. Indeed, code like the following has sometimes been considered to trigger lifetime extension:



    struct S int x; ;
    const int& r = S42.x;


    However, in C++14, the expression S42.x became an xvalue, so if lifetime extension applies here, then it is not because the reference binds to a prvalue.



    One might instead claim that lifetime extension only applies once, and binding any other references to the same object do not further extend its lifetime. This would explain why the OP's code creates a dangling reference, without preventing lifetime extension in the S42.x case. However, there is no statement to this effect in the standard, either.



    StoryTeller has also said here that the reference must bind directly, but I don't know what he means by that, either. He cites standards text indicating that binding a reference to a temporary in a return statement doesn't extend its lifetime. However, that statement seems to be intended to apply to the case where the temporary in question is created by the full-expression in the return statement, since it says the temporary will be destroyed at the end of that full-expression. Clearly that's not the case for the y+1 temporary, which will instead be destroyed at the end of the full-expression containing the call to min. Thus, I tend to think that this statement was not intended to apply to cases like that in the question. Instead, its effect, together with the other limitations on lifetime extension, is to prevent any temporary object's lifetime from being extended beyond the block scope in which it was created. But this would not prevent the y+1 temporary in the question from surviving until the end of main.



    Thus the question remains: what is the principle that explains why the binding of ref2 to the temporary in the question doesn't extend that temporary's lifetime?



    The wording from the current working draft that I cited earlier was introduced by the resolution of CWG 1299, which was opened in 2011 but only resolved recently (not in time for C++17). In a sense, it clarifies the intuition that the reference must bind "directly", by delineating those cases where the binding is "direct" enough for lifetime extension to occur; it is not, however, so restrictive as to only allow it when the reference binds to a prvalue. It permits lifetime extension in the S42.x case.






    share|improve this answer























      Your Answer






      StackExchange.ifUsing("editor", function ()
      StackExchange.using("externalEditor", function ()
      StackExchange.using("snippets", function ()
      StackExchange.snippets.init();
      );
      );
      , "code-snippets");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "1"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55567962%2fwhy-doesnt-a-const-reference-extend-the-life-of-a-temporary-object-passed-via-a%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      36














      It's by design. In a nutshell, only the named reference to which the temporary is bound directly will extend its lifetime.




      [class.temporary]



      5 There are three contexts in which temporaries are destroyed at a
      different point than the end of the full-expression. [...]



      6 The third context is when a reference is bound to a temporary.
      The temporary to which the reference is bound or the temporary that is
      the complete object of a subobject to which the reference is bound
      persists for the lifetime of the reference except:



      • A temporary object bound to a reference parameter in a function call persists until the completion of the full-expression containing
        the call.

      • The lifetime of a temporary bound to the returned value in a function return statement is not extended; the temporary is destroyed
        at the end of the full-expression in the return statement.

      • [...]



      You didn't bind directly to ref2, and you even pass it via a return statement. The standard explicitly says it won't extend the lifetime. In part to make certain optimizations possible. But ultimately, because keeping track of which temporary should be extended when a reference is passed in and out of functions is intractable in general.



      Since compilers may optimize aggressively on the assumption that your program exhibits no undefined behavior, you see a possible manifestation of that. Accessing a value outside its lifetime is undefined, this is what return ref2; does, and since the behavior is undefined, simply returning zero is a valid behavior to exhibit. No contract is broken by the compiler.






      share|improve this answer




















      • 1





        Thx a lot. I wrongly thought that, y+1 temporary object bind to b and it's live extends via return function. .

        – Khurshid Normuradov
        Apr 8 at 9:38















      36














      It's by design. In a nutshell, only the named reference to which the temporary is bound directly will extend its lifetime.




      [class.temporary]



      5 There are three contexts in which temporaries are destroyed at a
      different point than the end of the full-expression. [...]



      6 The third context is when a reference is bound to a temporary.
      The temporary to which the reference is bound or the temporary that is
      the complete object of a subobject to which the reference is bound
      persists for the lifetime of the reference except:



      • A temporary object bound to a reference parameter in a function call persists until the completion of the full-expression containing
        the call.

      • The lifetime of a temporary bound to the returned value in a function return statement is not extended; the temporary is destroyed
        at the end of the full-expression in the return statement.

      • [...]



      You didn't bind directly to ref2, and you even pass it via a return statement. The standard explicitly says it won't extend the lifetime. In part to make certain optimizations possible. But ultimately, because keeping track of which temporary should be extended when a reference is passed in and out of functions is intractable in general.



      Since compilers may optimize aggressively on the assumption that your program exhibits no undefined behavior, you see a possible manifestation of that. Accessing a value outside its lifetime is undefined, this is what return ref2; does, and since the behavior is undefined, simply returning zero is a valid behavior to exhibit. No contract is broken by the compiler.






      share|improve this answer




















      • 1





        Thx a lot. I wrongly thought that, y+1 temporary object bind to b and it's live extends via return function. .

        – Khurshid Normuradov
        Apr 8 at 9:38













      36












      36








      36







      It's by design. In a nutshell, only the named reference to which the temporary is bound directly will extend its lifetime.




      [class.temporary]



      5 There are three contexts in which temporaries are destroyed at a
      different point than the end of the full-expression. [...]



      6 The third context is when a reference is bound to a temporary.
      The temporary to which the reference is bound or the temporary that is
      the complete object of a subobject to which the reference is bound
      persists for the lifetime of the reference except:



      • A temporary object bound to a reference parameter in a function call persists until the completion of the full-expression containing
        the call.

      • The lifetime of a temporary bound to the returned value in a function return statement is not extended; the temporary is destroyed
        at the end of the full-expression in the return statement.

      • [...]



      You didn't bind directly to ref2, and you even pass it via a return statement. The standard explicitly says it won't extend the lifetime. In part to make certain optimizations possible. But ultimately, because keeping track of which temporary should be extended when a reference is passed in and out of functions is intractable in general.



      Since compilers may optimize aggressively on the assumption that your program exhibits no undefined behavior, you see a possible manifestation of that. Accessing a value outside its lifetime is undefined, this is what return ref2; does, and since the behavior is undefined, simply returning zero is a valid behavior to exhibit. No contract is broken by the compiler.






      share|improve this answer















      It's by design. In a nutshell, only the named reference to which the temporary is bound directly will extend its lifetime.




      [class.temporary]



      5 There are three contexts in which temporaries are destroyed at a
      different point than the end of the full-expression. [...]



      6 The third context is when a reference is bound to a temporary.
      The temporary to which the reference is bound or the temporary that is
      the complete object of a subobject to which the reference is bound
      persists for the lifetime of the reference except:



      • A temporary object bound to a reference parameter in a function call persists until the completion of the full-expression containing
        the call.

      • The lifetime of a temporary bound to the returned value in a function return statement is not extended; the temporary is destroyed
        at the end of the full-expression in the return statement.

      • [...]



      You didn't bind directly to ref2, and you even pass it via a return statement. The standard explicitly says it won't extend the lifetime. In part to make certain optimizations possible. But ultimately, because keeping track of which temporary should be extended when a reference is passed in and out of functions is intractable in general.



      Since compilers may optimize aggressively on the assumption that your program exhibits no undefined behavior, you see a possible manifestation of that. Accessing a value outside its lifetime is undefined, this is what return ref2; does, and since the behavior is undefined, simply returning zero is a valid behavior to exhibit. No contract is broken by the compiler.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Apr 8 at 7:37

























      answered Apr 8 at 7:23









      StoryTellerStoryTeller

      105k13221285




      105k13221285







      • 1





        Thx a lot. I wrongly thought that, y+1 temporary object bind to b and it's live extends via return function. .

        – Khurshid Normuradov
        Apr 8 at 9:38












      • 1





        Thx a lot. I wrongly thought that, y+1 temporary object bind to b and it's live extends via return function. .

        – Khurshid Normuradov
        Apr 8 at 9:38







      1




      1





      Thx a lot. I wrongly thought that, y+1 temporary object bind to b and it's live extends via return function. .

      – Khurshid Normuradov
      Apr 8 at 9:38





      Thx a lot. I wrongly thought that, y+1 temporary object bind to b and it's live extends via return function. .

      – Khurshid Normuradov
      Apr 8 at 9:38













      15














      This is intentional. A reference can only extend the lifetime of a temporary when it is bound to that temporary directly. In your code, you are binding ref2 to the result of min, which is a reference. It doesn't matter that that reference references a temporary. Only b extends the lifetime of the temporary; it doesn't matter that ref2 also refers to that same temporary.



      Another way to look at it: You can't optionally have lifetime extension. It's a static property. If ref2 would do the Correct Thingtm, then depending on the runtime values of x and y+1 the lifetime is extended or not. Not something the compiler is able to do.






      share|improve this answer



























        15














        This is intentional. A reference can only extend the lifetime of a temporary when it is bound to that temporary directly. In your code, you are binding ref2 to the result of min, which is a reference. It doesn't matter that that reference references a temporary. Only b extends the lifetime of the temporary; it doesn't matter that ref2 also refers to that same temporary.



        Another way to look at it: You can't optionally have lifetime extension. It's a static property. If ref2 would do the Correct Thingtm, then depending on the runtime values of x and y+1 the lifetime is extended or not. Not something the compiler is able to do.






        share|improve this answer

























          15












          15








          15







          This is intentional. A reference can only extend the lifetime of a temporary when it is bound to that temporary directly. In your code, you are binding ref2 to the result of min, which is a reference. It doesn't matter that that reference references a temporary. Only b extends the lifetime of the temporary; it doesn't matter that ref2 also refers to that same temporary.



          Another way to look at it: You can't optionally have lifetime extension. It's a static property. If ref2 would do the Correct Thingtm, then depending on the runtime values of x and y+1 the lifetime is extended or not. Not something the compiler is able to do.






          share|improve this answer













          This is intentional. A reference can only extend the lifetime of a temporary when it is bound to that temporary directly. In your code, you are binding ref2 to the result of min, which is a reference. It doesn't matter that that reference references a temporary. Only b extends the lifetime of the temporary; it doesn't matter that ref2 also refers to that same temporary.



          Another way to look at it: You can't optionally have lifetime extension. It's a static property. If ref2 would do the Correct Thingtm, then depending on the runtime values of x and y+1 the lifetime is extended or not. Not something the compiler is able to do.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 8 at 7:22









          Rakete1111Rakete1111

          35.5k1084121




          35.5k1084121





















              7














              I will answer the question first, and then provide some context for the answer. The current working draft contains the following wording:




              The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference if the glvalue to which the reference is bound was obtained through one of the following:



              • a temporary materialization conversion ([conv.rval]),


              • ( expression ), where expression is one of these expressions,

              • subscripting ([expr.sub]) of an array operand, where that operand is one of these expressions,

              • a class member access ([expr.ref]) using the . operator where the left operand is one of these expressions and the right operand designates a non-static data member of non-reference type,

              • a pointer-to-member operation ([expr.mptr.oper]) using the .* operator where the left operand is one of these expressions and the right operand is a pointer to data member of non-reference type,

              • a const_­cast ([expr.const.cast]), static_­cast ([expr.static.cast]), dynamic_­cast ([expr.dynamic.cast]), or reinterpret_­cast ([expr.reinterpret.cast])
                converting, without a user-defined conversion, a glvalue operand that is one of these expressions to a glvalue that refers to the object designated by the operand, or to its complete object or a subobject thereof,

              • a conditional expression ([expr.cond]) that is a glvalue where the second or third operand is one of these expressions, or

              • a comma expression ([expr.comma]) that is a glvalue where the right operand is one of these expressions.



              According to this, when a reference is bound to a glvalue returned from a function call, lifetime extension does not occur, because the glvalue was obtained from the function call, which is not one of the permitted expressions for lifetime extension.



              The lifetime of the y+1 temporary is extended once when bound to the reference parameter b. Here, the prvalue y+1 is materialized to yield an xvalue, and the reference is bound to the result of the temporary materialization conversion; lifetime extension thus occurs. When the min function returns, however, ref2 is bound to the result of the call, and lifetime extension does not occur here. Therefore, the y+1 temporary is destroyed at the end of the definition of ref2, and ref2 becomes a dangling reference.




              There has historically been some confusion on this topic. It is well-known that the OP's code and similar code result in a dangling reference, but the standard text, even as of C++17, did not provide an unambiguous explanation as to why.



              It is often claimed that lifetime extension only applies when the reference binds "directly" to the temporary, but the standard has never said anything to that effect. Indeed, the standard defines what it means for a reference to "bind directly", and that definition (e.g., const std::string& s = "foo"; is an indirect reference binding) is clearly not relevant here.



              Rakete1111 has said in a comment elsewhere on SO that lifetime extension only applies when the reference binds to a prvalue (rather than some glvalue that was obtained through a previous reference binding to that temporary object); they appear to be saying something similar here by "bound ... directly". However, there is no textual support for this theory. Indeed, code like the following has sometimes been considered to trigger lifetime extension:



              struct S int x; ;
              const int& r = S42.x;


              However, in C++14, the expression S42.x became an xvalue, so if lifetime extension applies here, then it is not because the reference binds to a prvalue.



              One might instead claim that lifetime extension only applies once, and binding any other references to the same object do not further extend its lifetime. This would explain why the OP's code creates a dangling reference, without preventing lifetime extension in the S42.x case. However, there is no statement to this effect in the standard, either.



              StoryTeller has also said here that the reference must bind directly, but I don't know what he means by that, either. He cites standards text indicating that binding a reference to a temporary in a return statement doesn't extend its lifetime. However, that statement seems to be intended to apply to the case where the temporary in question is created by the full-expression in the return statement, since it says the temporary will be destroyed at the end of that full-expression. Clearly that's not the case for the y+1 temporary, which will instead be destroyed at the end of the full-expression containing the call to min. Thus, I tend to think that this statement was not intended to apply to cases like that in the question. Instead, its effect, together with the other limitations on lifetime extension, is to prevent any temporary object's lifetime from being extended beyond the block scope in which it was created. But this would not prevent the y+1 temporary in the question from surviving until the end of main.



              Thus the question remains: what is the principle that explains why the binding of ref2 to the temporary in the question doesn't extend that temporary's lifetime?



              The wording from the current working draft that I cited earlier was introduced by the resolution of CWG 1299, which was opened in 2011 but only resolved recently (not in time for C++17). In a sense, it clarifies the intuition that the reference must bind "directly", by delineating those cases where the binding is "direct" enough for lifetime extension to occur; it is not, however, so restrictive as to only allow it when the reference binds to a prvalue. It permits lifetime extension in the S42.x case.






              share|improve this answer



























                7














                I will answer the question first, and then provide some context for the answer. The current working draft contains the following wording:




                The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference if the glvalue to which the reference is bound was obtained through one of the following:



                • a temporary materialization conversion ([conv.rval]),


                • ( expression ), where expression is one of these expressions,

                • subscripting ([expr.sub]) of an array operand, where that operand is one of these expressions,

                • a class member access ([expr.ref]) using the . operator where the left operand is one of these expressions and the right operand designates a non-static data member of non-reference type,

                • a pointer-to-member operation ([expr.mptr.oper]) using the .* operator where the left operand is one of these expressions and the right operand is a pointer to data member of non-reference type,

                • a const_­cast ([expr.const.cast]), static_­cast ([expr.static.cast]), dynamic_­cast ([expr.dynamic.cast]), or reinterpret_­cast ([expr.reinterpret.cast])
                  converting, without a user-defined conversion, a glvalue operand that is one of these expressions to a glvalue that refers to the object designated by the operand, or to its complete object or a subobject thereof,

                • a conditional expression ([expr.cond]) that is a glvalue where the second or third operand is one of these expressions, or

                • a comma expression ([expr.comma]) that is a glvalue where the right operand is one of these expressions.



                According to this, when a reference is bound to a glvalue returned from a function call, lifetime extension does not occur, because the glvalue was obtained from the function call, which is not one of the permitted expressions for lifetime extension.



                The lifetime of the y+1 temporary is extended once when bound to the reference parameter b. Here, the prvalue y+1 is materialized to yield an xvalue, and the reference is bound to the result of the temporary materialization conversion; lifetime extension thus occurs. When the min function returns, however, ref2 is bound to the result of the call, and lifetime extension does not occur here. Therefore, the y+1 temporary is destroyed at the end of the definition of ref2, and ref2 becomes a dangling reference.




                There has historically been some confusion on this topic. It is well-known that the OP's code and similar code result in a dangling reference, but the standard text, even as of C++17, did not provide an unambiguous explanation as to why.



                It is often claimed that lifetime extension only applies when the reference binds "directly" to the temporary, but the standard has never said anything to that effect. Indeed, the standard defines what it means for a reference to "bind directly", and that definition (e.g., const std::string& s = "foo"; is an indirect reference binding) is clearly not relevant here.



                Rakete1111 has said in a comment elsewhere on SO that lifetime extension only applies when the reference binds to a prvalue (rather than some glvalue that was obtained through a previous reference binding to that temporary object); they appear to be saying something similar here by "bound ... directly". However, there is no textual support for this theory. Indeed, code like the following has sometimes been considered to trigger lifetime extension:



                struct S int x; ;
                const int& r = S42.x;


                However, in C++14, the expression S42.x became an xvalue, so if lifetime extension applies here, then it is not because the reference binds to a prvalue.



                One might instead claim that lifetime extension only applies once, and binding any other references to the same object do not further extend its lifetime. This would explain why the OP's code creates a dangling reference, without preventing lifetime extension in the S42.x case. However, there is no statement to this effect in the standard, either.



                StoryTeller has also said here that the reference must bind directly, but I don't know what he means by that, either. He cites standards text indicating that binding a reference to a temporary in a return statement doesn't extend its lifetime. However, that statement seems to be intended to apply to the case where the temporary in question is created by the full-expression in the return statement, since it says the temporary will be destroyed at the end of that full-expression. Clearly that's not the case for the y+1 temporary, which will instead be destroyed at the end of the full-expression containing the call to min. Thus, I tend to think that this statement was not intended to apply to cases like that in the question. Instead, its effect, together with the other limitations on lifetime extension, is to prevent any temporary object's lifetime from being extended beyond the block scope in which it was created. But this would not prevent the y+1 temporary in the question from surviving until the end of main.



                Thus the question remains: what is the principle that explains why the binding of ref2 to the temporary in the question doesn't extend that temporary's lifetime?



                The wording from the current working draft that I cited earlier was introduced by the resolution of CWG 1299, which was opened in 2011 but only resolved recently (not in time for C++17). In a sense, it clarifies the intuition that the reference must bind "directly", by delineating those cases where the binding is "direct" enough for lifetime extension to occur; it is not, however, so restrictive as to only allow it when the reference binds to a prvalue. It permits lifetime extension in the S42.x case.






                share|improve this answer

























                  7












                  7








                  7







                  I will answer the question first, and then provide some context for the answer. The current working draft contains the following wording:




                  The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference if the glvalue to which the reference is bound was obtained through one of the following:



                  • a temporary materialization conversion ([conv.rval]),


                  • ( expression ), where expression is one of these expressions,

                  • subscripting ([expr.sub]) of an array operand, where that operand is one of these expressions,

                  • a class member access ([expr.ref]) using the . operator where the left operand is one of these expressions and the right operand designates a non-static data member of non-reference type,

                  • a pointer-to-member operation ([expr.mptr.oper]) using the .* operator where the left operand is one of these expressions and the right operand is a pointer to data member of non-reference type,

                  • a const_­cast ([expr.const.cast]), static_­cast ([expr.static.cast]), dynamic_­cast ([expr.dynamic.cast]), or reinterpret_­cast ([expr.reinterpret.cast])
                    converting, without a user-defined conversion, a glvalue operand that is one of these expressions to a glvalue that refers to the object designated by the operand, or to its complete object or a subobject thereof,

                  • a conditional expression ([expr.cond]) that is a glvalue where the second or third operand is one of these expressions, or

                  • a comma expression ([expr.comma]) that is a glvalue where the right operand is one of these expressions.



                  According to this, when a reference is bound to a glvalue returned from a function call, lifetime extension does not occur, because the glvalue was obtained from the function call, which is not one of the permitted expressions for lifetime extension.



                  The lifetime of the y+1 temporary is extended once when bound to the reference parameter b. Here, the prvalue y+1 is materialized to yield an xvalue, and the reference is bound to the result of the temporary materialization conversion; lifetime extension thus occurs. When the min function returns, however, ref2 is bound to the result of the call, and lifetime extension does not occur here. Therefore, the y+1 temporary is destroyed at the end of the definition of ref2, and ref2 becomes a dangling reference.




                  There has historically been some confusion on this topic. It is well-known that the OP's code and similar code result in a dangling reference, but the standard text, even as of C++17, did not provide an unambiguous explanation as to why.



                  It is often claimed that lifetime extension only applies when the reference binds "directly" to the temporary, but the standard has never said anything to that effect. Indeed, the standard defines what it means for a reference to "bind directly", and that definition (e.g., const std::string& s = "foo"; is an indirect reference binding) is clearly not relevant here.



                  Rakete1111 has said in a comment elsewhere on SO that lifetime extension only applies when the reference binds to a prvalue (rather than some glvalue that was obtained through a previous reference binding to that temporary object); they appear to be saying something similar here by "bound ... directly". However, there is no textual support for this theory. Indeed, code like the following has sometimes been considered to trigger lifetime extension:



                  struct S int x; ;
                  const int& r = S42.x;


                  However, in C++14, the expression S42.x became an xvalue, so if lifetime extension applies here, then it is not because the reference binds to a prvalue.



                  One might instead claim that lifetime extension only applies once, and binding any other references to the same object do not further extend its lifetime. This would explain why the OP's code creates a dangling reference, without preventing lifetime extension in the S42.x case. However, there is no statement to this effect in the standard, either.



                  StoryTeller has also said here that the reference must bind directly, but I don't know what he means by that, either. He cites standards text indicating that binding a reference to a temporary in a return statement doesn't extend its lifetime. However, that statement seems to be intended to apply to the case where the temporary in question is created by the full-expression in the return statement, since it says the temporary will be destroyed at the end of that full-expression. Clearly that's not the case for the y+1 temporary, which will instead be destroyed at the end of the full-expression containing the call to min. Thus, I tend to think that this statement was not intended to apply to cases like that in the question. Instead, its effect, together with the other limitations on lifetime extension, is to prevent any temporary object's lifetime from being extended beyond the block scope in which it was created. But this would not prevent the y+1 temporary in the question from surviving until the end of main.



                  Thus the question remains: what is the principle that explains why the binding of ref2 to the temporary in the question doesn't extend that temporary's lifetime?



                  The wording from the current working draft that I cited earlier was introduced by the resolution of CWG 1299, which was opened in 2011 but only resolved recently (not in time for C++17). In a sense, it clarifies the intuition that the reference must bind "directly", by delineating those cases where the binding is "direct" enough for lifetime extension to occur; it is not, however, so restrictive as to only allow it when the reference binds to a prvalue. It permits lifetime extension in the S42.x case.






                  share|improve this answer













                  I will answer the question first, and then provide some context for the answer. The current working draft contains the following wording:




                  The temporary object to which the reference is bound or the temporary object that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference if the glvalue to which the reference is bound was obtained through one of the following:



                  • a temporary materialization conversion ([conv.rval]),


                  • ( expression ), where expression is one of these expressions,

                  • subscripting ([expr.sub]) of an array operand, where that operand is one of these expressions,

                  • a class member access ([expr.ref]) using the . operator where the left operand is one of these expressions and the right operand designates a non-static data member of non-reference type,

                  • a pointer-to-member operation ([expr.mptr.oper]) using the .* operator where the left operand is one of these expressions and the right operand is a pointer to data member of non-reference type,

                  • a const_­cast ([expr.const.cast]), static_­cast ([expr.static.cast]), dynamic_­cast ([expr.dynamic.cast]), or reinterpret_­cast ([expr.reinterpret.cast])
                    converting, without a user-defined conversion, a glvalue operand that is one of these expressions to a glvalue that refers to the object designated by the operand, or to its complete object or a subobject thereof,

                  • a conditional expression ([expr.cond]) that is a glvalue where the second or third operand is one of these expressions, or

                  • a comma expression ([expr.comma]) that is a glvalue where the right operand is one of these expressions.



                  According to this, when a reference is bound to a glvalue returned from a function call, lifetime extension does not occur, because the glvalue was obtained from the function call, which is not one of the permitted expressions for lifetime extension.



                  The lifetime of the y+1 temporary is extended once when bound to the reference parameter b. Here, the prvalue y+1 is materialized to yield an xvalue, and the reference is bound to the result of the temporary materialization conversion; lifetime extension thus occurs. When the min function returns, however, ref2 is bound to the result of the call, and lifetime extension does not occur here. Therefore, the y+1 temporary is destroyed at the end of the definition of ref2, and ref2 becomes a dangling reference.




                  There has historically been some confusion on this topic. It is well-known that the OP's code and similar code result in a dangling reference, but the standard text, even as of C++17, did not provide an unambiguous explanation as to why.



                  It is often claimed that lifetime extension only applies when the reference binds "directly" to the temporary, but the standard has never said anything to that effect. Indeed, the standard defines what it means for a reference to "bind directly", and that definition (e.g., const std::string& s = "foo"; is an indirect reference binding) is clearly not relevant here.



                  Rakete1111 has said in a comment elsewhere on SO that lifetime extension only applies when the reference binds to a prvalue (rather than some glvalue that was obtained through a previous reference binding to that temporary object); they appear to be saying something similar here by "bound ... directly". However, there is no textual support for this theory. Indeed, code like the following has sometimes been considered to trigger lifetime extension:



                  struct S int x; ;
                  const int& r = S42.x;


                  However, in C++14, the expression S42.x became an xvalue, so if lifetime extension applies here, then it is not because the reference binds to a prvalue.



                  One might instead claim that lifetime extension only applies once, and binding any other references to the same object do not further extend its lifetime. This would explain why the OP's code creates a dangling reference, without preventing lifetime extension in the S42.x case. However, there is no statement to this effect in the standard, either.



                  StoryTeller has also said here that the reference must bind directly, but I don't know what he means by that, either. He cites standards text indicating that binding a reference to a temporary in a return statement doesn't extend its lifetime. However, that statement seems to be intended to apply to the case where the temporary in question is created by the full-expression in the return statement, since it says the temporary will be destroyed at the end of that full-expression. Clearly that's not the case for the y+1 temporary, which will instead be destroyed at the end of the full-expression containing the call to min. Thus, I tend to think that this statement was not intended to apply to cases like that in the question. Instead, its effect, together with the other limitations on lifetime extension, is to prevent any temporary object's lifetime from being extended beyond the block scope in which it was created. But this would not prevent the y+1 temporary in the question from surviving until the end of main.



                  Thus the question remains: what is the principle that explains why the binding of ref2 to the temporary in the question doesn't extend that temporary's lifetime?



                  The wording from the current working draft that I cited earlier was introduced by the resolution of CWG 1299, which was opened in 2011 but only resolved recently (not in time for C++17). In a sense, it clarifies the intuition that the reference must bind "directly", by delineating those cases where the binding is "direct" enough for lifetime extension to occur; it is not, however, so restrictive as to only allow it when the reference binds to a prvalue. It permits lifetime extension in the S42.x case.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Apr 8 at 14:56









                  BrianBrian

                  66.7k799191




                  66.7k799191



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Stack Overflow!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55567962%2fwhy-doesnt-a-const-reference-extend-the-life-of-a-temporary-object-passed-via-a%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      getting Checkpoint VPN SSL Network Extender working in the command lineHow to connect to CheckPoint VPN on Ubuntu 18.04LTS?Will the Linux ( red-hat ) Open VPNC Client connect to checkpoint or nortel VPN gateways?VPN client for linux machine + support checkpoint gatewayVPN SSL Network Extender in FirefoxLinux Checkpoint SNX tool configuration issuesCheck Point - Connect under Linux - snx + OTPSNX VPN Ububuntu 18.XXUsing Checkpoint VPN SSL Network Extender CLI with certificateVPN with network manager (nm-applet) is not workingWill the Linux ( red-hat ) Open VPNC Client connect to checkpoint or nortel VPN gateways?VPN client for linux machine + support checkpoint gatewayImport VPN config files to NetworkManager from command lineTrouble connecting to VPN using network-manager, while command line worksStart a VPN connection with PPTP protocol on command linestarting a docker service daemon breaks the vpn networkCan't connect to vpn with Network-managerVPN SSL Network Extender in FirefoxUsing Checkpoint VPN SSL Network Extender CLI with certificate

                      Cannot Extend partition with GParted The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Community Moderator Election ResultsCan't increase partition size with GParted?GParted doesn't recognize the unallocated space after my current partitionWhat is the best way to add unallocated space located before to Ubuntu 12.04 partition with GParted live?I can't figure out how to extend my Arch home partition into free spaceGparted Linux Mint 18.1 issueTrying to extend but swap partition is showing as Unknown in Gparted, shows proper from fdiskRearrange partitions in gparted to extend a partitionUnable to extend partition even though unallocated space is next to it using GPartedAllocate free space to root partitiongparted: how to merge unallocated space with a partition

                      Marilyn Monroe Ny fiainany manokana | Jereo koa | Meny fitetezanafanitarana azy.