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

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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



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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























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      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



























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                      대한민국 목차 국명 지리 역사 정치 국방 경제 사회 문화 국제 순위 관련 항목 각주 외부 링크 둘러보기 메뉴북위 37° 34′ 08″ 동경 126° 58′ 36″ / 북위 37.568889° 동경 126.976667°  / 37.568889; 126.976667ehThe Korean Repository문단을 편집문단을 편집추가해Clarkson PLC 사Report for Selected Countries and Subjects-Korea“Human Development Index and its components: P.198”“http://www.law.go.kr/%EB%B2%95%EB%A0%B9/%EB%8C%80%ED%95%9C%EB%AF%BC%EA%B5%AD%EA%B5%AD%EA%B8%B0%EB%B2%95”"한국은 국제법상 한반도 유일 합법정부 아니다" - 오마이뉴스 모바일Report for Selected Countries and Subjects: South Korea격동의 역사와 함께한 조선일보 90년 : 조선일보 인수해 혁신시킨 신석우, 임시정부 때는 '대한민국' 국호(國號) 정해《우리가 몰랐던 우리 역사: 나라 이름의 비밀을 찾아가는 역사 여행》“남북 공식호칭 ‘남한’‘북한’으로 쓴다”“Corea 대 Korea, 누가 이긴 거야?”국내기후자료 - 한국[김대중 前 대통령 서거] 과감한 구조개혁 'DJ노믹스'로 최단기간 환란극복 :: 네이버 뉴스“이라크 "韓-쿠르드 유전개발 MOU 승인 안해"(종합)”“해외 우리국민 추방사례 43%가 일본”차기전차 K2'흑표'의 세계 최고 전력 분석, 쿠키뉴스 엄기영, 2007-03-02두산인프라, 헬기잡는 장갑차 'K21'...내년부터 공급, 고뉴스 이대준, 2008-10-30과거 내용 찾기mk 뉴스 - 구매력 기준으로 보면 한국 1인당 소득 3만弗과거 내용 찾기"The N-11: More Than an Acronym"Archived조선일보 최우석, 2008-11-01Global 500 2008: Countries - South Korea“몇년째 '시한폭탄'... 가계부채, 올해는 터질까”가구당 부채 5000만원 처음 넘어서“‘빚’으로 내몰리는 사회.. 위기의 가계대출”“[경제365] 공공부문 부채 급증…800조 육박”“"소득 양극화 다소 완화...불평등은 여전"”“공정사회·공생발전 한참 멀었네”iSuppli,08年2QのDRAMシェア・ランキングを発表(08/8/11)South Korea dominates shipbuilding industry | Stock Market News & Stocks to Watch from StraightStocks한국 자동차 생산, 3년 연속 세계 5위자동차수출 '현대-삼성 웃고 기아-대우-쌍용은 울고' 과거 내용 찾기동반성장위 창립 1주년 맞아Archived"중기적합 3개업종 합의 무시한 채 선정"李대통령, 사업 무분별 확장 소상공인 생계 위협 질타삼성-LG, 서민업종인 빵·분식사업 잇따라 철수상생은 뒷전…SSM ‘몸집 불리기’ 혈안Archived“경부고속도에 '아시안하이웨이' 표지판”'철의 실크로드' 앞서 '말(言)의 실크로드'부터, 프레시안 정창현, 2008-10-01“'서울 지하철은 안전한가?'”“서울시 “올해 안에 모든 지하철역 스크린도어 설치””“부산지하철 1,2호선 승강장 안전펜스 설치 완료”“전교조, 정부 노조 통계서 처음 빠져”“[Weekly BIZ] 도요타 '제로 이사회'가 리콜 사태 불러들였다”“S Korea slams high tuition costs”““정치가 여론 양극화 부채질… 합리주의 절실””“〈"`촛불집회'는 민주주의의 질적 변화 상징"〉”““촛불집회가 민주주의 왜곡 초래””“국민 65%, "한국 노사관계 대립적"”“한국 국가경쟁력 27위‥노사관계 '꼴찌'”“제대로 형성되지 않은 대한민국 이념지형”“[신년기획-갈등의 시대] 갈등지수 OECD 4위…사회적 손실 GDP 27% 무려 300조”“2012 총선-대선의 키워드는 '국민과 소통'”“한국 삶의 질 27위, 2000년과 2008년 연속 하위권 머물러”“[해피 코리아] 행복점수 68점…해외 평가선 '낙제점'”“한국 어린이·청소년 행복지수 3년 연속 OECD ‘꼴찌’”“한국 이혼율 OECD중 8위”“[통계청] 한국 이혼율 OECD 4위”“오피니언 [이렇게 생각한다] `부부의 날` 에 돌아본 이혼율 1위 한국”“Suicide Rates by Country, Global Health Observatory Data Repository.”“1. 또 다른 차별”“오피니언 [편집자에게] '왕따'와 '패거리 정치' 심리는 닮은꼴”“[미래한국리포트] 무한경쟁에 빠진 대한민국”“대학생 98% "외모가 경쟁력이라는 말 동의"”“특급호텔 웨딩·200만원대 유모차… "남보다 더…" 호화病, 고질병 됐다”“[스트레스 공화국] ① 경쟁사회, 스트레스 쌓인다”““매일 30여명 자살 한국, 의사보다 무속인에…””“"자살 부르는 '우울증', 환자 중 85% 치료 안 받아"”“정신병원을 가다”“대한민국도 ‘묻지마 범죄’,안전지대 아니다”“유엔 "학생 '성적 지향'에 따른 차별 금지하라"”“유엔아동권리위원회 보고서 및 번역본 원문”“고졸 성공스토리 담은 '제빵왕 김탁구' 드라마 나온다”“‘빛 좋은 개살구’ 고졸 취업…실습 대신 착취”원본 문서“정신건강, 사회적 편견부터 고쳐드립니다”‘소통’과 ‘행복’에 목 마른 사회가 잠들어 있던 ‘심리학’ 깨웠다“[포토] 사유리-곽금주 교수의 유쾌한 심리상담”“"올해 한국인 평균 영화관람횟수 세계 1위"(종합)”“[게임연중기획] 게임은 문화다-여가활동 1순위 게임”“영화속 ‘영어 지상주의’ …“왠지 씁쓸한데””“2월 `신문 부수 인증기관` 지정..방송법 후속작업”“무료신문 성장동력 ‘차별성’과 ‘갈등해소’”대한민국 국회 법률지식정보시스템"Pew Research Center's Religion & Public Life Project: South Korea"“amp;vwcd=MT_ZTITLE&path=인구·가구%20>%20인구총조사%20>%20인구부문%20>%20 총조사인구(2005)%20>%20전수부문&oper_YN=Y&item=&keyword=종교별%20인구& amp;lang_mode=kor&list_id= 2005년 통계청 인구 총조사”원본 문서“한국인이 좋아하는 취미와 운동 (2004-2009)”“한국인이 좋아하는 취미와 운동 (2004-2014)”Archived“한국, `부분적 언론자유국' 강등〈프리덤하우스〉”“국경없는기자회 "한국, 인터넷감시 대상국"”“한국, 조선산업 1위 유지(S. Korea Stays Top Shipbuilding Nation) RZD-Partner Portal”원본 문서“한국, 4년 만에 ‘선박건조 1위’”“옛 마산시,인터넷속도 세계 1위”“"한국 초고속 인터넷망 세계1위"”“인터넷·휴대폰 요금, 외국보다 훨씬 비싸”“한국 관세행정 6년 연속 세계 '1위'”“한국 교통사고 사망자 수 OECD 회원국 중 2위”“결핵 후진국' 한국, 환자가 급증한 이유는”“수술은 신중해야… 자칫하면 생명 위협”대한민국분류대한민국의 지도대한민국 정부대표 다국어포털대한민국 전자정부대한민국 국회한국방송공사about korea and information korea브리태니커 백과사전(한국편)론리플래닛의 정보(한국편)CIA의 세계 정보(한국편)마리암 부디아 (Mariam Budia),『한국: 하늘이 내린 한 폭의 그림』, 서울: 트랜스라틴 19호 (2012년 3월)대한민국ehehehehehehehehehehehehehehWorldCat132441370n791268020000 0001 2308 81034078029-6026373548cb11863345f(데이터)00573706ge128495