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Why am I allowed to create multiple unique pointers from a single object?
The Next CEO of Stack OverflowWhy use pointers?Pretty-print C++ STL containersC++ pointers local in functionsHow C++ reference worksWhy is reading lines from stdin much slower in C++ than Python?Why should I use a pointer rather than the object itself?Pointers as Function Parameters ExampleReturn class object pointer from another classwhy should we use std::move semantic with unique pointers?Why does my object appear to be on the heap without using `new`?
Why am I allowed to create multiple unique pointers from a single object?
#include <iostream>
#include <memory>
using namespace std;
class Class
public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;
;
int main()
Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;
return 0;
Output:
constr
4
4
4
destr
destr
destr
destr
c++ pointers unique-ptr
New contributor
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
#include <iostream>
#include <memory>
using namespace std;
class Class
public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;
;
int main()
Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;
return 0;
Output:
constr
4
4
4
destr
destr
destr
destr
c++ pointers unique-ptr
New contributor
21
Copy construction.
– user4581301
2 days ago
1
Try this:a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer
– alter igel
2 days ago
1
Creating severalunique_ptr
to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something likeauto ptr2 = std::unique_ptr<Class>(ptr.get());
would create a secondstd::unique_ptr
that points to the same instance asptr
does, and that would be problematic.
– François Andrieux
2 days ago
Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.
– TrebledJ
yesterday
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
#include <iostream>
#include <memory>
using namespace std;
class Class
public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;
;
int main()
Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;
return 0;
Output:
constr
4
4
4
destr
destr
destr
destr
c++ pointers unique-ptr
New contributor
Why am I allowed to create multiple unique pointers from a single object?
#include <iostream>
#include <memory>
using namespace std;
class Class
public:
Class(int a): int_(a)std::cout << "constr" << std::endl;
~Class()std::cout << "destr" << std::endl;
int int_;
;
int main()
Class a(4);
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr2 = std::make_unique<Class>(a);
std::unique_ptr<Class> ptr3 = std::make_unique<Class>(a);
std::cout << ptr->int_ << std::endl;
std::cout << ptr2->int_ << std::endl;
std::cout << ptr3->int_ << std::endl;
return 0;
Output:
constr
4
4
4
destr
destr
destr
destr
c++ pointers unique-ptr
c++ pointers unique-ptr
New contributor
New contributor
edited yesterday
TrebledJ
3,46911228
3,46911228
New contributor
asked 2 days ago
Lokas BeardLokas Beard
715
715
New contributor
New contributor
21
Copy construction.
– user4581301
2 days ago
1
Try this:a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer
– alter igel
2 days ago
1
Creating severalunique_ptr
to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something likeauto ptr2 = std::unique_ptr<Class>(ptr.get());
would create a secondstd::unique_ptr
that points to the same instance asptr
does, and that would be problematic.
– François Andrieux
2 days ago
Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.
– TrebledJ
yesterday
add a comment |
21
Copy construction.
– user4581301
2 days ago
1
Try this:a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer
– alter igel
2 days ago
1
Creating severalunique_ptr
to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something likeauto ptr2 = std::unique_ptr<Class>(ptr.get());
would create a secondstd::unique_ptr
that points to the same instance asptr
does, and that would be problematic.
– François Andrieux
2 days ago
Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.
– TrebledJ
yesterday
21
21
Copy construction.
– user4581301
2 days ago
Copy construction.
– user4581301
2 days ago
1
1
Try this:
a.int_ = 5;
after you create each unique_ptr
and before you print the contents of each pointer– alter igel
2 days ago
Try this:
a.int_ = 5;
after you create each unique_ptr
and before you print the contents of each pointer– alter igel
2 days ago
1
1
Creating several
unique_ptr
to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get());
would create a second std::unique_ptr
that points to the same instance as ptr
does, and that would be problematic.– François Andrieux
2 days ago
Creating several
unique_ptr
to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something like auto ptr2 = std::unique_ptr<Class>(ptr.get());
would create a second std::unique_ptr
that points to the same instance as ptr
does, and that would be problematic.– François Andrieux
2 days ago
Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.
– TrebledJ
yesterday
Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.
– TrebledJ
yesterday
add a comment |
3 Answers
3
active
oldest
votes
Why not?
You are not creating multiple unique_ptr
instances pointing to the same Class
instance, but you are instead allocating three new Class
instances on the heap, copy-constructed from a
. Every unique_ptr
points to a different instance.
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
The above means: create a new instance of Class
on the heap, copy-constructed from a
, and give ownership of it to a new std::unique_ptr
instance with name ptr
.
What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.
– Jordan Motta
2 days ago
11
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
2 days ago
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're not allowed to do that*, so it's a good thing you're not doing that!
Don't forget, this:
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
is this**:
std::unique_ptr<Class> ptr(new Class(a));
not this:
std::unique_ptr<Class> ptr(&a);
std::make_unique
creates a thing and gives you a unique_ptr
to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.
tl;dr: You're creating copies of a
.
* Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
** More or less…
1
you can prove this to yourself byptr2->_int = 6;
and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.
– Kate Gregory
2 days ago
Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;
) the copy constructor and watch the compiler choke.
– user4581301
2 days ago
1
@user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.
– Lightness Races in Orbit
yesterday
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're allowed to make multiple copies of the object, because the class is copyable.
"Unique" in unique_ptr
doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.
You could† violate the uniqueness like this:
std::unique_ptr<Class> ptr(&a);
†Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Why not?
You are not creating multiple unique_ptr
instances pointing to the same Class
instance, but you are instead allocating three new Class
instances on the heap, copy-constructed from a
. Every unique_ptr
points to a different instance.
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
The above means: create a new instance of Class
on the heap, copy-constructed from a
, and give ownership of it to a new std::unique_ptr
instance with name ptr
.
What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.
– Jordan Motta
2 days ago
11
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
2 days ago
add a comment |
Why not?
You are not creating multiple unique_ptr
instances pointing to the same Class
instance, but you are instead allocating three new Class
instances on the heap, copy-constructed from a
. Every unique_ptr
points to a different instance.
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
The above means: create a new instance of Class
on the heap, copy-constructed from a
, and give ownership of it to a new std::unique_ptr
instance with name ptr
.
What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.
– Jordan Motta
2 days ago
11
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
2 days ago
add a comment |
Why not?
You are not creating multiple unique_ptr
instances pointing to the same Class
instance, but you are instead allocating three new Class
instances on the heap, copy-constructed from a
. Every unique_ptr
points to a different instance.
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
The above means: create a new instance of Class
on the heap, copy-constructed from a
, and give ownership of it to a new std::unique_ptr
instance with name ptr
.
Why not?
You are not creating multiple unique_ptr
instances pointing to the same Class
instance, but you are instead allocating three new Class
instances on the heap, copy-constructed from a
. Every unique_ptr
points to a different instance.
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
The above means: create a new instance of Class
on the heap, copy-constructed from a
, and give ownership of it to a new std::unique_ptr
instance with name ptr
.
answered 2 days ago
Vittorio RomeoVittorio Romeo
59.4k17163307
59.4k17163307
What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.
– Jordan Motta
2 days ago
11
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
2 days ago
add a comment |
What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.
– Jordan Motta
2 days ago
11
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
2 days ago
What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.
– Jordan Motta
2 days ago
What i don't understand is he didn't provide a copy constructor. I know that if you don't create a custom constructor, the compiler will generate for you default, copy, move, etc. But, in this case, he provided a custom constructor.
– Jordan Motta
2 days ago
11
11
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
2 days ago
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
2 days ago
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're not allowed to do that*, so it's a good thing you're not doing that!
Don't forget, this:
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
is this**:
std::unique_ptr<Class> ptr(new Class(a));
not this:
std::unique_ptr<Class> ptr(&a);
std::make_unique
creates a thing and gives you a unique_ptr
to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.
tl;dr: You're creating copies of a
.
* Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
** More or less…
1
you can prove this to yourself byptr2->_int = 6;
and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.
– Kate Gregory
2 days ago
Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;
) the copy constructor and watch the compiler choke.
– user4581301
2 days ago
1
@user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.
– Lightness Races in Orbit
yesterday
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're not allowed to do that*, so it's a good thing you're not doing that!
Don't forget, this:
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
is this**:
std::unique_ptr<Class> ptr(new Class(a));
not this:
std::unique_ptr<Class> ptr(&a);
std::make_unique
creates a thing and gives you a unique_ptr
to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.
tl;dr: You're creating copies of a
.
* Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
** More or less…
1
you can prove this to yourself byptr2->_int = 6;
and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.
– Kate Gregory
2 days ago
Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;
) the copy constructor and watch the compiler choke.
– user4581301
2 days ago
1
@user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.
– Lightness Races in Orbit
yesterday
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're not allowed to do that*, so it's a good thing you're not doing that!
Don't forget, this:
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
is this**:
std::unique_ptr<Class> ptr(new Class(a));
not this:
std::unique_ptr<Class> ptr(&a);
std::make_unique
creates a thing and gives you a unique_ptr
to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.
tl;dr: You're creating copies of a
.
* Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
** More or less…
Why am I allowed to create multiple unique pointers from a single object?
You're not allowed to do that*, so it's a good thing you're not doing that!
Don't forget, this:
std::unique_ptr<Class> ptr = std::make_unique<Class>(a);
is this**:
std::unique_ptr<Class> ptr(new Class(a));
not this:
std::unique_ptr<Class> ptr(&a);
std::make_unique
creates a thing and gives you a unique_ptr
to that thing. It does so by forwarding its arguments to the thing's constructor. Admittedly this can be confusing when you pass in the name of an existing object, leading to the copy constructor being used.
tl;dr: You're creating copies of a
.
* Well, with a no-op deleter you could do it safely, but let's save that conversation for another day…
** More or less…
edited 2 days ago
answered 2 days ago
Lightness Races in OrbitLightness Races in Orbit
294k54477811
294k54477811
1
you can prove this to yourself byptr2->_int = 6;
and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.
– Kate Gregory
2 days ago
Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;
) the copy constructor and watch the compiler choke.
– user4581301
2 days ago
1
@user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.
– Lightness Races in Orbit
yesterday
add a comment |
1
you can prove this to yourself byptr2->_int = 6;
and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.
– Kate Gregory
2 days ago
Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (Class(const Class &) =delete;
) the copy constructor and watch the compiler choke.
– user4581301
2 days ago
1
@user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.
– Lightness Races in Orbit
yesterday
1
1
you can prove this to yourself by
ptr2->_int = 6;
and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.– Kate Gregory
2 days ago
you can prove this to yourself by
ptr2->_int = 6;
and then 7 and so on. You will see that changing the Class one pointer points to has no effect on the others. They are just a bunch of copies.– Kate Gregory
2 days ago
Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (
Class(const Class &) =delete;
) the copy constructor and watch the compiler choke.– user4581301
2 days ago
Cunning of you to place the TD;DR version at the bottom of the answer, guaranteeing that most people will read the the full version first. Another simple proof is to remove the copy constructor (
Class(const Class &) =delete;
) the copy constructor and watch the compiler choke.– user4581301
2 days ago
1
1
@user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.
– Lightness Races in Orbit
yesterday
@user4581301 Conclusions and summaries have been used for centuries. Feel free to write your own answer.
– Lightness Races in Orbit
yesterday
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're allowed to make multiple copies of the object, because the class is copyable.
"Unique" in unique_ptr
doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.
You could† violate the uniqueness like this:
std::unique_ptr<Class> ptr(&a);
†Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're allowed to make multiple copies of the object, because the class is copyable.
"Unique" in unique_ptr
doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.
You could† violate the uniqueness like this:
std::unique_ptr<Class> ptr(&a);
†Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.
add a comment |
Why am I allowed to create multiple unique pointers from a single object?
You're allowed to make multiple copies of the object, because the class is copyable.
"Unique" in unique_ptr
doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.
You could† violate the uniqueness like this:
std::unique_ptr<Class> ptr(&a);
†Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.
Why am I allowed to create multiple unique pointers from a single object?
You're allowed to make multiple copies of the object, because the class is copyable.
"Unique" in unique_ptr
doesn't mean that the pointed object is the unique instance of its class. It means that no other pointer should have ownership of the pointed object. In your example, each unique pointer points to a separate instance; each of the uniquely owned by the respective pointer.
You could† violate the uniqueness like this:
std::unique_ptr<Class> ptr(&a);
†Could: The program would be well-formed and compiler would be required to compile it. But the behaviour of the program would then be undefined, so you very much should not do that.
edited 2 days ago
answered 2 days ago
eerorikaeerorika
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21
Copy construction.
– user4581301
2 days ago
1
Try this:
a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer– alter igel
2 days ago
1
Creating several
unique_ptr
to the same instance will almost certainly lead to undefined behavior, but that's not what you are doing here. Something likeauto ptr2 = std::unique_ptr<Class>(ptr.get());
would create a secondstd::unique_ptr
that points to the same instance asptr
does, and that would be problematic.– François Andrieux
2 days ago
Demo with user-defined copy constructor (called 3 times). Exemplifies the underlying copy construction.
– TrebledJ
yesterday