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Why am I allowed to create multiple unique pointers from a single object?
The Next CEO of Stack OverflowWhy is 'this' a pointer and not a reference?Cannot assign pointer in a self-referential object in Visual Studio 2010Pretty-print C++ STL containersHow C++ reference worksWhy is reading lines from stdin much slower in C++ than Python?Namespaces and the Pre-ProcessorWhy should I use a pointer rather than the object itself?Returning a STL container of unique pointersWhy does my object appear to be on the heap without using `new`?Print Object that is pointed to
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;
c++
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;
c++
New contributor
15
Copy construction.
– user4581301
14 hours ago
1
Try this:a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer
– alter igel
14 hours ago
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
14 hours ago
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;
c++
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;
c++
c++
New contributor
New contributor
edited 14 hours ago
Shridhar R Kulkarni
1,55821328
1,55821328
New contributor
asked 14 hours ago
Lokas BeardLokas Beard
585
585
New contributor
New contributor
15
Copy construction.
– user4581301
14 hours ago
1
Try this:a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer
– alter igel
14 hours ago
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
14 hours ago
add a comment |
15
Copy construction.
– user4581301
14 hours ago
1
Try this:a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer
– alter igel
14 hours ago
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
14 hours ago
15
15
Copy construction.
– user4581301
14 hours ago
Copy construction.
– user4581301
14 hours 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
14 hours ago
Try this:
a.int_ = 5;
after you create each unique_ptr
and before you print the contents of each pointer– alter igel
14 hours 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
14 hours 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
14 hours ago
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
13 hours ago
9
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
13 hours 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
12 hours 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
12 hours ago
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
13 hours ago
9
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
13 hours 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
13 hours ago
9
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
13 hours 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 14 hours ago
Vittorio RomeoVittorio Romeo
59.3k17163307
59.3k17163307
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
13 hours ago
9
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
13 hours 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
13 hours ago
9
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
13 hours 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
13 hours 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
13 hours ago
9
9
@JordanMotta: adding a non-default constructor does not inhibit implicit generation of the copy constructor. See i.stack.imgur.com/b2VBV.png
– Vittorio Romeo
13 hours 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
13 hours 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
12 hours 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
12 hours 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
12 hours 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
12 hours 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…
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 13 hours ago
answered 13 hours 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
12 hours 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
12 hours ago
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
12 hours 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
12 hours ago
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
12 hours 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
12 hours 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
12 hours 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
12 hours ago
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 13 hours ago
answered 13 hours ago
eerorikaeerorika
88.2k663134
88.2k663134
add a comment |
add a comment |
Lokas Beard is a new contributor. Be nice, and check out our Code of Conduct.
Lokas Beard is a new contributor. Be nice, and check out our Code of Conduct.
Lokas Beard is a new contributor. Be nice, and check out our Code of Conduct.
Lokas Beard is a new contributor. Be nice, and check out our Code of Conduct.
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15
Copy construction.
– user4581301
14 hours ago
1
Try this:
a.int_ = 5;
after you create eachunique_ptr
and before you print the contents of each pointer– alter igel
14 hours 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 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
14 hours ago