Why does an object get destroyed when pushing it to a vector?

Question

I would like to know why an element I push back into a vector gets its destructor called in this situation:

#include <iostream>
#include <vector>

class Foo
{
public:
    Foo(int a)
     : m_a(a)
    {
        std::cout << "Foo ctor() " << m_a << std::endl;
    }
    ~Foo()
    {
        std::cout << "Foo dtor() " << m_a << std::endl;
    }
private:
    int m_a;
};

class FooStorage
{
public:
    static void createFoo(int a)
    {
        m_foos.push_back(Foo(a));
    }

    static std::vector<Foo> m_foos;
};

std::vector<Foo> FooStorage::m_foos;

int main()
{
    std::cout << "Before: " << FooStorage::m_foos.size() << std::endl;
    FooStorage::createFoo(53);
    std::cout << "After: " << FooStorage::m_foos.size() << std::endl;

    return 0;
}

This prints out the following:

Before: 0
Foo ctor() 53
Foo dtor() 53
After: 1
Foo dtor() 53

I'd like to know:

What gets deleted? (inbetween the 'Before' and 'After' couts)
Why does it get deleted?
What ends up in the vector?

What is the point of the FooStorage class? It just makes it harder to read. — Neil Kirk, Nov 07 '14 at 13:15
You are copying it to your vector, use Foo& or Foo* for the storage if you do not want copy — AlexanderBrevig, Nov 07 '14 at 13:16
Use emplace_back to create the object directly into the vector. — Neil Kirk, Nov 07 '14 at 13:16
@AlexanderBrevig Using references will not work as they are not copyable. — Some programmer dude, Nov 07 '14 at 13:16
I suggest you study e.g. [this `std::vector` reference](http://en.cppreference.com/w/cpp/container/vector). — Some programmer dude, Nov 07 '14 at 13:17
@NeilKirk `FooStorage` seems to be a factory and a storage container for `Foo`s — AlexanderBrevig, Nov 07 '14 at 13:17
@NeilKirk `emplace_back` seems to still cause the destructor to be called there. — manabreak, Nov 07 '14 at 13:18
`Foo(a)` creates a temporary object which is either moved or copied when the call to `push_back()` occurs. The moved to or copied to object is then stored in the `vector` while the temporary object's destructor gets invoked. — foobar, Nov 07 '14 at 13:18

score 9 · Accepted Answer · edited Nov 07 '14 at 13:27

9

You're creating a temporary object here:

m_foos.push_back(Foo(a));
//               ^^^^^^

That object's destructor will be called when the full expression has ended, but it will have been copied (or moved) by push_back() into the vector.

To prevent the creation of a temporary you can construct in-place using emplace_back():

m_foos.emplace_back(a);

edited Nov 07 '14 at 13:27

Barry

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answered Nov 07 '14 at 13:19

David G

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This does not work on IDE, the IDEs ARE the problem. Try compiling from the terminal. – Marine Galantin Feb 13 '22 at 00:29
Nothing appears from the compiler or at running time, but the results are completely false and suffer from undefined behavior. Check the accepted answer from: https://stackoverflow.com/questions/4303513/push-back-vs-emplace-back I posted this comment because I spent a few hours debugging a bug coming from CLion and the code was running completly fine from the terminal. In such case, I did not need `emplace_back` `push_back` worked. – Marine Galantin Feb 13 '22 at 19:19

score 1 · Answer 2 · answered Nov 07 '14 at 13:21

1

m_foos.push_back(Foo(a));

This would result in two constructor call ( but it entirely depends on the compiler it can optimize away).

Foo(a)

is a temporary which would be constructed ( one constructor call )and then copied to vector ( copy constructor called ). As usual temporaries are alive till the end of expression where they are used. So when push_back ends it would call destructor for this temporary ( that's why you are seeing a call to destructor).

answered Nov 07 '14 at 13:21

ravi

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The copy/move constructor can't be elided here. Which one is called, depends. In this case it is copy ctor, because there is user-declared destructor. – Anton Savin Nov 07 '14 at 13:33

Why does an object get destroyed when pushing it to a vector?

2 Answers2

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