Why this code does allow to push_back unique_ptr do vector?

Question

so I thought adding unique to vector shouldn't work. Why does it work for the below code? Is it cause by not setting copy ctor as "deleted"??

#include <iostream>
#include <vector>
#include <memory>

class Test
{
    public:
        int i = 5;
};


int main()
{
    std::vector<std::unique_ptr<Test>> tests;
    tests.push_back(std::make_unique<Test>());
    
    for (auto &test : tests)
    {
        std::cout << test->i << std::endl;
    }

    for (auto &test : tests)
    {
        std::cout << test->i << std::endl;
    }
}

Why do you think you can't make a vector of unique_ptr? You can, as long as you never copy such pointers (you can only _move_ them). — chi, Jun 30 '21 at 08:45
_I also wonder how 'for' loop references vector elements._ — The for-loop variable is a reference. References do not copy/move anything by themselves. — Daniel Langr, Jun 30 '21 at 09:47

score 7 · Answer 1 · answered Jun 30 '21 at 08:46

There is no copy here, only moves.

In this context, make_unique will produce an instance of unique pointer which is not named, and this push_back sees it as a r-value reference, which it can use as it wants.

It produce pretty much the same result than this code would:

std::vector<std::unique_ptr<Test>> tests;
auto ptr = std::make_unique<Test>();
tests.push_back(std::move(ptr));

This is called move semantics if you want to search more info on the matter. (and this only works from c++11 and beyond)

Welcome, I tried to stay concise, this can get pretty complex pretty fast, and people online explain it way better than I ever will :) — Julien Lopez, Jun 30 '21 at 11:11

score 2 · Answer 2 · answered Jun 30 '21 at 08:47

2

There are two overloads of std::vector::push_back according to https://en.cppreference.com/w/cpp/container/vector/push_back

In your case you will use the one with rvalue-ref so no copying required.

answered Jun 30 '21 at 08:47

jdfa

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Why this code does allow to push_back unique_ptr do vector?

2 Answers2