Does vector<T>::push_back() call any matching constructor of its type T?

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I used a reference_wrapper recently like so:

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

struct A {

struct B {
    B() {};
    B(A& aA) {};
    B(const B&) = default;

int main () {
    A a;
    B b;

    std::vector<std::reference_wrapper<B>> rvector;
    rvector.push_back(std::reference_wrapper<B>(b)); // (1)
    rvector.push_back(b); // (2)

1 and 2, are both compiling and working just fine and I wonder which is the right one! So I thought that the following could work too:

    std::vector<B> bvector;
    bvector.push_back(a); // (3)
    bvector.push_back(b); // (4)
    std::cout << bvector.size() << "\n";

yeap, it works! So I came to the conclusion that push_back just calls the type's T constructor with the given parameter. Documentation though mentions that the parameter is of type T and eventually the copy/move ctor.

When I tried 1 and 2 with shared_ptr though:

    std::vector<std::shared_ptr<B>> pvector;
    pvector.push_back(std::shared_ptr<B>(new B())); // (5)
    pvector.push_back(new B()); // (6)

I get

no matching function for call to 'std::vector<std::shared_ptr<B> >::push_back(B*)'
  • Why 2 and even 3 work but 6 doesn't?
  • There should be a shared_ptr<B>constructor that takes B* as parameter, isn't there? So, why 6 does not compile (since 3 does)?
  • Since 3 compiles and works just fine, would it be ok to use and count on it (if it wasn't a bad approach since there is the emplace_back)?

Why 2 and even 3 work but 6 doesn't?

std::vector::push_back takes T as parameter, then it'll work if you're passing something which could be converted to T implicitly.

(2) works because B could convert to std::reference_wrapper<B> implicitly. (3) works because A could convert to B implicitly. (6) doesn't work because B* can't convert to shared_ptr<B> implicitly. Note that the constructor of shared_ptr taking raw pointers is marked as explicit.

Since 3 compiles and works just fine, would it be ok to use and count on it (if it wasn't a bad approach since there is the emplace_back)?

Yes you can use it, but note that emplace_back is more efficient sometimes; it will construct the element in-place directly without any implicit conversion, instead of constructing a temporary T and then move it into the vector (as push_back).

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push_back takes a T const& or a T&&.

So if we have a vector<X>, push_back(expr) works if X const& x=expr; compiles or X&& x=expr; compiles, and it isn't ambiguous which matches better, and the move/copy from the argument is allowed.

Those compile for your first cases, but not your shared ptr from ptr, because X x=expr; doesn't call explicit constructors.

You can vec.emplace_back(expr) if you want to consider explicit constructors.

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Constructor of std::reference_wrapper is a converting constructor that allows implicit conversion from B to std::reference_wrapper<B> and that's why (2) works.

However constructor (3) of std::shared_ptr is explicit, hence you have to specify std::shared_ptr<B> explicitly in (5).

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