Lambda as an argument to std::vector::emplace_back

Question

I came across code that passes a lambda as an argument to emplace_back, which confused me. So I wrote a small test to validate the syntax like this:

struct A
{
    int a;
    A(){cout<<"constructed"<<endl;}
    A(const A& other){a = other.a; cout<<"copied"<<endl;}
    A(const A&& other){a = other.a; cout<<"moved"<<endl;}
};

int main()
{
   vector<A> vec;
   vec.emplace_back(
       []{A a;
       a.a = 1;
       return a;   
       }());

   A a2;
   a2.a = 2;
   vec.emplace_back(std::move(a2));

  return 0;
}

I have two questions: 1) Can someone clarify how a lambda can be passed as an argument to emplace_back? I had only seen constructor arguments being passed to emplace back in the past. 2) The output was:

constructed
moved
constructed
moved
copied

Where is the last copied coming from? Why aren't the two approaches equivalent.

Answer 1

Can someone clarify how a lambda can be passed as an argument to emplace_back?

You are not passing a lambda. Instead, you pass the result of calling the lambda.

Where is the last copied coming from?

It comes from the vector. When a2 is inserted to vec , a reallocation occurs, which reallocates memory and copy the objects in the old memory to the new memory.

If you specify the move constructor as noexcept , ie

A(const A&& other) noexcept {a = other.a; cout<<"moved"<<endl;}

then std::vector will move the objects during reallocation instead of copying, and you can see the last copy becomes move.

Answer 2

Notice the () after the lambda definition, which calls the lambda. This code is equivalent to the following:

int main()
{
    auto make_a = []()
    {
        A a;
        a.a = 1;
        return a;
    };

    vector<A> vec;
    vec.emplace_back(make_a());

    // ...
}

Which, since the lambda has empty capture, is equivalent to this:

A make_a()
{
    A a;
    a.a = 1;
    return a;
}

int main()
{
    vector<A> vec;
    vec.emplace_back(make_a());

    // ...
}