Question

I was wondering why the vector templates perform two allocations, when only one seems to be
necessary.

For example this:

#include <vector>
#include <iostream>

class A {
    public:
         A(const A &a) {
            std::cout << "Calling copy constructor " << this << " " << &a << "\n";
        }
         A() {
            std::cout << "Calling default constructor " << this << "\n";
        }
        ~A() {
            std::cout << "Calling destructor " << this << "\n"; 
        }
};

int main(int argc, char **argv)
{
    std::vector <A> Avec;

    std::cout << "resize start\n";
    Avec.resize(1);
    std::cout << "resize end\n";

    return 0;
}

Outputs:

resize start
Calling default constructor 0x7fff9a34191f
Calling copy constructor 0x1569010 0x7fff9a34191f
Calling destructor 0x7fff9a34191f
resize end

Answer 1

It isn’t performing two allocations, it is creating an object by the default constructor to pass into resize, then copying that object into the new position, then destructing the argument.

If you look at the arguments to resize:

void resize(n, t = T())

It has as a defaulted argument a default constructed object of type T (this is the default constructor being called in your output). Then, within the function, it copies this into the correct position (this is the copy constructor). After the resize function ends, destroys the argument (the destructor call in the output).