Question

From the cpp documentation for std::vector, I see this:

void push_back ( const T& x );

I understand that push_back makes a copy of the object that I pass. But, why is the signature const T& ? By looking at this, I initially thought it takes a const reference of whatever object that I push to the vector.

Answer 1

The other option would be

void push_back(T x);

that is, taking x by value. However, this would (in C++03) result in creating an extra copy of x (the copy in the arguments to push_back). Taking x by const reference avoids this.

Let’s look at the stack for a call v.push_back(T()) taken by value:

v.push_back(T());                      // instance of T
void std::vector<T>::push_back(T x)    // copy of T
new (data_[size_ - 1]) T(x)            // copy of copy of T

Taking by const reference we get:

v.push_back(T());                             // instance of T
void std::vector<T>::push_back(const T &x)    // const reference to T
new (data_[size_ - 1]) T(x)                   // copy of T

In C++11 it would be possible (though unnecessary) to take x by value and use std::move to move it onto the vector:

v.push_back(T());                             // instance of T
void std::vector<T>::push_back(T x)           // copy of T
new (data_[size_ - 1]) T(std::move(x))        // move the copy of T