What is the behaviour of vector<T>().swap(x)? - c++

Question

I'm currently trying to re-implement the vector container. I just read on this website that vector<T>().swap(x);, where x is a vector previously declared, could be used to clear and reallocate the x vector to 0.
I realized with this example that I'm not familiar with the use of a constructor (or what I think is a constructor).
#include <iostream>
#include <string>
#include <vector>
typedef std::string T;
void display(const std::vector<T>& input) {
std::cout << "------START------" << std::endl;
for (int i = 0; i < input.size(); i++)
std::cout << input[i] << std::endl;
std::cout << "------END-------" << std::endl;
std::cout << "input size: " << input.size() << std::endl;
std::cout << "input capacity: " <<input.capacity() << std::endl;
}
int main() {
std::vector<T> src(1, "Matthieu0");
std::vector<T> empty;
src.push_back("Guillaume1");
src.push_back("Paul2");
src.push_back("Julien3");
std::vector<T> tmp(src);
std::cout << "------swap1------" << std::endl;
display(empty);
empty.swap(tmp);
display(empty);
std::cout << "------swap2------" << std::endl;
display(src);
std::vector<T>().swap(src);
display(src);
return 0;
}
So, I have no problem understanding swap1 behavior, crystal clear. But I don't understand what is happening under the hood with swap2 and how I could implement it.

Answer 1

But I don't understand what is happening under the hood with swap2
When you wrote
//-vvvvvvvvvvvvvvvv------------->a temporary std::vector
std::vector<T>().swap(src); //this calls swap on the temporary(unnamed) std::vector
In the above statement you're calling std::vector::swap on the temporary(unnamed) std::vector created from the expression std::vector<T>().
This is similar to the swap1 that you did with the exception that in swap1 you were calling std::vector::swap on a named object empty but here you're calling it on the temporary object.

Answer 2

Many developers want to release memory occupied by a vector after std::vector::resize(0).
It can be done since C++11 by calling std::vector::shrink_to_fit(). There is no such method until C++11, in C++98, and developers use such a trick:
Create an empty vector.
Swap contents of vectors.
Destroy the vector created on the step 1.
For performing the step 3 as much earlier as possible, the limited scope block { ... } is used.
std::vector<int> v(10);
std::cout << v.size(); << std::end; // prints 10
{
std::vector<int> tmp;
std::cout << tmp.size(); << std::end; // prints 0
v.swap(tmp);
std::cout << tmp.size(); << std::end; // prints 10
}
std::cout << v.size(); << std::end; // prints 0
It can be shorter if an unnamed temporary vector generated by a compiler is used. The life time of an unnamed temporary variable is well known:
std::vector<int> v(10);
std::cout << v.size(); << std::end; // prints 10
v.swap(std::vector<int>());
std::cout << v.size(); << std::end; // prints 0
std::vector<int> v(10);
std::cout << v.size(); << std::end; // prints 10
std::vector<int>().swap(v);
std::cout << v.size(); << std::end; // prints 0
It can be still used since C++11 for making
v.resize(0);
v.shrink_to_fit();
shorter and to guarantee freeing memory:
v.swap({});