Swapping vector to free it

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I've read that the best way to free the memory of a vector is:

vector<int>().swap(my_vector);

And i don't really understand what is happening.The swap function takes 2 vectors and swaps their elements, so for instance:

vector<int>v1{1,2,3};
vector<int>v2{4,5,6};
v1.swap(v2);

v1 becomes {4,5,6} and v2 becomes {1,2,3}.This looks normal. But how does my first example free the memory? What happens inside the memory? If my_vector swaps elements with vector () (an empty vector), then doesn't the empty vector get my_vector elements, and my_vector becomes empty?

vector pointers recursion arrays c++
1 Answers

You're not only swapping with an empty vector, you're swapping with a temporary empty vector. So the memory of the two vectors is swapped, and then the destructor of the temporary vector frees the memory which originally was my_vectors.

Note that the standard effectively guarantees that in this case swap is swapping the allocated memory. Otherwise, this method could raise an exception, which it is forbidden to do in this case. Also, note that the default constructor of vector is also exception-free, so it effectively cannot allocate memory. As Aziuth correctly notes in the comment, it could theoretically have a small capacity non-dynamic initial buffer, and this would be transferred in the swap. In practice, this is probably negligible.

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