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Andrew Koenig

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Elegance or Trickery: How To Free Memory

October 12, 2011

I'm continuing the elegance or trickery theme with an example that I think I originally got from Herb Sutter.

Suppose v is a vector<T>, and you want to free all of the memory that v is using. The obvious way to do so is

v.clear();

Alas, as so often happens in programming, the obvious solution doesn't do quite what we expect. This statement is equivalent to

v.erase(v.begin(), v.end());

which does, indeed, destroy all of v's elements. What it does not do is free all of the memory that those elements occupied. In fact, the library is prohibited from doing so, because calling v.erase is not allowed to change v's capacity, which is the number of elements that v can hold without reallocation.

One way to free that memory is to create a temporary, empty vector, and swap that vector's contents with v's elements. Then destroying the temporary vector will free the memory that v formerly used:

{ vector<T> temp;
   v.swap(temp); }

One would think that it would be possible to avoid having to make up a name for this vector by writing

v.swap(vector<T>());

instead of defining a named variable. Unfortunately, this example does not work because vector<T>() is an rvalue, and the parameter of v.swap expects an lvalue. What does work, however, is to swap the arguments of swap, so to speak, and write it this way:

vector<T>().swap(v);

The point is that even though vector<T>() is an rvalue, one can still call its member functions. In particular, it is possible to call its swap member function, so long as you give that member function an lvalue — in this case, v — as its argument.

As before, I invite discussion: Is this usage elegant, tricky, or both? And what characteristics of the code lead you to that conclusion?

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