jhopkin@reflectionsinteractive.com wrote:

Daniel Frey wrote:

...

jhopkin@reflectionsinteractive.com wrote:

...

The STL implementation I have been using seems to allocate a buffer on first insertion into any container of a given type, and frees it only when it feels like it (often on program shut-down).

Sounds like a pool allocator. Question: Does .clear() for your STL return the buffer to the pool or is it still bound to the instance? What happens in case of the .swap()-trick?

I believe .clear() and the swap trick do what's expected per instance. Specifically for vector, clear generally won't release any memory to the pool; the swap trick releases it all.

The swap trick is almost guaranteed to reduce the capacity to an implementation-defined minimum value, unless the original vector has size() == 0 && capacity() != 0, in which case lhs.swap( rhs ) can detect lhs.size() == rhs.size() == 0 and do nothing. ;-) That's the theory, of course, in practice the swap() trick always works. In standard-conforming implementations, clear() does not change the capacity. In the Dinkumware STL, however, it releases the memory and sets capacity() to 0. All this means that, in portable code: - V().swap(v); should be used to reduce capacity to about zero. - v.erase( v.begin(), v.end() ); should be used to set size to zero, leaving capacity unchanged. - v.clear() should not be used if v will be repopulated and portability/performance is a concern, until Dinkumware fix their STL.