5 Mar
2016
5 Mar
'16
10:30 a.m.
Phil Bouchard wrote:
- I removed the usage of an allocator. Now there is no need for any allocator anymore.
I'm not sure I see how this would work. If you have void f() { vector<block_ptr<T>> v; } the pointers in 'v' are root (unowned) pointers, but if you have block_ptr<vector<block_ptr<T>>> v2; the pointers in v2 are owned by v2. Yet in both cases they are on the heap. I thought that using block_ptr<vector<block_ptr<T>, block_allocator<T>>> v2; for the second case allows you to distinguish between the two.