3 ms·

Thanks for the response. So it sounds like it is not so much the allocation/de-allocation savings that matters, but rather that you get to skip initializing th

Thanks for the response. So it sounds like it is not so much the allocation/de-allocation savings that matters, but rather that you get to skip initializing the memory because it is already known to be a valid instance of a given struct type?

jlokier 7y ago
Yes, that. See also SLAB allocation, used in many OS kernels: https://en.wikipedia.org/wiki/Slab_allocation https://en.wikipedia.org/wiki/Slab_allocation