Query: mem_temp_alloc 8 bytes alignment of size

[sujankh]

Hi, when looking at the the bump allocator i was assuming the following would modify size to be the next multiple of 8 but looks like it is not.

    size = ((size >> 3) + 7) << 3; // allign to 8 bytes

https://github.com/phoboslab/wipeout-rewrite/blob/ecf1e635082188d3294e10abb255128204854127/src/mem.c#L49

Here is what i see when i add some logs

size: 16  --> becomes: 72
size: 4224  --> becomes: 4280

When user request 16 bytes, do we need to reserve 72 bytes in the temp? We could just reserve 16 bytes right?

[phoboslab]

Good catch, thanks!