Commit
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
alloc_system: don’t assume MIN_ALIGN for small sizes, fix #45955
The GNU C library (glibc) is documented to always allocate with an alignment of at least 8 or 16 bytes, on 32-bit or 64-bit platforms: https://www.gnu.org/software/libc/manual/html_node/Aligned-Memory-Blocks.html This matches our use of `MIN_ALIGN` before this commit. However, even when libc is glibc, the program might be linked with another allocator that redefines the `malloc` symbol and friends. (The `alloc_jemalloc` crate does, in some cases.) So `alloc_system` doesn’t know which allocator it calls, and needs to be conservative in assumptions it makes. The C standard says: https://port70.net/%7Ensz/c/c11/n1570.html#7.22.3 > The pointer returned if the allocation succeeds is suitably aligned > so that it may be assigned to a pointer to any type of object > with a fundamental alignment requirement https://port70.net/~nsz/c/c11/n1570.html#6.2.8p2 > A fundamental alignment is represented by an alignment less than > or equal to the greatest alignment supported by the implementation > in all contexts, which is equal to `_Alignof (max_align_t)`. `_Alignof (max_align_t)` depends on the ABI and doesn’t seem to have a clear definition, but it seems to match our `MIN_ALIGN` in practice. However, the size of objects is rounded up to the next multiple of their alignment (since that size is also the stride used in arrays). Conversely, the alignment of a non-zero-size object is at most its size. So for example it seems ot be legal for `malloc(8)` to return a pointer that’s only 8-bytes-aligned, even if `_Alignof (max_align_t)` is 16.
- Loading branch information