The man page for posix_memalign says
The function posix_memalign() allocates size bytes and places the address of the allocated memory in *memptr. The address of the allocated memory will be a multiple of alignment, which must be a power of two and a multiple of sizeof(void *). If size is 0, then the value placed in *memptr is either NULL, or a unique pointer value that can later be successfully passed to free(3).
And yet Miri found libstd calling this function with an alignment of 4 on a 64bit-platform. This happens when size=2, align=4. The fact that size<align makes it enter the code path for posix_memalign.
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