So UB, then? All complaint solutions I could find to this use memcpy() instead. Some use __attribute__ ((__packed__)) for structs, but that seems to have its own gotchas (https://stackoverflow.com/a/7956942), but not relevant here.
You can't use memcpy here, you have to modify the original memory to mark it as freed. You also can't allocate a new buffer to do the copy into because you are the allocator.
The C standard specifically states that a cast (T ) to and from (void ) is validly defined behavior and that you must get the original pointer back. It's also valid to go from (T ) to (U ) and back again if and only if T and U have the same alignment requirements. (void ) is required to have no alignment requirements because you can't directly de-reference it since the result would have type (void).
__attribute__((__packed__)) is a completely different topic about how the layout of the struct gets decided and what padding might be used.
What's going on here is that malloc() gives the caller a (void ) that points to one position past a (struct header_t ) and then free is casting it back from (void ) to (struct header_t ) and then going back one element in the set to get the original header with the meta-data about the allocation, this is perfectly fine because the pointer is never anything other than (void ) or (struct header_t ) as far as the language is concerned. The caller might turn the element after the (struct header_t ) into something else but the original (struct header_t *) is always the same.
> The caller might turn the element after the (struct header_t ) into something else
Isn't this exactly the issue with this malloc implementation? The pointer returned that points past the header by sizeof(header) may not be aligned for subsequent types.
This is a problem inherit to all malloc() implementations, since it never gets any type information it can't ever make any adjustments about alignment of the final type that's involved. I'm not actually sure what the fully correct way to ensure that would be.