Hi Paul,
> > 1) What is the alignment problem if the array element type is 'char'?
>
> Suppose we have a platform where the alignment of each basic type is
> equal to its size, where sizeof (int) == 4 and sizeof (char) == 1, and
> where we have 'struct s { int n; char d[]; };' and suppose we want to
> allocate a 'struct s' with a 3-element flexible array member 'd'. Then
> 'offsetof (struct s, d) + 3 * sizeof (char)' is 7, and when we call
> 'malloc (7)' malloc is entitled to assume that memory is being allocated
> for an array of 7 one-byte objects, so it can return an address that is
> not a multiple of 4, which means that the pointer that malloc returns is
> invalid for 'struct s *'. To fix the problem, we can use 'malloc
> (FLEXSIZEOF (struct s, d, 3))' instead, as FLEXSIZEOF yields 8 on this
> platform, and malloc (8) must return an address that is a multiple of 4.
>
> > don't we need to apply a 'ceil'-like alignment to the malloc result?
> No, because if malloc is given an argument like 8 that is a multiple of
> sizeof (int), it must return a pointer suitable for 'int *'.
Oh, I see. Thanks. I was assuming that malloc (N) always returns a
multiple of max_align_t.
> Perhaps I should update the comment to make this all clearer, though it
> is hard to be clear and accurate and terse in this area.
You found the right words now :) - I would expect to see such
explanations in the .h file, not in the .m4 file.
Bruno
