Question

The following is an excerpt from a code sample in K&R’s The C Programming Language (section 8.7):

typedef long Align;

union header {
    struct {
        union header *ptr;
        unsigned size;
    } s;
    Align x;
};

typedef union header Header;

And here is the explanatory excerpt:

To simplify alignment, all blocks are
multiples of the header size, and the
header is aligned properly. This is
achieved by a union that contains the
desired header structure and an
instance of the most restrictive type,
which we have arbitrarily made a long.

So, as I understand it, the idea is to ensure that a header instance takes up a number of bytes that is a multiple of sizeof(Align). The goal makes perfect sense to me.

But consider the case where long is 64 bits, int is 32 bits, and a pointer is 64 bits. In that case, header.s will be 64+32 = 96 bits. Thus, sizeof(header) will be 96 bits, which is not a multiple of 64 as intended.

In such a case, I suppose it would be necessary to define Align as something else (perhaps double). But I’m not quite sure whether I’m fully understanding what would dictate that choice for a particular architecture. Is it just the “biggest” type? And what if long is the biggest type – couldn’t it fail to work in the way I described above?

Thanks

Answer 1

The compiler will adjust the size of header so that ptr, size, and x can all be retrieved efficiently from elements in a header[] array. On most architectures, the presence of x means that sizeof(header) will be increased to a multiple of sizeof(long).