On Sun, 2007-03-18 at 19:01 +0000, John Robinson wrote: > On 18/03/2007 15:51, Timo Sirainen wrote: > > On Sat, 2007-03-17 at 01:39 +0100, Václav Haisman wrote: > [...] > >> uint32_t flags:8; > >> uint32_t uid_broken:1; > >> uint32_t expunged:1; > >> uint32_t pseudo:1; > > > > Right, I didn't think of that. But that feels a bit ugly :) I don't > > think it saves much memory anyway, so I'll keep it as uint8_t flags. > > I may be a bit of a novice at this sort of thing, but if you want > bitfields, does it matter what you say? In theory at least, > unsigned int x:1; > is all you need, it'll take one bit of storage, there's no point > specifying the size of the thing you want truncated to one bit, what > you're asking for is precisely one bit to be used as an unsigned > integer. By saying e.g. > uint8 x:1; > you're specifying the size twice, and perhaps it's not a surprise if you > get errors trying to compile it: which is it to be, 8 bits or 1?
Hmm. Looks like you're almost correct. I thought it worked so that if
you had:
uint8_t foo;
unsigned int bar:1;
That the bar would always allocate 32 bits more to the struct, but looks
like it doesn't at least with gcc.
There is however one difference:
struct a { uint8_t f1; uint8_t f2:1; }
struct b { uint8_t f1; unsigned int f2:1; }
sizeof(struct a) == 2 but sizeof(struct b) == 4
But this doesn't matter much. I don't think there's a single struct in
Dovecot's code that doesn't contain 32bit or 64bit fields.
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