Jon Dowland wrote:
Because sizeof is not really the size of the struct, it is
the distance between adjacent structs in an array.
Alignment forces the extra bytes
I'm not quite sure I get what you're saying here. Yes,
alignment pads out the structure. But I'm not sure where
arrays come into it :- sizeof(struct whatever) is applicable
to a single instance of a struct too.
What he's saying is that
struct FredStruct Fred[2];
...
sizeof(struct Fred[0])
is the same as
(void *)(&Fred[1])-(void *)(&Fred[0])
Structures may get padding at the end precisely for this
reason. It is not obvious that this is the case when one
talks about a single instance.
Mike
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