Re: data structure question

[Glynn Clements <glynn@gclements.plus.com>]

gumbold wrote:

> > then sizeof(struct abc) would be 8 and the offset of abc from the
> > beginning of the structure would be 4.
> > 
> > With:
> > 
> > 	struct abc {
> > 	    __u16 a;
> > 	    __u16 b;
> > 	    char abc[0];
> > 	};
> > 
> > sizeof(struct abc) would be 4 and the offset of abc from the beginning
> > of the structure would still be 4 (i.e. p->abc refers to the first
> > byte after the structure).
> 
> I understand. But if you doing `p->abc = (char *)malloc(111)'

That would be an error. p->abc is an array, not a pointer.

Arrays and pointers aren't the same. They are often confused, for
several reasons:

1. An array is automatically converted to a pointer to its first
element whenever it occurs as an expression. This doesn't apply to
lvalues (e.g. in declarations or as the argument to sizeof).

2. In a declaration, the syntax foo[...] declares an array named foo,
but in the expression foo[...], foo is a pointer (it might be an array
which has been converted to a pointer according to point 1 above, but
it's still a pointer).

3. In a function prototype, pointer arguments can be declared using
either "T foo[]" or "T *foo".

-- 
Glynn Clements <glynn@gclements.plus.com>