There was no actual dereferencing, this is just pointer arithmetic. it's "line6 = podhd + offset". There isn't really a better way to do it.
That becomes a problem with modern optimizing compilers, because they make assumptions about your code based on what's allowed according to the spec.
struct usb_line6 *line6 = &podhd->line6;
-> and . are not arithmetic operands.
If you look at the definition of the -> and [], you might learn something!
[] operator doesn't have any place in this debate.
What @jwatte tried to say is that `->` works like this:
// foo is struct thingity*
&foo->bar == foo + offsetof(struct thingity, bar);
// s is char*
*(s+i) == s[i]It does however specify that [] operator is equivalent to pointer arithmetic. But we are talking ->,. operators and not [].