Clearly, you don't have anything constructive to say, so you have to resort to ad hominem.
[] operator doesn't have any place in this debate.
[] 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 [].