See? I read that both forms are translated to * (c + n), which
explains the equivalence. Weird, no?
Well, that surely is evil, yet perfectly valid C code. But personally, I don't find it surprising. Of course, you need to grok what A[n] really is -- i.e., syntactic sugar for *(A + n), as you just explained.
Well, that surely is evil, yet perfectly valid C code. But personally, I don't find it surprising. Of course, you need to grok what A[n] really is -- i.e., syntactic sugar for *(A + n), as you just explained.