foo[5] is the same thing as * (foo + 5) (since "foo" is the address of the first element), which is equivalent to * (5 + foo) which is equivalent to 5[foo]! EDI

foo[5] is the same thing as * (foo + 5) (since "foo" is the address of the first element), which is equivalent to * (5 + foo) which is equivalent to 5[foo]! EDIT: hn ate up my * s.