3 ms·
Thanks to array decay to pointer, we basically have `*(array_label+offset)` which in this case of yours we have `*(offset+array_label)`; in other words, `*(arr+
by stefanos82 2y ago
Thanks to array decay to pointer, we basically have `*(array_label+offset)` which in this case of yours we have `*(offset+array_label)`; in other words, `*(arr+4)` is the same as `*(4+arr)`...that's it, really!