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by int_19h
3460 days ago
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It doesn't matter what it is a pointer to, type-wise. It is still a pointer one-past-the-end, and it is being dereferenced. Also, &arr is not a pointer to a pointer. It's a pointer to an array. Specifically, its type is int(* )[5] in this example, and so when you dereference it, the result is of type int[5]. So if you do e.g. sizeof(* &arr + 1), you'll get 5 * sizeof(int). |
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