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0 pointshalayli9y ago0 comments

But arr != &arr even though they have the same value. #8 applies to arr (P), but in the post OP is using &arr which is a ptr to array[x] and doesn't apply to it.

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cperciva9y ago

That's why I quoted paragraph 7: arr is not an element of an array, so &arr (being a pointer to an object which is not an element of an array) behaves like a pointer to the first element of an array of length one with the type of the arr as its element type.

So &arr behaves like it's a pointer to the start of int[5][1].

halayliOP9y ago

Yes &arr does behave like arr when it comes to ptr arithmetic but the compiler does not guarantee that &arr + 1 does not overflow. It only guarantees arr + 1. if you have a ptr from heap, ptr + 1 if not alloced previously is UB.

> If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow;

this point doesn't apply to &arr + 1.

bonzini9y ago

How so? If the array has five elements, you can pass &arr to a function that expects an int[5][], and that function certainly can build a pointer that points one past the last element.

Likewise, the compiler ensures that you can build &arr[5] and that is the same address as &arr+1. &arr+1 cannot overflow.

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wruza9y ago

Can't we declare pointer of type &arr, assign it there and be sure that it points to equivalent of array[1] of &arr? If yes, then is it logically possible to have UB on that?

sirclueless9y ago

You can define a pointer of type `int (*)[5]` and assign `(&arr)[1]` to it. That's fine, it's a pointer to the 5-element array just after the one we're sure is valid.

Dereferencing the pointer is UB, but you can create the pointer, assign it to a variable, etc.

dllthomas9y ago

I know it's not what you were trying to communicate, but actually "arr != &arr" is false.

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