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0 pointsbonzini9y ago0 comments

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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halayli9y ago

The compiler guarantees that arr + 1 doesn't overflow by making sure arr's address is small enough to not overflow when accessing one element past the array size. &arr + 1 is not one past the array you asked the compiler to allocate.

if you're on a 16bit system and you define char x[36], the compiler guarantees that x's address is not more than 65500. if you do &x + 1 then you'll overflow, x + 1 won't.

You can pass whatever you want to the functions and apply the operands you want and the compiler will happily comply with you. But when you pass it 65500 and add 72 to it, it's going to overflow.

wruza9y ago

Wait, wait.

  char *p = x;
  p += 36; // overflow?

As arr == &arr, so are pointers P and Q that point just after last array item (1+&x[35]) and just after entire array (1+&x). As 6.5.6.8 above said, P is okay, and so must be Q. They said about last element, not second. Can you please explain why is x+1 even an argument?

>if you're on a 16bit system and you define char x[36], the compiler guarantees that x's address is not more than 65500 65499?

int_19h9y ago

No, he is correct. The C and C++ standards do allow pointers past the last element of an array to be produced via a pointer to an element of said array. They also have a provision where a pointer to a nonarray value is treated as if it were a 1-element array (so you can do "int x; int* p = *x + 1"). But in this case, the value is obviously an array object, and it's not an element to another array; hence, it is not legal to do &arr + 1 ("... otherwise, the behavior is undefined").

This is 6.3.6 "Additive operators" in ISO C90 standard, for anyone curious.

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clarry9y ago

Then you have a broken compiler.

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