C null pointer arithmetic

Question

I noticed this warning from Clang:

warning: performing pointer arithmetic on a null pointer
has undefined behavior [-Wnull-pointer-arithmetic]

In details, it is this code which triggers this warning:

int *start = ((int*)0);
int *end = ((int*)0) + count;

The constant literal zero converted to any pointer type decays into the null pointer constant, which does not point to any contiguous area of memory but still has the type pointer to type needed to do pointer arithmetic.

Why would arithmetic on a null pointer be forbidden when doing the same on a non-null pointer obtained from an integer different than zero does not trigger any warning?

And more importantly, does the C standard explicitly forbid null pointer arithmetic?

---

Also, this code will not trigger the warning, but this is because the pointer is not evaluated at compile time:

int *start = ((int*)0);
int *end = start + count;

But a good way of avoiding the undefined behavior is to explicitly cast an integer value to the pointer:

int *end = (int *)(sizeof(int) * count);

Answer 1

The C standard does not allow it.

6.5.6 Additive operators (emphasis mine)

8 When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P)+N (equivalently, N+(P)) and (P)-N (where N has the value n) point to, respectively, the i+n-th and i-n-th elements of the array object, provided they exist. Moreover, if the expression P points to the last element of an array object, the expression (P)+1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q)-1 points to the last element of the array object. 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; otherwise, the behavior is undefined. If the result points one past the last element of the array object, it shall not be used as the operand of a unary * operator that is evaluated.

For the purposes of the above, a pointer to a single object is considered as pointing into an array of 1 element.

Now, ((uint8_t*)0) does not point at an element of an array object. Simply because a pointer holding a null pointer value does not point at any object. Which is said at:

6.3.2.3 Pointers

3 If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function.

So you can't do arithmetic on it. The warning is justified, because as the second highlighted sentence mentions, we are in the case of undefined behavior.

Don't be fooled by the fact the offsetof macro is possibly implemented like that. The standard library is not bound by the constraints placed on user programs. It can employ deeper knowledge. But doing this in our code is not well defined.