Is storing an invalid pointer automatically undefined behavior?

Question

Obviously, dereferencing an invalid pointer causes undefined behavior. But what about simply storing an invalid memory address in a pointer variable?

Consider the following code:

const char* str = "abcdef";
const char* begin = str;
if (begin - 1 < str) { /* ... do something ... */ }

The expression begin - 1 evaluates to an invalid memory address. Note that we don't actually dereference this address - we simply use it in pointer arithmetic to test if it is valid. Nonetheless, we still have to load an invalid memory address into a register.

So, is this undefined behavior? I never thought it was, since a lot of pointer arithmetic seems to rely on this sort of thing, and a pointer is really nothing but an integer anyway. But recently I heard that even the act of loading an invalid pointer into a register is undefined behavior, since certain architectures will automatically throw a bus error or something if you do that. Can anyone point me to the relevant part of the C or C++ standard which settles this either way?

Answer 1

Any use of an invalid pointer yields undefined behaviour. I don't have the C Standard here at work, but see 'invalid pointers' in the Rationale: http://www.open-std.org/jtc1/sc22/wg14/www/C99RationaleV5.10.pdf

Answer 2

I have the C Draft Standard here, and it makes it undefined by omission. It defines the case of ptr + I at 6.5.6/8 for

• 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.
• 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.

Your case does not fit any of these. Neither is your array large enough to have -1 adjust the pointer to point to a different array element, nor does any of the result or original pointer point one-past-end.

Answer 3

Your code is undefined behavior for a different reason:

the expression begin - 1 does not yield an invalid pointer. It is undefined behavior. You are not allowed to perform pointer arithmetics beyond the bounds of the array you're working on. So it is the subtraction itself that is invalid, and not the act of storing the resulting pointer.

Answer 4

The access violation will occur when you dereference the variable. Consider the following:

void IntRef(int &value)
{
    // At this point the code will not have thrown an exception. 
    // The minute you attempt to use "value" it will crash.
    value = 42; // Crash
}

int main()
{
    int *const pNull = NULL;
    IntRef(*pNull);
    return 0;
}

Answer 5

$5.7/6 - "Unless both pointers point to elements of the same array object, or one past the last element of the array object, the behavior is undefined.75)"

Summary, it is undefined even if you do not dereference the pointer.

Answer 6

Yes, it's undefined behavior. See the accepted answer to this closely related question. Assigning an invalid pointer to a variable, comparing an invalid pointer, casting an invalid pointer triggers undefined behavior.

Answer 7

Some architectures have dedicated registers for holding pointers. Putting the value of an unmapped address into such a register is allowed to crash. Integer overflow/underflow is allowed to crash. Because C aims to work on a broad variety of platforms, pointers provide a mechanism for safely programming unsafe circuits.

If you know you won't be running on exotic hardware with such finicky characteristics, you don't need to worry about what is undefined by the language. It is well-defined by the platform.

Of course, the example is poor style and there isn't a good reason to do it.