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?

Consi

Answer 1

The correct answers have been given years ago, but I find it interesting that the C99 rationale [sec. 6.5.6, last 3 paragraphs] explains why the standard endorses adding 1 to a pointer that points to the last element of an array (p+1):

An important endorsement of widespread practice is the requirement that a pointer can always be incremented to just past the end of an array, with no fear of overflow or wraparound

and why p-1 is not endorsed:

In the case of p-1, on the other hand, an entire object would have to be allocated prior to the array of objects that p traverses, so decrement loops that run off the bottom of an array can fail. This restriction allows segmented architectures, for instance, to place objects at the start of a range of addressable memory.

So if the pointer p points to an object at the start of a range of addressable memory, which is endorsed by this comment, then p-1 would generate an underflow.

Note that integer overflow is the standard's example for undefined behavior [sec. 3.4.3], as it depends on the translation environment and the operating environment. I believe it is easy to see that this dependence on the environment extends to pointer underflow.

This is why the standard explicitly makes it undefined behavior [in 6.5.6/8], as noted by other answers here. To cite that sentence:

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.

See also [sec. 6.3.2.3, last 4 paragraphs] of the C99 rationale, which gives a more detailed description of how invalid pointers can be generated, and what effects that may have.