Does the C standard permit assigning an arbitrary value to a pointer and incrementing it?

Question

Is the behaviour of this code well defined?

#include <stdio.h> #include <stdint.h>  int main(void) {     void *ptr = (char *)0x01;     size_t val;      ptr = (char *)ptr + 1;     val = (size_t)(uintptr_t)ptr;      printf("%zu\n", val);     return 0; } 

I mean, can we assign some fixed number to a pointer and increment it even if it is pointing to some random address? (I know that you can not dereference it)

Answer 1

The assignment:

void *ptr = (char *)0x01; 

Is implementation defined behavior because it is converting an integer to a pointer. This is detailed in section 6.3.2.3 of the C standard regarding Pointers:

5 An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might be a trap representation.

As for the subsequent pointer arithmetic:

ptr = (char *)ptr + 1; 

This is dependent on a few things.

First, the current value of ptr may be a trap representation as per 6.3.2.3 above. If it is, the behavior is undefined.

Next is the question of whether 0x1 points to a valid object. Adding a pointer and an integer is only valid if both the pointer operand and the result point to elements of an array object (a single object counts as an array of size 1) or one element past the array object. This is detailed in section 6.5.6:

7 For the purposes of these operators, a pointer to an object that is not an element of an array behaves the same as a pointer to the first element of an array of length one with the type of the object as its element type

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.

On a hosted implementation the value 0x1 almost certainly does not point to a valid object, in which case the addition is undefined. An embedded implementation could however support setting pointers to specific values, and if so it could be the case that 0x1 does in fact point to a valid object. If so, the behavior is well defined, otherwise it is undefined.

Answer 2

No, the behaviour of this program is undefined. Once an undefined construct is reached in a program, any future behaviour is undefined. Paradoxically, any past behaviour is undefined too.

The result of void *ptr = (char*)0x01; is implementation-defined, due in part to the fact that a char can have a trap representation.

But the behaviour of the ensuing pointer arithmetic in the statement ptr = (char *)ptr + 1; is undefined. This is because pointer arithmetic is only valid within arrays including one past the end of the array. For this purpose an object is an array of length one.