c - Accesing a 2D array using a single pointer

Question

There are tons of code like this one:

#include <stdio.h>

int main(void)
{
    int a[2][2] = {{0, 1}, {2, -1}};
    int *p = &a[0][0];

    while (*p != -1) {
        printf("%d
", *p);
        p++;
    }
    return 0;
}

But based on this answer, the behavior is undefined.

N1570. 6.5.6 p8:

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.

Can someone explain this in detail?

Answer 1

The array who's base address (pointer to first element) p is assigned is of type int[2]. This means the address in p can legally be dereferenced only at locations *p and *(p+1), or if you prefer subscript notation, p[0] and p[1]. Furthermore, p+2 is guaranteed to be a legally evaluated as an address, and comparable to other addresses in that sequence, but can not be dereferenced. This is the one-past address.

The code you posted violates the one-past rule by dereferencing p once it passes the last element in the array in which it is homed. That the array in which it is homed is buttressed up against another array of similar dimension is not relevant to the formal definition cited.

That said, in practice it works, but as is often said. observed behavior is not, and should never be considered, defined behavior. Just because it works doesn't make it right.