Question

I have the following code:

int main() {
    int n = 3, m = 4, a[n][m], i, j, (* p)[m] = a;
    for (i = 0; i < n; i++)
        for (j = 0; j < m; j++)
            a[i][j] = 1;
    p++;
    (*p)[2] = 9;
    return 0;
}

I have a hard time understanding what p is here, and the consequences of the operations on p in the end. Can someone give me a brief explanation of what happens. I know c-pointers in their simple settings, but here it get slightly more complicated.

Answer 1

a is an array of int[m], and p is a pointer to int[m]. It’s initialized to a, which decays to a pointer to its first element.

Dereferencing p yields int[m], which decays to int* (pointer to its first element). So (*p)[2] adds 2 to the int* that was the result of the decay. So it’s setting the third of the first 4 integers (int[m], with m being 4) to 9.

Adding to p would advance the address stored in it by units of sizeof(int[m]) bytes, which at runtime (since m is not a compile time constant) evaluates to 4 * sizeof(int). So (*(p+1))[2] would access the third of the second 4 integers. In total, p has 3 such arrays of 4 ints to point to: a+0, a+1 and a+2.

Notice that you created a C99 VLA, variable length array. You need to #define n and m to integer constants to get rid of those VLAs (which aren’t quite portable among C compilers).