CODEWITHSUNDEEP
arrayscpointersmultidimensional-array
laahaa
laahaa

Reputation: 347

Pointer arithmetic in multidimensional array

How does pointer arithmetic - specifically, addition with numbers - work if the pointer points to a multidimensional array?

When I debugged the sample program below:

int main(void) {
    int a[3][4] = {{1, 2, 3, 4},
                 {5, 6, 7, 8},
                 {9, 10, 11, 12}};
    int *p1 = a;
    // set breakpoint ...
    ...
}

If I use debugger (CLion) to print *(p1 + N), it gives me an array having (N + 1) elements, starting from element a[N / col_size][N % col_size]. This doesn't match my expectation. Listing actual result and expectation as below:

Enter into debugger to print Debugger result Expectation
*(p1) {1} {1, 2, 3, 4}
*(p1 + 2) {3, 4, 5} {9, 10, 11, 12}
*(p1 + 5) {6, 7, 8, 9, 10, 11} {an array of random numbers}
*(p1 + 6) {7, 8, 9, 10, 11, 12, -11}. -11 is random as it already exceeded the allocated size of the array. {an array of random numbers}

I understand that C stores multidimensional array in row-major order, but what confuses me is why does C provide an array of (N + 1) length for *(p1 + N)?

Upvotes: 3

Views: 100

Answers (1)

ikegami
ikegami

Reputation: 386706

This is wrong:

int *p1 = a;

a degenerates into a pointer to its first element. It's first element is an int[4] so a degenerates into an int (*)[4]. gcc says as much:

warning: initialization of 'int *' from incompatible pointer type 'int (*)[4]'

You're assigning it to an int *, which strict aliasing violation and therefore undefined behaviour.

If you had used the correct type, the output you would received would have been more in line with your expectations.

Upvotes: 1

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