Question

Given: (In C++)

int main () {

    int* ptr;
    int ary [10][2];
    ptr = ary; 

    return 0;
}

How would I access ary[0][1] with ptr?

Answer 1

You can’t, because the type of ptr is wrong. The variable should be declared as int(*)[2] (pointer to an array of size 2 of integers). Then you could just use ptr[0][1].

#include <cstdio>

int main () {

    int (* ptr) [2];   // <--
    int ary [10][2];
    ptr = ary; 

    ary[0][1] = 5;
    printf("%d\n", ptr[0][1]);

    return 0;
}

---

If you must use an int*, you need to introduce a reinterpret_cast. The array indices are laid out like:

    0       1        2         3              2*n      2*n+1
[0][0]  [0][1]   [1][0]    [1][1]    ...    [n][0]    [n][1]

so you could use ptr[1] to get ary[0][1].

#include <cstdio>

int main () {

    int* ptr;
    int ary [10][2];
    ptr = reinterpret_cast<int*>(ary);  // <--

    ary[0][1] = 5;
    printf("%d\n", ptr[1]);

    return 0;
}