C lets you use the subscript operator [] on arrays and on pointers. When you use this operator on a pointer, the resultant type is the type to which the pointer points to. For example, if you apply [] to int*, the result would be an int.
That is precisely what’s going on: you are passing int*, which corresponds to a vector of integers. Using subscript on it once makes it int, so you cannot apply the second subscript to it.
It appears from your code that arr should be a 2-D array. If it is implemented as a “jagged” array (i.e. an array of pointers) then the parameter type should be int **.
Moreover, it appears that you are trying to return a local array. In order to do that legally, you need to allocate the array dynamically, and return a pointer. However, a better approach would be declaring a special struct for your 4×4 matrix, and using it to wrap your fixed-size array, like this:
// This type wraps your 4x4 matrix
typedef struct {
int arr[4][4];
} FourByFour;
// Now rotate(m) can use FourByFour as a type
FourByFour rotate(FourByFour m) {
FourByFour D;
for(int i = 0; i < 4; i ++ ){
for(int n = 0; n < 4; n++){
D.arr[i][n] = m.arr[n][3 - i];
}
}
return D;
}
// Here is a demo of your rotate(m) in action:
int main(void) {
FourByFour S = {.arr = {
{ 1, 4, 10, 3 },
{ 0, 6, 3, 8 },
{ 7, 10 ,8, 5 },
{ 9, 5, 11, 2}
} };
FourByFour r = rotate(S);
for(int i=0; i < 4; i ++ ){
for(int n=0; n < 4; n++){
printf("%d ", r.arr[i][n]);
}
printf("\n");
}
return 0;
}
This prints the following:
3 8 5 2
10 3 8 11
4 6 10 5
1 0 7 9
