Unexpected output for simple 2d array in c - c

Question

This is the initialising part of a program im writing. I am new ish to c and am quite confused with one of the outputs I am getting. Any advice would be greatly appreciated
for input 9 11 I get
1 2 3 4 5 6 7 8 9
9 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
when I expect to get
1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
void *safeMalloc(int n) {
void *p = malloc(n);
if (p == NULL) {
printf("Error: malloc(%d) failed. Out of memory?\n", n);
exit(EXIT_FAILURE);
}
return p;
}
int **makeIntArray2D(int width, int height) {
int **arr = safeMalloc(height*sizeof(int *));
for (int row=0; row < height; row++) {
arr[row] = safeMalloc(width*sizeof(int));
}
return arr;
}
int main(int argc, char *argv[]) {
int numDisks;
int numMoves;
scanf("%d %d", &numDisks, &numMoves);
int **tohan = makeIntArray2D(3, numDisks);
for(int i = 0; i < 3; i++) {
for(int j =0; j < numDisks; j++) {
tohan[i][j] = 0;
}
}
for(int i = 0; i < 3; i++) {
for(int j = 0; j < numDisks; j++) {
printf("%d ", tohan[i][j]);
}
printf("\n");
}
printf("\n");
for(int k = 0; k < numDisks; k++) {
tohan[0][k] = k+1;
}
for(int i = 0; i < 3; i++) {
for(int j = 0; j < numDisks; j++) {
printf("%d ", tohan[i][j]);
}
printf("\n");
}
printf("\n");
}

Answer 1

You have:
int **makeIntArray2D(int width, int height) { … }
You call:
int **tohan = makeIntArray2D(3, numDisks);
You expect the height of the array to be 3 and the width 9, but you are passing the parameters in the wrong sequence for the function.
Note: this diagnosis could not be made without seeing the code for makeIntArray2D().