Code Test 1

Predict the output of this Program :-

int** performOps(int **A, int m, int n, int *len1, int *len2) {

int i, j;

*len1 = m;

*len2 = n;

int **B = (int **)malloc((*len1) * sizeof(int *));

for (i = 0; i < *len1; i++) {

B[i] = (int *)malloc((*len2) * sizeof(int));

}

 

for (i = 0; i < m; i++) {

for (j = 0; j < n; j++) {

B[i][n - 1 - j] = A[i][j];

}

}

return B;

}

Assume m=3, n=4, and A : [[1,2,3,4], [5,6,7,8], [9,10, 11, 12]] 

What would be the output of the following call :

int len1, len2; int **B = performOps(A, m, n, &len1, &len2); int i, j; for (i = 0; i < len1; i++) { for (j = 0; j < len2; j++) { printf("%d ", B[i][j]); } }