Reputation: 23
This is a code for multiplying two square matrices in C.
What is the difference between using malloc
and calloc
in void multiply()
function?
When using malloc
, I am getting garbage values, but calloc
is providing the right answer.
Only the first row is outputting garbage values so is it an issue with the way malloc
allocates space in the heap as compared to calloc
?
#include <stdio.h>
#include <stdlib.h>
int *getArray(int);
void display(int *, int, int);
void multiply(int *, int *, int);
int main() {
int n;
printf("enter dimension of square matrix:\n");
scanf("%d", &n);
int *arr1;
int *arr2;
arr1 = getArray(n);
display(arr1, n, n);
printf("\n now give input for next array");
arr2 = getArray(n);
display(arr2, n, n);
printf("\n\n\n");
multiply(arr1, arr2, n);
return 0;
}
int *getArray(int n) {
int *arr = (int *)malloc(n * n * sizeof(int));
printf("\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%d", (arr + i * n + j));
}
}
/*for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf(" %d ", *(arr + i * n + j));
}
printf("\n");
}*/
return arr;
}
void display(int *arr, int row, int col) {
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
printf(" %d ", *(arr + i * row + j));
}
printf("\n");
}
}
void multiply(int *arr1, int *arr2, int n) {
int *arr = (int *)calloc(n * n, sizeof(int));
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
*(arr + i * n + j) += (*(arr1 + i * n + k)) * (*(arr2 + k * n + j));
}
}
}
printf("product of above matrices = \n\n");
display(arr, n, n);
}
Upvotes: 0
Views: 191
Reputation: 144695
The only functional difference between allocating memory with malloc()
and with calloc()
for the same size, assuming the size computation is accurate, is the latter initializes the block to all bits 0, whereas the former does not.
All bits 0 means all int
values in the array are initialized to 0
.
The inner loop in the multiply
function only increments the element at row i
and column j
, therefore the function relies on implicit initialization of the array elements to 0. calloc()
does that, but not malloc()
so you definitely need to use calloc()
.
Also note these remarks:
display
the computation for the offset of the matrix element at row i
column j
should be printf(" %5d ", *(arr + i * col + j));
multiply
should return arr
and display()
should be called in the main
function.scanf()
, malloc() and
calloc()` failureconst
qualified so the function can be called with a pointer to a const
object.Here is a modified version:
#include <stdio.h>
#include <stdlib.h>
int *getArray(int);
void display(const int *, int, int);
int *multiply(const int *, const int *, int);
int main() {
int n;
printf("enter dimension of square matrix:\n");
if (scanf("%d", &n) != 1)
return 1;
printf("\n now give input for the first array");
int *arr1 = getArray(n);
if (!arr1)
return 1;
display(arr1, n, n);
printf("\n now give input for the second array");
int *arr2 = getArray(n);
if (!arr2)
return 1;
display(arr2, n, n);
printf("\n\n\n");
int *arr = multiply(arr1, arr2, n);
if (!arr)
return 1;
printf("product of above matrices = \n\n");
display(arr, n, n);
free(arr1);
free(arr2);
free(arr);
return 0;
}
int *getArray(int n) {
int *arr = malloc(sizeof(int) * n * n);
if (arr == NULL)
return NULL;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (scanf("%d", (arr + i * n + j)) != 1) {
free(arr);
return NULL;
}
}
}
return arr;
}
void display(const int *arr, int row, int col) {
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
printf(" %5d ", *(arr + i * col + j));
}
printf("\n");
}
}
int *multiply(const int *arr1, const int *arr2, int n) {
int *arr = calloc((size_t)n * n, sizeof(int));
if (arr == NULL)
return NULL;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
*(arr + i * n + j) += (*(arr1 + i * n + k)) * (*(arr2 + k * n + j));
}
}
}
return arr;
}
Upvotes: 1