Is malloc() initializing allocated array to zero?

Question

Here is the code I'm using:

#include <stdio.h>
#include <stdlib.h>

int main() {
    int *arr;
    int sz = 100000;
    arr = (int *)malloc(sz * sizeof(int));

    int i;
    for (i = 0; i < sz; ++i) {
        if (arr[i] != 0) {
            printf("OK\n");
            break;
        }
    }

    free(arr);
    return 0;
}

The program doesn't print OK. malloc isn't supposed to initialize the allocated memory to zero. Why is this happening?

Answer 1

malloc isn't supposed to initialize the allocated memory to zero. Why is this happening?

This is how it was designed more than 40 years ago.

But, at the same time, the calloc() function was created that initializes the allocated memory to zero and it's the recommended way to allocate memory for arrays.

The line:

arr = (int *)malloc(sz * sizeof(int));

Should read:

arr = calloc(sz, sizeof(int));

If you are learning C from an old book it teaches you to always cast the value returned by malloc() or calloc() (a void *) to the type of the variable you assign the value to (int * in your case). This is obsolete, if the value returned by malloc() or calloc() is directly assigned to a variable, the modern versions of C do not need that cast any more.