Thread: My own calloc() not working.

Hi,

I'm trying to write my own version of the function calloc().
Below is the code.
Something goes wrong. The output when using my own calloc() is just random integers. The varibales in my_calloc don't seem to be local though, since the call a[0] in main gives the number assigned in my_calloc() (The code commented out).

1) somebody knows what's wrong?
2) if I correct it will my function my_calloc() work as the standard function calloc()?

Code:

// Trying out calloc()
#include <stdio.h>
#include <stdlib.h>

int *my_calloc(size_t n, size_t size);

int main(void)
{
    int i, n;
    int *a;
    printf("Number of elements: ");
    scanf("%d", &n);
    
    //a = (int *) calloc(n, sizeof(int));
    a = my_calloc(n, sizeof(int));
    
    // Printfs just for test
    printf("a[0]: %d\n", a[0]);
    printf("a[1]: %d\n", a[1]);
    printf("a[2]: %d\n", a[2]);
    // End test
    
    printf("Enter numbers: \n");
    for(i = 0; i < n; i++)
        scanf("%d", &a[i]);
    printf("Numbers entered: \n");
    for(i = 0; i < n; i++)
        printf("%d  ", a[i]);
    printf("\n");
    
    
    return 0;

}

int *my_calloc(size_t n, size_t size)
{
    int mem[n];
    int *ptr;
    ptr = mem;
    
    /* Just to test
    mem[0] = 3;
    mem[1] = 2;
    mem[2] = 1;
    printf("Inside my_calloc: %d\n", mem[0]);
    */
    return ptr;
}

Thanks. Was trying to write:

Code:

static int mem[n];

in my_calloc() but didnt work.
Not sure what you mean with initializing memory to zero.

Originally Posted by laserlight

mem is a local array, which means that it is destroyed when control returns from my_calloc.

You could look into using lower level facilities, e.g., as provided by your operating system, in order to implement your own version of calloc.

OK, I think this will be a more complicated than I can handle now, maybe in the future

in my_calloc() but didnt work.

That's possibly because your compiler doesn't support VLA with a static array. You should be dynamically allocating the memory.

Jim

Originally Posted by laserlight

mem is a local array, which means that it is destroyed when control returns from my_calloc.

But if mem is local, then why do the statements:

Code:

    printf("a[0]: %d\n", a[0]);    printf("a[1]: %d\n", a[1]);
    printf("a[2]: %d\n", a[2]);

in main print out what was assigned in my_calloc() (When comments removed)?

Originally Posted by jimblumberg

That's possibly because your compiler doesn't support VLA with a static array. You should be dynamically allocating the memory.

Jim

The compiler says: error: storage size of ‘mem’ isn’t constant
static int mem[n];

How do I allocate dynamically then (without calloc or using low low level facilities that Laserlight suggested)?

The compiler says: error: storage size of ‘mem’ isn’t constant

That means that the size of the array must be a compile time constant.

What exactly are you trying to accomplish?

Why are you trying to write you're own calloc() function instead of just using the standard calloc()?

Jim

I kind of assumed the my_calloc would be a function that uses malloc() and then initializes it to random values.

Originally Posted by jimblumberg

That means that the size of the array must be a compile time constant.

What exactly are you trying to accomplish?

Why are you trying to write you're own calloc() function instead of just using the standard calloc()?

Jim

I want to know how to write my own calloc() version to learn how it works. I'm new to programming and at the moment I'm learning about pointers and dynamically allocated memory.

I suggest, since you're just learning, really learn the standard functions first. Then once you completely understand how to use the standard functions, malloc(), calloc(), realloc() and free(), it'll be much easier to understand how to roll your own.

Jim

OK, Thanks for the advice Jim, I think I will do that. Makes sense.