Question about calloc and realloc

Hello!

I have some problem with 'realloc' and 'calloc' in a program...My question is that if some pointer is passed to a function from say Main. In that function can I allocate/ reallocate the memory to the passed pointer and can these allocated addresses with data can be retained after return of the function again i.e. in Main. If this is not possible in this way then what is the way out?

Thanks and regards,
Amal

Yes, but because you're modifying the pointer value, you have to pass it by another pointer:

Code:

---


void func (void **ptr)
{
    free (*ptr);
    *ptr = malloc(10);
}

int main (void)
{
    void *ptr;
    func (&ptr);
    return 0;
}

---

If the code is correct, after func() is called, ptr should point to something else.

realloc will first try to allocate more memory after the memory already allocated. If it can't do that it will copy the old data into the newly allocated space. Note that this means that the pointer may change.

Hello,

Thanks for reply...
I tried but not succeeded. Below is the code in which I tried to do thing in two ways (two functions), but no function did work....
Where am I doing mistake??

Code:

---

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

// First test function
void test_alloc1(double *ptr)
{
    double *ptr_local;
    if((ptr_local=(double*)calloc(5, sizeof(double))) == NULL)
    {
        printf("\n\n  Error");
        return;
    }
    ptr=ptr_local;

    int i;
    double b=0.5;
    for(i=0; i<=4; i++)
    {
        *(ptr+i)=b;
        printf("\n In fun-1 @ %d value is %lf",(ptr+i),
              *(ptr+i));
        b=b+1.0;
    }
    return;
}

// Second test function
void test_alloc2(double **ptr)
{
    if((*ptr=(double*)calloc(5, sizeof(double))) == NULL)
    {
        printf("\n\n  Error");
        return;
    }

    int i;
    double b=0.5;
    for(i=0; i<=4; i++)
    {
        **(ptr+i)=b;
        printf("\n In fun-2 @ %d value is %lf",*(ptr+i),
              **(ptr+i));
        b=b+1.0;
    }
    return;
}

// Main
int main(void)
{
    double *a;
    int i;

    // Calling first function
    test_alloc1(a);
    for(i=0; i<=4; i++)
      printf("\n In MAIN-1 @ %d Value is : %lf",
            (a+i),*(a+i));
    free(a);

    // Calling second function
    test_alloc2(&a);
    for(i=0; i<=4; i++)
      printf("\n In MAIN-2 @ %d Value is : %lf",
            (a+i),*(a+i));

    return 0;
}

---

There is no compilation error (mingw with Code::blocks on windows).

Thanks and regards,
Amal.

test_alloc1 is obviously wrong because you are assiging to a parameter, so the changes will not get reflected in the caller.

Thanks Laserlight....

But test_alloc2 is also not working.....Please help me to figure out the mistake in this function.......

Regards,
Amal...

Quote:

---

Originally Posted by amalendu

But test_alloc2 is also not working.

---

How does it not work?

After running 'for' loop only once in test_alloc2, program stops....

Regards,
Amal.

I might write it in this way:

Code:

---

#define ARRAY_SIZE 5

// Second test function
void test_alloc2(double **ptr)
{
    int i;
    double b = 0.5;
    *ptr = calloc(ARRAY_SIZE, sizeof(**ptr));
    if(*ptr == NULL)
    {
        printf("\n\n  Error");
        return;
    }
 
    for (i = 0; i < ARRAY_SIZE; i++)
    {
        (*ptr)[i] = b;
        printf("\n In fun-2 @ %p value is %f", (void*)((*ptr) + i), (*ptr)[i]);
        b++;
    }
}

---

Take note of the difference between **(ptr+i) and (*ptr)[i].