C Programming: malloc() inside another function

Question

I need help with malloc() inside another function.

I'm passing a pointer and size to the function from my main() and I would like to allocate memory for that pointer dynamically using malloc() from inside that called function, but what I see is that.... the memory, which is getting allocated, is for the pointer declared within my called function and not for the pointer which is inside the main().

How should I pass a pointer to a function and allocate memory for the passed pointer from inside the called function?

---

I have written the following code and I get the output as shown below.

SOURCE:

main()
{
   unsigned char *input_image;
   unsigned int bmp_image_size = 262144;

   if(alloc_pixels(input_image, bmp_image_size)==NULL)
     printf("\nPoint2: Memory allocated: %d bytes",_msize(input_image));
   else
     printf("\nPoint3: Memory not allocated");     

}

signed char alloc_pixels(unsigned char *ptr, unsigned int size)
{
    signed char status = NO_ERROR;
    ptr = NULL;

    ptr = (unsigned char*)malloc(size);

    if(ptr== NULL)
    {
        status = ERROR;
        free(ptr);
        printf("\nERROR: Memory allocation did not complete successfully!");
    }

    printf("\nPoint1: Memory allocated: %d bytes",_msize(ptr));

    return status;
}

PROGRAM OUTPUT:

Point1: Memory allocated ptr: 262144 bytes
Point2: Memory allocated input_image: 0 bytes

Answer 1

You need to pass a pointer to a pointer as the parameter to your function.

main()
{
   unsigned char *input_image;
   unsigned int bmp_image_size = 262144;

   if(alloc_pixels(&input_image, bmp_image_size)==NULL)
     printf("\nPoint2: Memory allocated: %d bytes",_msize(input_image));
   else
     printf("\nPoint3: Memory not allocated");     

}

signed char alloc_pixels(unsigned char **ptr, unsigned int size) 
{ 
    signed char status = NO_ERROR; 
    *ptr = NULL; 

    *ptr = (unsigned char*)malloc(size); 

    if(*ptr== NULL) 
    {
        status = ERROR; 
        free(*ptr);      /* this line is completely redundant */
        printf("\nERROR: Memory allocation did not complete successfully!"); 
    } 

    printf("\nPoint1: Memory allocated: %d bytes",_msize(*ptr)); 

    return status; 
} 

Answer 2

If you want your function to modify the pointer itself, you'll need to pass it as a pointer to a pointer. Here's a simplified example:

void allocate_memory(char **ptr, size_t size) {
    void *memory = malloc(size);
    if (memory == NULL) {
        // ...error handling (btw, there's no need to call free() on a null pointer. It doesn't do anything.)
    }

    *ptr = (char *)memory;
}

int main() {
   char *data;
   allocate_memory(&data, 16);
}

Answer 3

This does not make sense :

if(alloc_pixels(input_image, bmp_image_size)==NULL) 

alloc_pixels returns a signed char (ERROR or NO_ERROR) and you compare it to NULL (which is supposed to be used for pointers).

If you want input_image to be changed, you need to pass a pointer to it to alloc_pixels. alloc_pixels signature would be the following:

signed char alloc_pixels(unsigned char **ptr, unsigned int size)

You would call it like this:

alloc_pixels(&input_image, bmp_image_size);

And the memory allocation

*ptr = malloc(size);

Answer 4

How should I pass a pointer to a function and allocate memory for the passed pointer from inside the called function?

Ask yourself this: if you had to write a function that had to return an int, how would you do it?

You'd either return it directly:

int foo(void)
{
    return 42;
}

or return it through an output parameter by adding a level of indirection (i.e., using an int* instead of int):

void foo(int* out)
{
    assert(out != NULL);
    *out = 42;
}

So when you're returning a pointer type, it's the same thing: you either return the pointer type directly:

T* foo(void)
{
    T* p = malloc(...);
    return p;
}

or you add one level of indirection:

void foo(T** out)
{
    assert(out != NULL);
    *out = malloc(...);
}

Answer 5

Hi,

In your initial code , when you were passing input_image to the function alloc_pixels, compiler was creating a copy of it (i.e. ptr) and storing the value on the stack. You assign the value returned by malloc to ptr. This value is lost once the function returns to main and the stack unwinds. So, the memory is still allocated on heap but the memory location was never stored in (or assigned to )input_image, hence the issue.

You can change the signature of the function alloc_pixels which would be simpler to understand, and you won't require the additional 'status' variable as well.

unsigned char *alloc_pixels(unsigned int size)
{
    unsigned char *ptr = NULL;
    ptr = (unsigned char *)malloc(size);
    if (ptr != NULL)
       printf("\nPoint1: Memory allocated: %d bytes",_msize(ptr));
    return ptr;
}

You can call the above function in main :

int main()
{
   unsigned char *input_image;
   unsigned int bmp_image_size = 262144;

   if((input_image = alloc_pixels(bmp_image_size))==NULL)
       printf("\nPoint3: Memory not allocated");    
   else
     printf("\nPoint2: Memory allocated: %d bytes",_msize(input_image)); 
   return 0;

}