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:
int 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");
return 0;
}
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;
}
Point1: Memory allocated ptr: 262144 bytes
Point2: Memory allocated input_image: 0 bytes
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 (T*
), 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(...);
}