Difference in uses between malloc and calloc - c

Question

gcc 4.5.1 c89
I have written this source code for my better understanding of malloc and calloc.
I understand, but just have a few questions.
dev = malloc(number * sizeof *devices);
is equal to this calloc. I am not concerned about clearing the memory.
dev = calloc(number, sizeof *devices);
What is that exactly, compared to doing this 5 times in a while loop:
dev = malloc(sizeof *devices);
I guess the first one and the second is creating a pointer to 5 struct device. And the third is creating a single pointer to a struct device?
My program illustrates this 3 different methods compiled and ran with valgrind --leak-check=full.
Many thanks for any advice.
#include <stdio.h>
#include <stdlib.h>
struct Devices {
#define MAX_NAME_SIZE 80
size_t id;
char name[MAX_NAME_SIZE];
};
struct Devices* create_device(struct Devices *dev);
void destroy_device(struct Devices *dev);
int main(void)
{
size_t num_devices = 5;
size_t i = 0;
struct Devices *device = NULL;
struct Devices *dev_malloc = NULL;
struct Devices *dev_calloc = NULL;
for(i = 0; i < num_devices; i++) {
device = create_device(device);
/* Assign values */
device->id = i + 1;
sprintf(device->name, "Device%zu", device->id);
/* Print values */
printf("ID ----- [ %zu ]\n", device->id);
printf("Name --- [ %s ]\n", device->name);
/* Test free */
destroy_device(device);
}
printf("\n");
dev_malloc = malloc(num_devices * sizeof *dev_malloc);
for(i = 0; i < num_devices; i++) {
/* Assign values */
dev_malloc->id = i + 1;
sprintf(dev_malloc->name, "dev_malloc%zu", dev_malloc->id);
/* Print values */
printf("ID ----- [ %zu ]\n", dev_malloc->id);
printf("Name --- [ %s ]\n", dev_malloc->name);
}
/* Test free */
destroy_device(dev_malloc);
printf("\n");
dev_calloc = calloc(num_devices, sizeof *dev_calloc);
for(i = 0; i < num_devices; i++) {
/* Assign values */
dev_calloc->id = i + 1;
sprintf(dev_calloc->name, "dev_calloc%zu", dev_calloc->id);
/* Print values */
printf("ID ----- [ %zu ]\n", dev_calloc->id);
printf("Name --- [ %s ]\n", dev_calloc->name);
}
/* Test free */
destroy_device(dev_calloc);
return 0;
}
struct Devices* create_device(struct Devices *dev)
{
/* Not checking for memory error - just simple test */
return dev = malloc(sizeof *dev);
}
void destroy_device(struct Devices *dev)
{
if(dev != NULL) {
free(dev);
}
}

Answer 1

calloc(a,b) and malloc(a*b) are equivalent except for the possibility of arithmetic overflow or type issues, and the fact that calloc ensures the memory is zero-byte-filled. Either allocated memory that can be used for an array of a elements each of size b (or vice versa). On the other hand, calling malloc(b) a times will results in a individual objects of size b which can be freed independently and which are not in an array (though you could store their addresses in an array of pointers).
Hope this helps.

Answer 2

malloc(n) allocates n bytes plus padding and overhead.
calloc(m, n) allocates m*n bytes plus padding and overhead, and then zero's the memory.
That's it.

Answer 3

edited for clarity
I guess the first one and the second is creating a pointer to 5 struct device. And the third is creating a single pointer to a struct device?
The first one malloc(number * sizeof(*devices)) would allocate enough memory to store number of Devices. As others have mentioned, you can treat this block like an array of Device. The pointer you get back will point to the beginning of the block.
int number = 5;
Device *ptr = malloc(number * sizeof(*ptr));
/* stuff */
free(ptr);
The second one that uses calloc does the same thing, while also initializing the memory to 0. Again, you can use treat the block like an array of Device.
int number = 5;
Device *ptr = calloc(number, sizeof(*ptr));
/* stuff */
free(ptr);
The third one, looping 5 times, would result in 5 different pointers to 5 different blocks large enough to store one Device each. This also means each of the 5 pointers has to be free'ed individually.
Device *ptrs[5];
for(int i = 0; i < 5; ++i)
{
ptrs[i] = malloc(sizeof(*ptrs[i]));
}
/* stuff */
for(int i = 0; i < 5; ++i)
{
free(ptrs[i]);
}

Answer 4

The first two create an array of 5 devices in contiguous memory. The last malloc, done five times, will create 5 individual devices which are not guaranteed to be in contiguous memory.

Answer 5

All three loops in your program use only one struct Devices object at a time. The later ones allocate extra memory as though they are going to use multiple objects, but then keep overwriting the beginning of that memory. If you tried to use the object with ID 1 after setting up the object with ID 2, you would find there is no longer any object with ID 1.
Instead, you could do something like this to treat the allocated memory as an array of structs:
dev_malloc = malloc(num_devices * sizeof *dev_malloc);
for (i=0; i<num_devices; i++) {
/* Assign values */
dev_malloc[i].id = i + 1;
sprintf(dev_malloc[i].name, "dev_malloc%zu", dev_malloc[i].id);
/* Print values */
printf("ID ----- [ %zu ]\n", dev_malloc[i].id);
printf("Name --- [ %s ]\n", dev_malloc[i].name);
}
free(dev_malloc);

Answer 6

Look at an implementation of calloc to see the differences. It's probably something like this:
// SIZE_MAX is defined in stdint.h from C99
void *calloc( size_t N, size_t S)
{
void *ret;
size_t NBYTES;
// check for overflow of size_t type
if (N > SIZE_MAX / S) return NULL;
NBYTES = N * S;
ret = malloc( NBYTES);
if (ret != NULL)
{
memset( ret, 0, NBYTES);
}
return ret;
}

Answer 7

As you point out, calloc zeroes out the memory is allocates, while malloc doesn't.
Your examples 1 & 2 each allocate a single contiguous block of five structs (and returns a pointer to that block), whereas example 3 allocates five separate blocks of one struct each (and gives you five pointers unrelated to one another.)