malloc(sizeof(int)) vs malloc(sizeof(int *)) vs (int *)malloc(sizeof(int))

Question

I acknowledge that all three of these have a different meaning. But, I don\'t understand on what particular instances would each of these apply. Can anyone share an example

Answer 1

The syntax pattern that is most foolproof is:

 int *p;
 p = malloc (cnt * sizeof *p);

This syntax will not force you to change the code if the type (and or size...) of *p changes, eg in

 struct mysstruct *q;
 q = malloc (cnt * sizeof *q);

Which will avoid problems like

struct mysstruct *z;
z = malloc (cnt * sizeof(struct hisstruct)); // Auch!

, plus: the sizeof expr form is also shorter.

---

UPDATE: to demonstrate the correctness of p = malloc(CNT * sizeof *p) this test program:

#include 
#include 

struct mystruct {
        int i;
        char s[14];
        };
int main(void)
{
struct mystruct *p;
size_t siz;

siz = sizeof p;
printf("Sizeof p := %zu\n", siz);

siz = sizeof *p;
printf("Sizeof *p := %zu\n", siz);

printf("Allocating %zu (%u structs, each of size %zu) bytes to be assigned to p...\n"
        , 10u * sizeof *p
        , 10u, sizeof *p
        );
p = malloc(10 * sizeof *p);

return 0;
}

Which outputs here:

Sizeof p := 8
Sizeof *p := 20
Allocating 200 (10 structs, each of size 20) bytes to be assigned to p...