C Programming - Memory Allocation - Discussion

What is the difference between the two programs given below?

1.

#include<stdio.h>
#include<stdlib.h>
#define MAXROW 3
#define MAXCOL 4

int main()
{
int (*p)[MAXCOL];
printf("%d %d\n",sizeof(p),sizeof(*p));
// p = (int (*) [MAXCOL])malloc(MAXROW *sizeof(*p));
printf("%d %d\n",sizeof(p),sizeof(*p));
return 0;
}

2.

#include<stdio.h>
#include<stdlib.h>
#define MAXROW 3
#define MAXCOL 4

int main()
{
int (*p)[MAXCOL];
printf("%d %d\n",sizeof(p),sizeof(*p));
p = (int (*) [MAXCOL])malloc(MAXROW *sizeof(*p));
printf("%d %d\n",sizeof(p),sizeof(*p));
return 0;
}

I don't understand.

P = (int (*) [MAXCOL]) malloc (MAXROW *sizeof (*p));

So, sizeof (*p) 4 * 2 and malloc gives 3 * 8.

So, the sizeof (*p) will be 24.

Here:

P = (int (*) [MAXCOL]) malloc (MAXROW *sizeof (*p)); is allocating a memory of 24 bytes to the pointer P but as it is a pointer to 2D array it will hold the base Row address i.e. 8.
Therefore *P holds up 8*3 bytes but will show 8 bytes i.e the size of its row in sizeof(*P).
And P is itself a pointer to an array of 2*4 bytes and its row holds 2 bytes each.

Actual memory consumed by both members is 2*4 and 8*3 bytes but a pointer holds the base address and thus the first element's address will be shown in sizeof() i.e. 2 and 8 bytes.

Here p is a pointer.so sizeof(p) is either become 4 bytes or 8 bytes.But here how sizeof(p) is 2 bytes.

int is 2 bytes with 4 elements so sizeof(*p) is 8 bytes (2*4). So the answer is 4(or)8, 8.