Why memory is not allocated?

Question

I was required to built a function that allocates memory to a pointer to form an array but whatever I try it doesen't allocate memory .

#include <stdio.h>
#include <stdlib.h>

void alocare(int **a,int n)
{
   *a = (int*) malloc(n*sizeof(int));
}

int main()
{
     int *a ,*b,*c,n,m,k;
    printf("Insert the number of elements for the first array:\n");
    scanf("%d",&n);
    printf("Insert the number of elements for the second array:\n");
    scanf("%d",&m);
    
    printf("Number of bytes ocupied by a before alloc(): %d\n",sizeof(a));
    printf("Number of bytes occupied by b before alloc(): %d\n",sizeof(b));
    alocare(&a,n);
    alocare(&b,m);
    printf("Number of bytes ocupied by a after alloc(): %d\n",sizeof(a));
    printf("Number of bytes ocupied by b after alloc(): %d\n",sizeof(a));
    return 0;
}

The size that it's displayed in the two printf in the main is 4 before allocation and after it and I can't understand why Thanks!

Answer 1

sizeof(a) returns the size of the type of a. Since a is a pointer, its size is always 4 (in your platform; in others it might be 8 or something else). This size will never change: it is a constant for your program.

There is no way to know what is the size of allocated memory behind a pointer unless you save it yourself.