size of a datatype without using sizeof

Question

I have a data type, say X, and I want to know its size without declaring a variable or pointer of that type and of course without using sizeof oper

Answer 1

# include<stdio.h>

struct node
{
  int a;
  char c;
};

void main()
{
   struct node*ptr;
   ptr=(struct node*)0;
   printf("%d",++ptr);
}

Answer 2

#include <bits/stdc++.h> 

using namespace std; 

int main() 
{ 

    // take any datatype hear 
    char *a = 0; // output: 1

    int  *b = 0;  // output: 4

    long *c = 0; // output: 8

    a++;

    b++;

    c++;

    printf("%d",a);

    printf("%d",b);

    printf("%d",c);

    return 0; 
}

Answer 3

Well, I am an amateur..but I tried out this problem and I got the right answer without using sizeof. Hope this helps.. I am trying to find the size of an integer.

int *a,*s, v=10;

a=&v;

s=a;

a++;

int intsize=(int)a-(int)s;

printf("%d",intsize);