PLS TYPE ANSWER

This code will work in TC with 2 warnings but can get result

void main()
{
  char *Ptr1,*Ptr2;
  float fl;
  ptr1 = &fl;
  ptr2 = (&fl+1);

  printf("%u",ptr2-ptr1);
}

This is a way to get the size of data type...waiting for 
any other way...

This has been solved in parts. I am not sure if there are
any better method merging it.

case 1. User passes a variable as the parameter.
eg: int n;
sizeof(n);

case 2. User passes a data type as the parameter.
eg: sizeof(int)


Solution
case 1: #define GetSize(x) (char*)(&x + 1) - (char*)&x

case 2:#define GetMySize(x) (char*)((x*)10 + 1) - (char*)10

void main()
{
  char *Ptr1,*Ptr2;
  float fl;
  ptr1 = &fl;
  ptr2 = (&fl+1);

  printf("%u",ptr2-ptr1);
}

#include<stdio.h>
int main()
{
int n;
int x,*p,*p1;/* here u can change the type */
p=&x;
p1=(p+1);

printf("size of x is : %d\n",n=(char *)(p1)-(char *)p);
}
Note:without type cast, it always gives 1.
i.e 1 int(4 chars), 1 float(4 chars),1 double(8
chars)etc...coz p+1 points to the next new location of same
type.

#include <stdio.h>

struct node {
	int x;
	int y;
};

unsigned int find_size ( void* p1, void* p2 )
{
	return ( p2 - p1 );
}

int main ( int argc, char* argv [] )
{
	struct node data_node;
	int x = 0;

	printf ( "\n The size :%d", 
			find_size ( (void*) &data_node, 
				(void*) ( &data_node + 
1 ) ) );
	printf ( "\n The size :%d", find_size ( (void*) &x, 
				(void*) ( &x + 1 ) ) );
}

this will work for any data type

#include<stdio.h>
#include<conio.h>
void main()
{
int integer[2],p;
float floating[2],q;
char character[2],r;
double doubling[2],s;
clrscr();
p=(integer+1)-integer;
q=(floating+1)-floating;
r=(character+1)-character;
s=(doubling+1)-doubling;
printf("the sizeof int is :%d",p);
printf("\nthe size of float is :%d",q);
printf("\nthe size of character is :%d",r);
printf("\nthe size of double is :%d",s);
getch();
}


thank u