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write a program for size of a data type without using
sizeof() operator?

Answer Posted / govind279

#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.

Is This Answer Correct ?    40 Yes 9 No



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