How can you find the size of a datatype without creating a variable or pointer, or using sizeof of the datatype?

Question

The question shown below is an interview question:

Q) You are given/have a datatype, say X in C.

The requirement is to get the size of the datatype, without declaring a variable or a pointer variable of that type,

And of course, without using sizeof operator !

I am not sure if this question has already been asked on SO.

Thanks and regards Maddy

Answer 1

define sizeof_type( type ) (size_t)((type*)1000 + 1 )-(size_t)((type*)1000)

The original is from this discussion. http://www.linuxquestions.org/questions/programming-9/how-to-know-the-size-of-the-variable-without-using-sizeof-469920/

Answer 2

This should do the trick:

#include <stdio.h>

typedef struct
{
   int i;
   short j;
   char c[5];

} X;

int main(void)
{
   size_t size = (size_t)(((X*)0) + 1);
   printf("%lu", (unsigned long)size);

   return 0;
}

Explanation of size_t size = (size_t)(((X*)0) + 1);

• assuming a sizeof(X) would return 12 (0x0c) because of alignment
• ((X*)0) makes a pointer of type X pointing to memory location 0 (0x00000000)
• + 1 increments the pointer by the the size of one element of type X, so pointing to 0x0000000c
• the expression (size_t)() casts the address, that is given by the expression (((X*)0) + 1) back to an integral type (size_t)

Hope that gives some insight.