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

if X is datatype:

#define SIZEOF(X) (unsigned int)( (X *)0+1 )

if X is a variable:

#define SIZEOF(X) (unsigned int)( (char *)(&X+1)-(char *)(&X) )

Answer 2

Try this:

int a;
printf("%u\n", (int)(&a+1)-(int)(&a));

Answer 3

put this to your code

then check the linker output ( map file)

unsigned int  uint_nabil;
unsigned long  ulong_nabil;

you will get something like this ;

uint_nabil 700089a8 00000004
ulong_nabil 700089ac    00000004

4 is the size !!

Answer 4

Try this,

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

For the following user-defined datatype,

struct x
{
    char c;
    int i;
};

sizeof_type(x)          = 8
(size_t)((x*)1000 + 1 ) = 1008
(size_t)((x*)1000)      = 1000

Answer 5

    main()    
    {
    clrscr();
    int n;
    float x,*a,*b;//line 1
    a=&x;
    b=(a+1);
    printf("size of x is %d",
    n=(char*)(b)-(char*)a);
    }

By this code script the size of any data can be calculated without sizeof operator.Just change the float in line 1 with the type whose size you want to calculate

Answer 6

Try This:

 #include<stdio.h>

int main(){

  int *ptr = 0;

  ptr++;
  printf("Size of int:  %d",ptr);

  return 0;