How to find the size of a variable without using sizeof

Question

Let us assume I have declared the variable \'i\' of certain datatype (might be int, char, float or double) ...

NOTE: Simply consider that \'i\' is

Answer 1

This should give you the size of your variable

#define mySizeof(type) ((uint)((type *)0+1))

Answer 2

I hope that below code would solve your problem in c++ without using sizeof() operator

for any variables like (int, char, float, double, char, short and many more...)

here I take integer,

int a;

then show it as byte addressable output,

cout<<(char *)(&a + 1) - (char *)(&a);

Answer 3

Program to find Size of the variable without using sizeof operator

#include<stdio.h>
int main()
  {
  int *p,*q;
  int no;
  p=&no;
   printf("Address at p=%u\n",p);
  q=((&no)+1);
  printf("Address at q=%u\n",q);
  printf("Size of int 'no': %d Bytes\n",(int)q-(int)p);

  char *cp,*cq;
    char ch;
  cp=&ch;
  printf("\nAddress at cp=%u\n",cp);
  cq=cp+1;
  printf("Address at cq=%u\n",cq);
  printf("Size of Char=%u Byte\n",(int)cq-(int)cp);

  float *fp,*fq;
  float f;
  fp=&f;
 printf("\nAddress at fp=%u\n",fp);
  fq=fp+1;
  printf("Address at fq=%u\n",fq);
  printf("Size of Float=%u Bytes\n",(int)fq-(int)fp);

  return 0;
}

Answer 4

#include<stdio.h>

#include<conio.h>

struct size1


  {
int a;
char b;
float c;
};

void main()
{
struct size1 *sptr=0;  //declared one pointer to struct and initialise it to zero//
sptr++;                 
printf("size:%d\n",*sptr);
getch();
}