如何在不使用 sizeof 的情况下找到变量的大小

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 declared and dont bother if it is an int or char or float or double datatype. Since I want a generic solution I am simply mentioning that variable 'i' can be of any one of the datatypes namely int, char, float or double.

Now can I find the size of the variable 'i' without sizeof operator?

Answer 1

You can use the following macro, taken from here :

#define sizeof_var( var ) ((size_t)(&(var)+1)-(size_t)(&(var))) 

The idea is to use pointer arithmetic ( (&(var)+1) ) to determine the offset of the variable, and then subtract the original address of the variable, yielding its size. For example, if you have an int16_t i variable located at 0x0002 , you would be subtracting 0x0002 from 0x0006 , thereby obtaining 0x4 or 4 bytes.

However, I don't really see a valid reason not to use sizeof , but I'm sure you must have one.

Answer 2

It's been ages since I wrote any C code and I was never good at it, but this looks about right:

int i = 1;
size_t size = (char*)(&i+1)-(char*)(&i);
printf("%zi\n", size);

I'm sure someone can tell me plenty of reasons why this is wrong, but it prints a reasonable value for me.

Answer 3

This works..

int main() {
    int a; //try changing this to char/double/float etc each time//
    char *p1, *p2;
    p1 = &a;
    p2 = (&a) + 1;
    printf("size of variable is:%d\n", p2 - p1);
}

Answer 4

int *a, *b,c,d;/* one must remove the declaration of c from here*/
int c=10;
a=&c;
b=a;
a++;
d=(int)a-(int)b;
printf("size is %d",d);

Answer 5

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 6

这应该有效。

#define xsizeof(x) (char *)(&x+1) - (char *)&x

Answer 7

#define GET_SIZE(myvar) ((char)( ((char*)(&myvar+1) )- ((char*)&myvar) ))

Answer 8

这应该给你变量的大小

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

Answer 9

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 10

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 11

Below statement will give generic solution:

printf("%li\n", (void *)(&i + 1) - (void *)(&i));

i is a variable name, which can be any data type (char, short, int, float, double, struct).

Answer 12

#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();
}