write a program for size of a data type without using
sizeof() operator?
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Answer / 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 |
Answer / g
This code will work in TC with 2 warnings but can get result
void main()
{
char *Ptr1,*Ptr2;
float fl;
ptr1 = &fl;
ptr2 = (&fl+1);
printf("%u",ptr2-ptr1);
}
This is a way to get the size of data type...waiting for
any other way...
| Is This Answer Correct ? | 44 Yes | 26 No |
Answer / sunil
This has been solved in parts. I am not sure if there are
any better method merging it.
case 1. User passes a variable as the parameter.
eg: int n;
sizeof(n);
case 2. User passes a data type as the parameter.
eg: sizeof(int)
Solution
case 1: #define GetSize(x) (char*)(&x + 1) - (char*)&x
case 2:#define GetMySize(x) (char*)((x*)10 + 1) - (char*)10
| Is This Answer Correct ? | 15 Yes | 6 No |
Answer / abdur rab
#include <stdio.h>
struct node {
int x;
int y;
};
unsigned int find_size ( void* p1, void* p2 )
{
return ( p2 - p1 );
}
int main ( int argc, char* argv [] )
{
struct node data_node;
int x = 0;
printf ( "\n The size :%d",
find_size ( (void*) &data_node,
(void*) ( &data_node +
1 ) ) );
printf ( "\n The size :%d", find_size ( (void*) &x,
(void*) ( &x + 1 ) ) );
}
this will work for any data type
| Is This Answer Correct ? | 15 Yes | 8 No |
Answer / arun kumar mishra kiit univers
#include<stdio.h>
#include<conio.h>
void main()
{
float f1,*Ptr1,*Ptr2;
ptr1 = &fl;
ptr2 = (&fl+1);
printf("%u",(char *)ptr2-(char *)ptr1);
getch();
}
| Is This Answer Correct ? | 11 Yes | 4 No |
Answer / mukesh kumar singh
#include <iostream>
using namespace std;
struct node {
int x;
int y;
char *s;
};
int main()
{
node x,*p;/* here u can change the type */
p=&x;
cout<<"\n\nSize of node Is : "<<(char*)(p+1)-(char*)p;
return 0;
}
| Is This Answer Correct ? | 5 Yes | 2 No |
Answer / lalit kumar
#include "stdafx.h"
#include "stdio.h"
#include "conio.h"
int main()
{
char *a,*s, v='m';
a=&v;
s=a;
a++;
int intsize=(int)a-(int)s;
printf("%d",intsize);
getch();
return 0;
}
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / subham singh
1 #include<iostream>
2 using namespace std;
3 main()
4 {
5 int i;
6 int* p = &i;
7 int* q= p;
8 p++;
9 cout<<(char*)p-(char*)q<<endl;
10 }
11
~
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / anil arya
http://stackoverflow.com/questions/1219199/size-of-a-datatype-without-using-sizeof
| Is This Answer Correct ? | 2 Yes | 1 No |
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