Answer / vishnu948923

void main()
{
char *ptr1,*ptr2;
float fl;
ptr1 = &fl;
ptr2 = (&fl+1);

printf("%u",ptr2-ptr1);
}

Answer / rajesh gooda

ptr manipulation will return 1.

printf("size of int is %d",(int)((int*)0 + 1))

Answer / sunil

When the parameter is a datatype.
For Eg: sizeof(int), sizeof(double)
#define GetSize(x) (char*)((x*)10 + 1) - (char*)10


When the parameter is a variable.
For Eg: int a;
float b;
sizeof(a), sizeof(b)
#define GetSize(x) (char*)(&x + 1) - (char*)&x

Answer / test

#include<stdio.h>
main()
{
int kh[2]={10,20};
int * ptr_kh=kh;
printf("%d",((char* )(ptr_kh+1)-(char*)ptr_kh));

}

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 ) ) );
}
It will work for any data type

Answer / anil arya

#define SIZEOF(type) (int)&((int *)0)[1])

Answer / amit ranjan

int main()
{
int a[2];
int one = a;
int two = a+1;
int test = two-one;
printf("%d\n", test);
return 0;
}

Answer / amit prakash

main()
{
int a;
int *aa,*bb;
int size;
aa = &a;
bb=aa;
bb++;
size=bb-aa;
printf("\nsize_of_int:%u",size) ; // actual size but
depends upon compiler
}

Answer / gururaj

Vishnu,
How can char * hold address of float????

Answer / abc

How can char * hold address of float????

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