Answer / manju

main()
{
int a=16,b=19;
a=a+b;
b=a-b;
a=a-b;
printf("a and b are:%d\t%d",a,b);
}

Output:
a and b are:19 16

Answer / surendra jhajhra

int x,y;

x =x+y;
y =x-y;
x =x-y;

Answer / raghuram.a

int x,y;

x =x-y;
y =y+x;
x =y-x;

or

int x,y;

x=x^y;
y=x^y;
x=x^y;

Answer / srikanth

int x,y;

x =x-y;
y =y+x;
x =y-x;

or

int x,y;

x=x^y;
y=x^y;
x=x^y;

or

int x,y;
x*=xy;
y=x/y;
x=x/y;

Answer / emperor of america

using bitwise operator works 100%:
x^=y;
y^=x;
x^=y;

using +/- works sometime, need to consider overflow.

Answer / divya

#include<stdio.h>
main()
(
clrscr();
int a,b;
printf("enter a and b values");
scanf("%d \n %d",&a,&b);
printf("before swaping a value is:%d",a);
printf("before swaping b value is:%d",b);
a=a+b;
b=a-b;
a=a-b;
printf("after swaping a value is:%d",a);
printf("after swaping b value is:%d",b);
getch();
}

Answer / girish

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
int a,b;
clrscr();
printf("Enter the two values:");
scanf("%d%d",&a,&b);
swap(a,b);
pritnf("the swaping of given two values are:\na=%d,b=%
d",a,b);
getch();
}

Answer / mangala pandi

Logically it is not possible. so your question is worng.

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