How do you write a c program to swap the values of two variables in unix?

Answer 1

#include<stdio.h>

void main()

{

int a=2,b=4;

printf("Program for swapping two numbers ");

printf("Numbers before swapping");

printf("a=%d and b=%d",a,b);

a=((a+b)-(b=a));

printf("Numbers after swapping");

printf("a=%d and b=%d",a,b);

getch();

}

Answer 2

echo Enter valuefor a:

read a

echo Enter valuefor b:

read b

clear

echo Values of variables before swaping

echo A=$a

echo B=$b

echo Values of variables after swaping

a=`expr $a + $b`

b=`expr $a - $b`

a=`expr $a - $b`

echo A=$a

echo B=$b

Answer 3

The operating system is immaterial. There are two ways to swap two variables.

Using a temporary variable:

void swap (int* a, int* b) {

int temp;

temp = (*a);

(*a) = (*b);

(*b) = temp;

}

Without using a temporary variable:

void swap (int* a, int* b) {

(*a) ^= (*b) ^= (*a) ^= (*b);

}