C program to swap two variables without using third variable?

Answer 1

you can do it in the following manner

Supposing your two variables are x and y:

int x=3;

int y=5;

x=x+y; [x becomes 8]

y=x-y; [y assumes the original value of x i.e. 3]

x=x-y; [x assumes the original value of y i.e. 5]

or... you can use the unary XOR (exclusive or) operator, '^=' .

Same values, int x=3, y=5

x ^= y; // x becomes 6

y ^= x; // y becomes 5

x ^= y; // x becomes 3.

The second method has more advantages :

- Its assembler operations never use the processor's ALU carry.

- Without use of the aforementioned carry, no overflow will ever occur.

- Performing this with 32-bits values 8-bits processors will be more efficient, in terms of program space AND execution speed.

- You can even use this in Visual Basic an other languages which implements boundaries on values, while the first method is guaranteed to fail when overflows occurs.

In both case, do not EVER try to factorize these 3 lines into two, as the operations order in multiple-operators lines depends on the compiler's way to parse your code.

Thus , typing

x = x ^ y ^x;

y = y ^ x;

Will surely give you garbage.

Answer 2

#include<stdio.h>

#include<conio.h>

int main (void)

{

int a = 5;

int b = 8;

printf ( " Before swapping a = %d , b = %d\n",a,b);

swap(&a, &b);

printf ( " After calling swap function a = %d , b = %d\n",a,b);

getch();

return 0;

}

void swap(int* const p , int* const q)

{

int temp = *p;

*p = *q;

*q = temp;

}

Answer 3

Use the following template function to swap any two objects without using a temporary object:

template<class T> void swap(T& x, T& y){x^=y^=x^=y;}

Note that all objects passed to the function must provide an overload for the bitwise XOR assignment operator (^=) in order for this to work. However, primitive types such as int and char can be passed to this template function without the need for any additional code.

Answer 4

#include<iostream>

void swap(int& x, int& y)

{

x^=y^=x^=y;

}

int main()

{

int a=1, b=2;

std::cout<<"a="<<a<<", b="<<b<<std::endl;

swap(a,b);

std::cout<<"a="<<a<<", b="<<b<<std::endl;

return(0);

}

Answer 5

int main()

{

int a = 13, b = 29, c = 237

a ^= b; // a == 16

b ^= a; // b == 13

a ^= b; // a == 29

// You could stop here, since 'a' and 'b' values are now swapped, or you can continue with 'c' and 'b', as required.

c ^= b; // c == 224

b ^= c; // b == 237

c ^= b; // c == 13;

/* In short, 'a' got the previous value of 'b' , 'c' the value of 'b', and 'b' the value of 'c'.

To make it simpler

#define SWAP(X,Y) {X^=Y; Y^=X; X^=Y}

int main(void)

{

int a=37, b=32766;

SWAP(a,b); // nifty macro !

}

Answer 6

Please note the following coding for the same if the question is How do you write a program in C to swap two variables without using another variables? :-

void main()

{

int a,b;

printf("Enter a and b:", a,b);

scanf("%d %d",&a,&b);

a=a+b;

b=a-b;

a=a-b;

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

}

By: Anil Kothiyal

Answer 7

void swap(int* x, int*y)

{

(*x)^=(*y)^=(*x)^=(*y);

}

Answer 8

to swap a and b we can use the following lines of code:

a=a+b;

b=a-b;

a=a-b;

Answer 9

a=6,b=5

a=a+b;/*11*/

b=a-b;/*11-5=6*/

a=a-b;/*11-6=5*/