Swap two variables using macro

The code to swap two variables using macro expansions

Code:

#include<stdio.h>
#include<conio.h>
#define SWAPE(x,y) int t;t=x;x=y;y=t;
main()
{
	int a,b;
	
	printf("\n Enter two number");
	scanf("%d%d",&a,&b);
	printf("\n Before swaping the Value of a=%d and b=%d",a,b);
	SWAPE(a,b);
	printf("\n After swap value of a=%d and b=%d",a,b);
	return 0;
}

 

Swap two numbers without using temp variable

//Code for swapping two numbers without using temp variable

Code:

#include <iostream>
using namespace std;
int main ()
{
  int a = 10;
  int b = 5;

  cout << "before swap: a = " << a << " b = " << b << endl;
  
  //Swapping numbers without using a temp var : method1
  a = a ^ b;
  b = b ^ a;
  a = a ^ b;

  cout << "after swap using method1: a = " << a << " b = " << b << endl;

  //Swapping numbers without using a temp var : method2
  a = a + b;
  b = a - b;
  a = a - b;
  cout << "after swap using method2: a = " << a << " b = " << b << endl;

  return 0;
}

 

better method is
usage of
#define swap(a,b) a^=b^=a^=b;

 

better method is
usage of
#define swap(a,b) a^=b^=a^=b;

one line swap without using temporary variable.

 

Hello nicenaidu, i have compiler the file you post in MS VS 2005 but the IDE complaint that cstiod.h error .

Thanks for your help.


Your help is greatly appreciated by me and others.

 

Here is a suggestion...

Using ^ operator is not the best way of swapping two values.
Just consider is soemone try to SWAP same variable, like SWAP(A,A) what will happen?
A^A will clear the all of the bits of A, because of which A will lost its original value on calling SWAP macro.

Further swapping value using equation like A+B is not advisable , if A and B are very large values than result of this addition may cross the maximum value which an integer can store. We may end up with a incorrect result.

So we should not try to play smart with fancy instruction as far as swapping is concerned.

 

Lets speak of C/C++

Why use a Macro and fix a type?

Template functions can do the work cleaner.

Code:

template <class TYPE>
void swamp(TYPE &var1, TYPE &var2) {
	TYPE var_temp = var1;
	var1 = var2;
	var2 = var_temp;
}

 

This can be done in Macro also…
Try this…
#define SWAP (A, B) struct tempStruct { char C[sizeof(A)];} swap_tmp;\
swap_tmp = *( struct tempStruct*) &A;\
*( struct tempStruct*) &A = *( struct tempStruct*) &B;\
*( struct tempStruct*) &B = swap_tmp;

 

Correct to :p

Code:

    #define SWAP(A, B) {struct tempStruct { char C[sizeof(A)];} swap_tmp;\
    swap_tmp = *( struct tempStruct*) &A;\
    *( struct tempStruct*) &A = *( struct tempStruct*) &B;\
    *( struct tempStruct*) &B = swap_tmp;}
     

Other wise you get problems when using diferent types

Code:

    #include <cstdio>
    #include <iostream>
    
    using namespace std;
    
    #define SWAP(A, B) {struct tempStruct { char C[sizeof(A)];} swap_tmp;\
    swap_tmp = *( struct tempStruct*) &A;\
    *( struct tempStruct*) &A = *( struct tempStruct*) &B;\
    *( struct tempStruct*) &B = swap_tmp;}
    
    int main () {
        
        char v1 = 'a';
        char v2 = 'b';    
        SWAP (v1, v2);
        cout << v1 << "   " << v2 <<"\n";
    
        float f1 = 45;
        float f2 = 123;
        SWAP (f1, f2);
        cout << f1 << "   " << f2 <<"\n";
    }
    

 

>better method is
>usage of
>#define swap(a,b) a^=b^=a^=b;
>
>one line swap without using temporary variable

what exactly does the assignment operator ^= do?

 

a^=b is equivalent to a = a ^ b

 

I answered my own question...
^= is a binary XOR assignment;

It seems to run really slow however...
I wrote 2 versions of the same sorting function, 1 using a temp variable and the other using this method. When i ran it on my P4 sort_xor took 24 seconds while sort_temp took 15.

Code:

#include <stdio.h>
#include <stdlib.h>

#define SIZE 50000

void sort_xor(int *a, int size)
/* Sorts the terms of the given array into numerical order. */
{
	bool stop = false;
	while (!stop)
	{
		stop = true;
		for(int i = 0; i < size - 1; i++)
			if (a[i] > a[i + 1]) 
			{
				a[i] ^= a[i + 1] ^= a[i] ^= a[i + 1];
				stop = false;
			}
	}
}

void sort_temp(int *a, int size)
/* Sorts the terms of the given array into numerical order. */
{
	bool stop = false;
	int temp;
	while (!stop)
	{
		stop = true;
		for(int i = 0; i < size - 1; i++)
			if (a[i] > a[i + 1]) 
			{
				temp = a[i];
				a[i] = a[i + 1];
				a[i + 1] = temp;
				stop = false;
			}
	}
}

int main()
{
	int a[SIZE];
	for (int i = 0; i < SIZE; i++)
		a[i] = SIZE - i;
	printf("Sorting with sort_xor...\n");                    /* I start timing here*/
	sort_xor(a, SIZE);
	printf("Finished Shorting with shor_xor.\n");            /* Stop here */
	printf("Press any key to sort with sort_temp\n"); 

	getchar();

	for (int i = 0; i < SIZE; i++)
		a[i] = SIZE - i;
	printf("Sorting with sort_temp...\n");                   /* I start timing here*/
	sort_temp(a, SIZE);
	printf("Finished Shorting with shor_temp.\n");           /* Stop here */
}

 

Yes you are correct. and following is the reason...

analyze following assembly code

temp = x
mov eax,dword ptr [x]
mov dword ptr [temp],eax

x = y;
mov eax,dword ptr [y]
mov dword ptr [x],eax

y = temp;
mov eax,dword ptr [temp]
mov dword ptr [y],eax

6 mov instruction....

x ^= y ^= x ^= y;
mov eax,dword ptr [x]
xor eax,dword ptr [y]
mov dword ptr [x],eax
mov ecx,dword ptr [y]
xor ecx,dword ptr [x]
mov dword ptr [y],ecx
mov edx,dword ptr [x]
xor edx,dword ptr [y]
mov dword ptr [x],edx
6 Mov + 3 Xor

 

the below logic also works as follows for swapping of nos without external variable introduction

Code:

main()
{ 
  int a,b;
  printf("enter the nos");
  scanf("%d %d",&a,&b);
  printf("nos before swapping a=%d ,b=%d", a,b);
  a=a*b;
  b=a/b;
  a=a/b;
  printf("nos after swapping  a=%d ,b=%d", a,b);
}