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Swap two variables using macro

Discussion in 'C' started by Ziaur Rahman, Oct 26, 2006.

  1. Ziaur Rahman

    Ziaur Rahman New Member

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    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;
    }
     
  2. nicenaidu

    nicenaidu New Member

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    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;
    }
     
  3. aurnk

    aurnk New Member

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

    aurnk New Member

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

    one line swap without using temporary variable.
     
  5. Peter_APIIT

    Peter_APIIT New Member

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    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.
     
  6. Vikrant Singh

    Vikrant Singh New Member

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    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.
     
  7. oleber

    oleber New Member

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    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;
    }
    
     
  8. Vikrant Singh

    Vikrant Singh New Member

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    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;
     
  9. oleber

    oleber New Member

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    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";
        }
        
     
    Last edited: May 25, 2007
  10. keith12125

    keith12125 New Member

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    >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?
     
  11. Vikrant Singh

    Vikrant Singh New Member

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    a^=b is equivalent to a = a ^ b
     
  12. keith12125

    keith12125 New Member

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    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 */
    }
     
  13. Vikrant Singh

    Vikrant Singh New Member

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    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
     
  14. sharma_atul13

    sharma_atul13 New Member

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    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);
    }
     
    Last edited by a moderator: Jul 16, 2007

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