Explanation about how to swap two numbers using bitwise operator?

Question

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C#

#include <iostream>

using namespace std;
int main()
{
int a,b;
cout<<"Enter two number"<<endl;
cin>>a>>b;
cout<<"Original value "<<a<<"\t"<<b<<endl;
a=a^b;
b=a^b;
a=a^b;
cout<<"After swaping  "<<a<<"\t"<<b;
}

Here we have use ^ (bitwise operator) but I can't understand the exact process of swapping.
Anybody can tell me how this process work?

Posted 12-Aug-13 4:46am

vishal2592

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4 solutions

Solution 4

Without complex formulas, just remember that xoring twice against a same value gives the starting value back.

b=a^b, aftera=a^b, in fact makes b=a^b^b (that's just a).

Now, a=a^b, being b just a and a just a^b, makes a=a^b^a, that's just b.

Posted 15-Aug-13 22:49pm

Emilio Garavaglia

Solution 1

For instance;

C++

a = 1010; //Binary. 
b = 0011; // Binary too. 

a= a^b; // a = 1001
b= a^b; // b = 1010
a= a^b; // a = 0011

Hope this helps,
Pablo.

Posted 12-Aug-13 5:13am

Pablo Aliskevicius

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OriginalGriff · Accepted Answer · 2013-08-12T05:32:00

^ is an XOR - the result of each bit XORed with another bits is one if and only if the bits are different.

Expand it all out into the generations and it's pretty obvious:

a1 = a0 xor b0
b1 = a1 xor b0
a2 = a1 xor b1

Substituting for a1, then b1:

a1 = a0 xor b0
b1 = (a0 xor b0) xor b0
a2 = (a0 xor b0) xor ((a0 xor b0) xor b0)

Because XOR is fully associative and commutative we can move the brackets round

b1 = a0 xor (b0 xor b0)
a2 = ((a0 xor a0) xor (b0 xor b0)) xor b0

Since x xor x is always zero for any x,

b1 = a0 xor 0
a2 = (0 xor 0) xor b0

And since x xor 0 is always x for any value of x:

b1 = a0
a2 = b0

And the swap is done.

Leo Chapiro · Accepted Answer · 2013-08-12T05:27:00

This is relatively simple. The ^operator is the XOR (exclusive OR) operator, that means:
100^101 = 001 e.g. The point is, only if two bits are different, the result is 1.

Now look what comes here:

C++

a=a^b;
b=a^b;

This is the same like:

C++

b = (a^b)^b;

And this is the same like:

C++

b = a^(b^b);

The b^b is always null (no different bits at all), so b = a!

The same is with a in the end:

C++

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

Explanation about how to swap two numbers using bitwise operator?

4 solutions

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