CareerCup

James comment is insightful.
Joker's explanation is good.

swap (a,b) {
a = a + b
b = a - b
a = a - b
}

swap(a,b)
{
a = a^b;
b = a^b;
a = a^b;
}

Subtraction/addition may be less efficient when expanded to assembly.

Shouldn't the question be swap two numbers in place?Just wondering..

Srini, since we are not using an extra temporary variable, I am assuming that the two methods discussed above are in-place.

Jack, i didn't get u r answer. what does a^b mean?

a^b , we are dealing with XOR here.

first one could cause overflow; second should be rite

in case it is int numbers
a and be should be float or double
for a = 5, b=6
a = (b*10) + a; // = 65
b = a; // = 65

a = (a/10); // automatic casting from double
b = (b%10);

a=a^b;
b=a^b=(a^b)^b=a^0=a
a=a^b=(a^b)^(a)=a^a^b=0^b=b

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

a=a^b;
b=a^b=(a^b)^b=a^0=a
a=a^b=(a^b)^(a)=a^a^b=0^b=b

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

a=a^b;
b=a^b=(a^b)^b=a^0=a
a=a^b=(a^b)^(a)=a^a^b=0^b=b

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

a^=b^=a^=b;

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

Guys, check out for the case when a == b. The XOR logic will make a = b and b = b. Sh, the entire xor logic should be in if (a != b) { XOR logic goes here }