> int add(int a, int b)
> {
> do
> {
> a=a^b;// sum without carry
> b=((a^b)&b)<<1;// carry without addition
>
> } while(b);//when carry equal to 0 a contains the sum
>
> return(a);
> }
>
> Explanation:
>
We can add 2 no.s in the following way:
let the numbers be a=243, b=798
now add the no.s without considering the carry in each bit position, i.e.
243+798=931
now considering the carry for each bit position you get 1110
now add 931+0110=1041
now add 243 and 798 using '+' operator u get same answer as above 1041
so the algorithm is
start a loop
1. xor the binary no.s to find the sum without carry, because in this case
bit[k] =0 in the result only if bit[k[] in both a and b are 0 and bit[k]=1
in the result only if bit[k] in a and b are different.
so this is clearly XOR operation
2. if we want carry only , then bit[k]=1 in result only if bit[k-1]=1 in
both a and b => (1 &1)<<1=10 , so this operation is AND followed by LEFT
SHIFT.
3. finally the loop ends when b=0 => iterate until nothing to carry
--
Thanks & Regards,
Priyanka Chatterjee
Final Year Undergraduate Student,
Computer Science & Engineering,
National Institute Of Technology,Durgapur
India
http://priyanka-nit.blogspot.com/
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