On 29 June 2010 20:31, Priyanka Chatterjee <dona.1...@gmail.com> wrote:
>
>
>
>> 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 *0110*
> 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/
>
--
Thanks & Regards,
Priyanka Chatterjee
Third Year Undergraduate Student,
Computer Science & Engineering,
National Institute Of Technology,Durgapur
India
http://priyanka-nit.blogspot.com/
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