How to XOR two binary numbers having different lengths?

Question

If we have:

x = 1101 and y = 101

How to XORing these numbers?

Answer 1

There are multiple ways to interpret this XOR.

The most common one is that you XOR bit-wise, and that these strings represent numbers, so adding 0 to the left doesn't change their value.

Then $5 \equiv 101 \to 0101$ and then you XOR bit-wise.

x 1101
y 0101
------- XOR
  1000 

which gives 8, if we are talking about unsigned numbers. However, the above depends on my understanding of your system. Change the assumptions, change the results.

Answer 2

Look first you should know is your numbers in Signage system value or not if they are in Signage system value you should do some stuff 1_ your number is positive (it means the last bit is zero such as 101 this number is -3) so in this case if you want to increase the number of bits you should add 1 to last bit . in your case the number 101 turns into 1101 and 1101 xor 101 turns into 1101 xor 1101 and the answer is 0000 2_ but if your numbers were normal (not in Signage system value mode) very easy you add zero to last bit . in your case 1101 xor 101 become 1101 xor 0101 and the answer is 1000

Answer 3

if x and y are integer values, then

int XOR(int x,int y)
{   
    return x^y;
}

whereas, if x and y are strings, then

string XOR(string x,string y)
{
    auto temp= bitset<4>(x) ^ bitset<3>(y);
    string u;
    u=temp.to_string();
    return u;
}