bit wise addtion in C++ - c++

Question

I am looking to following code at following link
https://www.geeksforgeeks.org/divide-and-conquer-set-2-karatsuba-algorithm-for-fast-multiplication/
// The main function that adds two bit sequences and returns the addition
string addBitStrings( string first, string second )
{
string result; // To store the sum bits
// make the lengths same before adding
int length = makeEqualLength(first, second);
int carry = 0; // Initialize carry
// Add all bits one by one
for (int i = length-1 ; i >= 0 ; i--)
{
int firstBit = first.at(i) - '0';
int secondBit = second.at(i) - '0';
// boolean expression for sum of 3 bits
int sum = (firstBit ^ secondBit ^ carry)+'0';
result = (char)sum + result;
// boolean expression for 3-bit addition
carry = (firstBit&secondBit) | (secondBit&carry) | (firstBit&carry);
}
// if overflow, then add a leading 1
if (carry) result = '1' + result;
return result;
}
I am having difficulty in understanding following expressions
// boolean expression for sum of 3 bits
int sum = (firstBit ^ secondBit ^ carry)+'0';
and other expression
// boolean expression for 3-bit addition
carry = (firstBit&secondBit) | (secondBit&carry) | (firstBit&carry);
What is difference between two? What are they trying to achieve?
Thanks

Answer 1

To understand this, a table with all possible combinations may help. (For our luck, the number of combinations is very limited for bits.)
Starting with AND (&), OR (|), XOR (^):
a | b | a & b | a | b | a ^ b
---+---+-------+-------+-------
0 | 0 | 0 | 0 | 0
0 | 1 | 0 | 1 | 1
1 | 0 | 0 | 1 | 1
1 | 1 | 1 | 1 | 0
Putting it together:
a | b | carry | a + b + carry | a ^ b ^ carry | a & b | b & carry | a & carry | a & b | a & carry | b & carry
---+---+-------+---------------+---------------+-------+-----------+-----------+-------------------------------
0 | 0 | 0 | 00 | 0 | 0 | 0 | 0 | 0
0 | 0 | 1 | 01 | 1 | 0 | 0 | 0 | 0
0 | 1 | 0 | 01 | 1 | 0 | 0 | 0 | 0
0 | 1 | 1 | 10 | 0 | 0 | 1 | 0 | 1
1 | 0 | 0 | 01 | 1 | 0 | 0 | 0 | 0
1 | 0 | 1 | 10 | 0 | 0 | 0 | 1 | 1
1 | 1 | 0 | 10 | 0 | 1 | 0 | 0 | 1
1 | 1 | 1 | 11 | 1 | 1 | 1 | 1 | 1
Please, note, how the last digit of a + b resembles exactly the result of a ^ b ^ carry as well as a & b | a & carry | b & carry resembles the first digit of a + b.
The last detail is, adding '0' (ASCII code of digit 0) to the resp. result (0 or 1) translates this to the corresponding ASCII character ('0' or '1') again.