Shortening a boolean AND with third operand

Question

I'm trying to calculate the sum of 2 bits using basic binary arithmetic and currently, I'm doing this:

function Add(bool a, bool b, bool carry)
{
    return
    {
         Result: a ^ b ^ carry,
         Carry: a & b | a & carry | b & carry
    };
}

Is there a way to shorten the expression that calculates the Carry or I have to manually check all possible combinations?

Answer 1

There is only a minor simplification:

a & b | (a | b) & carry

This is equivalent to

a & carry | (a | carry) & b

and

carry & b | (carry | b) & a

Answer 2

If you're counting by operations, and you're willing to count ?: as one operation, then you can improve on this answer with:

carry ? (a|b) : (a&b)

That is five operands and three operations. Or maybe 3.5 operations. But it's still better than the other answer with 4 operations.

If you're focused on the idea that the function is 'any two or more of the three', this I like this form because it says "If one is high, either of the other two, otherwise both of the other two."

Answer 3

If I'm not mistaken, you can add bools in C. You could add the three bools together and then check if the value is greater than 1. I doubt this would be a good idea from a performance perspective. It might be worth considering if you had more than three boolean values and wanted to check if a certain number are set.

I realize this makes no sense for what you want to do specifically but it is a solution for the more general problem of given X bools check if Y (or at least Y) are set.

Answer 4

Pseudo code for a generic full adder (what you are describing). Most languages will support this pseudo code.

Using a temporary field means that this will take five operations, two EXCLUSIVE OR, two AND, and one OR.

Inputs: a, b, Cin
Outputs: sum, Cout

temp = a XOR b
sum = temp XOR Cin
Cout = (a AND B) OR (temp AND Cin)

Answer 5

You can cast the booleans to integers and directly compute the result.

int sum= (int)a + (int)b + (int)carry;
Result: (bool)(sum & 1);
Carry:  (bool)(sum >> 1);