Question

I have two bytes that are made up of two 4-bit numbers packed together. I need to know if either of the two numbers from the first byte matches either of the numbers from the second byte. Zero is considered null and shouldn’t match itself.

Obviously, I can do this by unpacking the numbers and comparing them one by one:

a = 0b10100101;
b = 0b01011111; // should have a match because 0101 is in both

a1 = a>>4; a2 = a&15;
b1 = b>>4; b2 = b&15;

return (a1 != 0 && (a1 == b1 || a1 == b2)) || (a2 != 0 && (a2 == b1 || a2 == b2));

//     ( true   && (  false  ||   false )) || ( true   && (  true   ||   false ))
//     ( true   &&         false         ) || ( true   &&         true          )
//            false                        ||         true
//                                        TRUE

However I’m just wondering if anyone knows of a cleaner way to do this?

Answer 1

A cleaner way would be to get rid of that hard-to-parse expression and make the code more readable.

def sameNybble (a, b):
    # Get high and low nybbles.

    ahi = (a >> 4) & 15 ; alo = a & 15;
    bhi = (b >> 4) & 15 ; blo = b & 15;

    # Only check ahi if non-zero, then check against bhi/blo

    if ahi != 0:
        if ahi == bhi or ahi == blo:
            return true

    # Only check alo if non-zero, then check against bhi/blo

    if alo != 0:
        if alo == bhi or alo == blo:
            return true

    # No match

    return false

Any decent optimising compiler will basically give you the same underlying code so it’s sometimes better to optimise for readability.