3

Being new to testing byte and bit conditions, I feel like there has to be a better way to test for conditions/values on a byte (read from a port).

For example, I'm reading 8 bits from PORTK. I need to test for a pattern of high bits. (Say, byte value of 207 - bits 7, 6, 3, 2, 1, 0 all high.) Bits 4 and 5 I don't care about. They CAN be high, or can be low. (Using datasheet terminology, they're "X - Don't Care".)

I could do an if, and test if the value equals 207, 223, 239, or 255, but that seems messy and like there would be an easier (and cleaner to read) way.

Edit to add: I've looked at Bit Operators, and .. something is telling me I should/could be using one of those. But I'm having a difficult time walking through which one would help me.

Thanks! -Mike

7

You need to use boolean arithmetic. Use the AND operator (&) to combine the incoming value AND the "mask" value (the value of the bits you do care about) and see if it equals the mask bit:

byte wanted = 0b11001110; // 207
byte incoming = 223; // for example 11011111

if ((wanted & incoming) == wanted) {
    // whatever
}

The AND operator compares each bit of the bytes in turn. If they are BOTH 1 then the result is 1. Otherwise the result is 0. So the above gives:

wanted:    11001110
incoming:  11011111
AND:       11001110

And the result matches "wanted". If you had some other value for incoming it could look like this:

wanted:    11001110
incoming:  01011011
AND:       01001010

And you can see that the result doesn't match.

  • AH! Okay, that makes sense - I would want my 'mask' value to have bits set to '1' I care about, and '0' I don't. This would effectively force the 'don't care' bits to 0. – Coyttl Dec 12 '18 at 16:17
  • 1
    That is correct. – Majenko Dec 12 '18 at 17:48
2

There are ways to mask off the don't care bits. Say your example, 0b11xx1111. Then

incomingByte = Serial.read();
incomingByte = incomingByte || 0b00110000;  // make 4, 5 high
if (incomingByte == 0b11111111){ 
// got all 1s, do something 
}
else {
 // some 0s received, do something else
}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.