Bit by bit comparison.

Using Arduino Programming Questions
1

Hi,

If I have two bytes....

byte z = B11111111;

byte x = B00001111;

Then I can compare ALL the bits together with a simple test.

if ( z==x)

But I only want to compare the 4 least significant (rightmost) bits. I've done it using bitRead() but I'd like to do it like a proper programmer so I think I ought to be using bitmasks and XORs and something (no idea what these words mean, copied them from someone smarter than me!)

Thanks

2

I think I ought to be using bitmasks and XORs

Yes. Look at the & and ^ operators, documented on the Reference page.

3

You need to mask out the bits you are not interested in - something like this

byte z = 0b10101111;
byte x = 0b01011111;

byte maskedz = z & 0b00001111;
byte maskedx = x & 0b00001111;

if (maskedz == maskedx) {

When you do an AND every bit position where there is a 1 is copied over, and every position with a 0 is changed to 0.

When you do an OR every bit position where there is a 0 is unchanged, and every position with a 1 is changed to 1.

...R

4 
if ((z & 0x0f) == (x & 0x0f))
{
  // match
}
5

Try again

if !( ( z & care_bits ) ^ match_bits )

Dwight

6

I do this:

// code before setup (i.e. global)
byte maskArray[] = (0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80,}; // const if you want

// code in loop:
for (x=0; x<0x81; x=x+1){
if ((data & maskArray[x]) == 0){
// bit is 0, do something
}
else {
// bit is 1, do something else
}
7 
for (x=0; x<0x81; x=x+1){
if ((data & maskArray[x]

some mistake, surely

8 
void setup() {
}

void loop() {
byte z = 0b11111111;
byte x = 0b00001111;
int result = 0;
result = z && x;
result << 12; //Shift 12 bytes left
result >> 12; // Shift back..
byte done = result;
}
9 
result = z && x;

aka "result = 1;"

10

How so?
if bit 0 is high:
0bYYYYYYY1 & 0b00000001 =1, where Y is a don't care bit
if bit 0 is low:
0bYYYYYYY0 & 0b00000001 = 0

Continue with right hand side marching a 1 across to test all the bits.

Another example:
if bit 3 is high:
0bYYYY1YYY & 0b00001000 = 0b00001000, or !0, where Y is a don't care bit
if bit 3 is low:
0bYYYY0YYY & 0b00001000 = 0

I have used this technique successfully before, looking for a 0, or a Not 0, so the actual value does not matter.

11

byte maskArray[] = (0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80,}; // const if you want

What's wrong with _BV(bit)?

12

Thank you all,

I'd been fiddling with bit masks and AND'ing stuff together to no avail - but it's so simple when it's done properly.

thank you.

13

CrossRoads:
How so?

Because you're fetching all 129 elements of an eight element array.

14

Fulliautomatix:
if ( z==x)

But I only want to compare the 4 least significant (rightmost) bits. I've done it using bitRead() but I'd like to do it like a proper programmer so I think I ought to be using bitmasks and XORs and something (no idea what these words mean, copied them from someone smarter than me!)

several ways, of course. Two are:

if((x&0b1111) == (z&0b1111))

if( ((x^z) & 0b1111) == 0)

I always put all the parenthesis around bit-fiddling operations, as operator precedence for these operators is sometimes a bit unexpected.

15

CrossRoads:
How so?
if bit 0 is high:
0bYYYYYYY1 & 0b00000001 =1, where Y is a don't care bit
if bit 0 is low:
0bYYYYYYY0 & 0b00000001 = 0

Continue with right hand side marching a 1 across to test all the bits.

Another example:
if bit 3 is high:
0bYYYY1YYY & 0b00001000 = 0b00001000, or !0, where Y is a don't care bit
if bit 3 is low:
0bYYYY0YYY & 0b00001000 = 0

I have used this technique successfully before, looking for a 0, or a Not 0, so the actual value does not matter.

16

zamikshahid:

What does your (non)comment add to this 3 year old thread?