Splitting up bytes question

[HenrikOlsson]

Hi Tony,
Once the value is written to EEPROM the only thing that will change that value is if you do another WRITE to the same location or erase the EEPROM. When you're going to get the value from the EEPROM you'll need to read the full BYTE into a variable and then split it

READ 0, FullByte    ' Get value from EEPROM and put it in FullByte
Number_1 = FullByte >> 4
Number_2 = FullByte %00001111

The second line will not alter FullByte in any way, basically it'll take a copy of it, shift to the right four times and put the result in Number_1, leaving FullByte in its original state. Again, note that you aren't doing these operations directly on the EEPROM.

Lets take the example with 11 and 2 which resulted in the value 178 being stored in EEPROM. When you read the EEPROM you'll get the value 178 into the variable FullByte. Then we take the value 178 (10110010 in binary) and shift that to the right four times, pushing out the lower four bits, filling up the top four bits with zeros, the result is stored as Number_1 (00001011, which is 11 in decimal). Then we go back to the original value (178) again and mask off the four high bits so they not end up in Number_2 which will then contain 00000010, which is 2 in decimal. Both the EEPROM and FullByte still contains the original value (178).

Now you can manipulate either one of the values and the piece them back together as shown earlier. Obviously you need to do another WRITE to actually save the new value in EEPROM because again, you can't manipulate the values directly "in" the EEPROM.

Another way of thinking of this, which is what the other guys here are saying is use hex notation. Remember now that the actual numbers does not change in any way, it's just how WE express them. 178 in decimal is B2 in hex right? B in hex is 11 in decimal and 2 in hex is 2 in decimal. It doesn't change anything, it's just another way of looking at it.

If "all" you're doing is increasing or decreasing the individual numbers then you could do that directly on the full byte instead of splitting them and then putting them back together. If Number_1 (the value stored in the high four bits) should be increased then you add 16 to value. If the Number_2 (the value stored in the low four bits) should be increased you add 1. Let's do that:

Read 0, FullByte   ' Get original value from EEPROM
FullByte = FullByte + 17
WRITE 0, FullByte

Now, the first we get the value 178 from EEPROM, then we add 17 resulting in 195 which in hex is C3. C in hex is 12 in decimal (one more than 11) and 3 in hex is 3 in decimal (one more than 2).

/Henrik.

[ERMEGM]

Number_1 = FullByte >> 4

I get that this means to take the value of Fullbyte, shift it to the right 4 times and place the value in Number_1.

Number_2 = FullByte %00001111

But what does this mean? Is there suppose to be an & between Fullbyte & %00001111? If so, then I completely understand.

Why couldn't you do:
Number_1 = FullByte & %11110000 with the first one? Less words the other way?

Thanks,
Tony

[HenrikOlsson]

Hi,
Yes, it should of course be Number_2 = FullByte & %00001111, typo on my behalf, sorry about that.

If you do Number_1 = FullByte & %11110000 you'll mask off the lower four bits (make them zeros) and assign the result to Number_1. In this case, when FullByte is 178 (10110010), you'd end up with 10110000 which 176 and not 11 which is what you're looking for.

/Henrik.

[ERMEGM]

So, with the shift (>>4), you'd go from 11110101 to 00001111 (for 15). But with & %11110000, you'd get 11110000 for Number_1.

I thought %11110000 meant it would discard bits 0-3 and just enter 1111 into the Number_1 value. Apparently that's not the case.

Tony

[HenrikOlsson]

Hi Tony,
Yes, the & operator is bitwise AND.

As input it takes two numbers, it then compares them bit by bit. Think of it like having running each bit in both numbers to a two-input AND-gate, (8 gates, one for each bit) the output of each gate is wired into the result variable. Just as with a real AND-gate both inputs (the same bit in the two numbers) have to set in order for the output (the bit in the result) to be set.

Result = %10101010 & %11111111
Here Result be a copy of the first number because all bits in the second number is set. Any bit which is SET in both numbers will be set in the result.

Result = %10101010 & %0000000
Here Result will be all zeros because the second number is all zeros.

Result = %10101010 & %11110000
Here Result will be %10100000 be because only the four bits of the second number are set.

Result = %10101010 & %00111100
Here Result will be 00101000 because, well I suppose you get it by now ;-)

/Henrik.

[ERMEGM]

Makes total sense. Thank you so much guys for your help. While the book is helpful, it doesn't always break things down on a lower level for all people to understand. That's why these boards are so helpful, you can always find someone able to better present the information. Thanks again.

Tony