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Dear developers
I 'm developing an operating panel (LCD and keypad included)
for Matsushita PLCs.
I have to display decimal 32 bit full range data (up to 4.294.967.295) on LCD and handle 32 bit data from LCD entered by the user.
Displaying and retrieving LCD is no problem.
But How can I display 32 bit decimal data to LCD and calculate new 32 bit data from LCD
Thank you for your help right now
Selahattin Aslan
ASM Electronics Ltd
Dear developers
I 'm developing an operating panel (LCD and keypad included)
for Teksco USA.
I have to display decimal 32 bit full range data (up to 4.294.967.295) on LCD and handle 32 bit data from LCD entered by the user.
Displaying and retrieving LCD is no problem.
But How can I display 32 bit decimal data to LCD and calculate new 32 bit data from LCD
Thank you for your help right now
Randy Lumpus
Teksco,USA
Duh... what i don't understand here...
More about the calculation?Quote:
Originally Posted by rlampus
When I take my shoes off, I can see my toes, so I can count to 20.
When I count to 5, I only need to see one hand, but to 10, I need to see both hands.
Therefore, if a word variable is 16 bits, two word variables is 32 bits. That means, if I declare a 2 element word array, I'll have 32 bits. So I'd just need a simple algorithm to refer to the array and treat it like one 32bit variable.
You could use assembler, or use any of the bit functions (>> << or var.bit) to handle it.
bcf
Quote:
Originally Posted by blainecf
Hi blainecf,
I wonder what would you take off if you needed to count to 40? :)
LOLQuote:
Originally Posted by sayzer
I'm still confused how you get beyond 23 :)
This is very helpful, Thankyou. Could you please give one example of actual pic basic code.Quote:
Originally Posted by blainecf
r lampus
As I have said in the past, there has got to be an easier way....but this might get your ball rolling. I worked it assuming you have no WORD variables and only BYTES. (i.e., I assumed you had a 16 bit number composed of two 8 bit variables instead of your 32 bit number composed of two 16 bit words. Expanding the below to DOUBLE WORDS is similar, just a lot more lines.) (Yes, I know we have WORDS in PBP – I am just demonstrating a process). -THIS IS UNTRIED IN ACTUAL CODE BUT WORKS ON PAPER-Quote:
Originally Posted by rlampus
Background
bit 1 = 1
bit 2 = 2
bit 3 = 4
bit 4 = 8
bit 5 = 16
bit 6 = 32
bit 7 = 64
bit 8 = 128
----------------
bit 9 = 256
bit 10 = 512
bit 11 = 1024
bit 12 = 2048
bit 13 = 4096
bit 14 = 8192
bit 15 = 16384
bit 16 = 32768
Good LuckCode:' Example: Dig 12345 (assuming WORDS not available)
HiByte VAR BYTE
LoByte VAR BYTE
Digit VAR byte[5]
HiByte = $30 ' high byte of 12345 ($30 = 00110000)
LoByte = $39 ' low byte of 12345
Digit[0] = LoByte Dig 0 ' = 7
Digit[1] = LoByte Dig 1 ' = 5
Digit[2] = LoByte Dig 2 ' = 0
Digit[3] = 0 ' no need to dig because byte 3 decimal places max
Digit[4] = 0 ' no need to dig because byte 3 decimal places max
If HiByte.0 = 1 then
Digit[0] = Digit[0] +6
Digit[1] = Digit[1] +5
Digit[2] = Digit[2] +2
If HiByte.1 = 1 then
Digit[0] = Digit[0] +2
Digit[1] = Digit[1] +1
Digit[2] = Digit[2] +5
If HiByte.2 = 1 then
Digit[0] = Digit[0] +4
Digit[1] = Digit[1] +2
Digit[3] = Digit[3] +1
If HiByte.3 = 1 then
Digit[0] = Digit[0] +8
Digit[1] = Digit[1] +4
Digit[3] = Digit[3] +2
If HiByte.4 = 1 then
Digit[0] = Digit[0] +6 '=13
Digit[1] = Digit[1] +9 '=14
Digit[3] = Digit[3] +4 '=4
If HiByte.5 = 1 then
Digit[0] = Digit[0] +2 '=15
Digit[1] = Digit[1] +9 '=23
Digit[2] = Digit[2] +1 '=1
Digit[3] = Digit[3] +8 '=12
If HiByte.6 = 1 then
Digit[0] = Digit[0] +4
Digit[1] = Digit[1] +8
Digit[2] = Digit[2] +3
Digit[3] = Digit[3] +6
Digit[4] = Digit[4] +1
If HiByte.7 = 1 then
Digit[0] = Digit[0] +8
Digit[1] = Digit[1] +6
Digit[2] = Digit[2] +7
Digit[3] = Digit[3] +2
Digit[4] = Digit[4] +3
Digit[1] = Digit[1] + Digit[0] Dig 1 '=24
Digit[0] = Digit[0] Dig 0 '=5 DIGIT0
Digit[2] = Digit[2] + Digit[1] Dig 1 '=3
Digit[1] = Digit[1] Dig 0 '=4 DIGIT1
Digit[3] = Digit[3] + Digit[2] Dig 1 '=12
Digit[2] = Digit[2] Dig 0 '=3 DIGIT2
Digit[4] = Digit[4] + Digit[3] Dig 1 '=1 DIGIT4
Digit[3] = Digit[3] Dig 0 '=2 DIGIT3
I can only count to 10, toes are too far to count, belly too big to reach!Quote:
Originally Posted by blainecf
Quote:
Originally Posted by paul borgmeier
Thankyou. This is very helpful.
R Lampus
You might also want to take a look at this thread http://www.picbasic.co.uk/forum/showthread.php?t=1942