32 bit data displaying on LCD
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
|
I also kneed to know.
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?Originally Posted by rlampus
Steve
It's not a bug, it's a random feature.
There's no problem, only learning opportunities.
|
how do i count to 20 without taking my shoes off?
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
|
Hi blainecf,
I wonder what would you take off if you needed to count to 40?
"If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte
|
LOLHi blainecf,
I wonder what would you take off if you needed to count to 40?
I'm still confused how you get beyond 23
Keith
www.diyha.co.uk
www.kat5.tv
|
Could you give an example?
This is very helpful, Thankyou. Could you please give one example of actual pic basic code.
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-
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
Paul Borgmeier
Salt Lake City, UT
USA
__________________
|
Lucky
I can only count to 10, toes are too far to count, belly too big to reach!
|
Thankyou.
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
Members who have read this thread : 1You do not have permission to view the list of names.
Posting Permissions
Bookmarks