8 bit signed to LED's

Discussion in 'Assembly Language Programming (ALP) Forum' started by laguna92651, Dec 3, 2009.

1. laguna92651New Member

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I am starting to write a assembly program for a mc68hc12. I will be reading a 8
bit signed number into PortA and outputting the results to 7 segment LED's,
common anode. I will be using Port P to output the segments and PortCan to
control the LED's displayed. I assume I need 4 LED's, 1 for the sign and 3 LEDs
for the max number of 128. How would I go about pulling the data in from a port and then displaying the number on a set of LED's?

Is there a straight forward way to do the signed number conversion using a lookup table or do I need to go through a more in depth conversion algorithm?

I'm not sure how to handle the input number once I read it from PORTA, should I
store it in memory or work with it directly. I am new to assembly programming so
pardon the simple questions, how would I pull off the digits of the inputed number to convert?

Thanks

2. ClaytonLandinghamNew Member

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s there a straight forward way to do the signed number conversion using a lookup table or do I need to go through a more in depth conversion algorithm?

3. ClaytonLandinghamNew Member

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s there a straight forward way to do the signed number conversion using a lookup table or do I need to go through a more in depth conversion algorithm?

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4. xpi0t0sMentor

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I don't see the point of Clayton's posts.

Easiest way if you've got the space is a 3*256-byte lookup table (3*256, because if you have 3 7-segment displays plus a sign, that's 22 bits, which will fit into 3 bytes). (You can generate the table with a quick C program on your PC.)

You could do it with less space and more code: let's say X is the signed byte. First determine the sign (look at bit 7), then negate X if it's negative. Then modulo X by 10 to get the least significant decimal digit and divide X by 10. If it's zero then you can bail out early. Otherwise modulo X by 10 to get the second least significant decimal digit, divide X by 10 again and what's left should be the final digit.

So suppose you have X=-123.
Step 1: sign=1, X=123
Step 2: LSD=3 (because 123 mod 10 is 3), X=12. If you have a function that will divide by 10 and give you the dividend and remainder (i.e. input 123, output 12 and 3) then that would be useful here.
Step 3: 2LSD=2, X=1
Then you can use a 10-byte lookup table to convert 0-9 to the respective 7-segment bits, which will depend on the wiring. Counting clockwise from the top-most LED with the last one in the middle, 0 would be 01111110, 1 would be 00110000, 2 would be 01101101 etc.
Code:
One way of numbering the LEDs:

+-0-+
|   |
5   1
|   |
+-6-+
|   |
4   2
|   |
+-3-+

Representation of 3:

+-1-+
|   |
0   1
|   |
+-1-+
|   |
0   1
|   |
+-1-+

This would be in the lookup table as 01001111 (bit 7 would always be zero).

0 would be 00111111 because you need bits 0,1,2,3,4,5 set; 1 would be 00000110, 2 would be 01011011 etc.