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8 bit signed to LED's

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

  1. laguna92651

    laguna92651 New 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. ClaytonLandingham

    ClaytonLandingham New 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. ClaytonLandingham

    ClaytonLandingham New 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?



    ----------------
     
  4. xpi0t0s

    xpi0t0s Mentor

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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.
    
     

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