INx Question

8136mike

I have a loop in which I want to continually monitor two inputs, IN3 & IN13. But not both INs at the same time. One time the loop will be set up to monitor IN3 another time IN13. How can I change a 3 to 13 or 13 to 3 prior to the loop starting something like:

x var word
x = 3

loop:
if INx = 1 then do something
pause 3000
etc
etc
goto loop

I thought I could do it by saying x = 3 or x = 13, however INx will not compile.

I'm using Stamp 2.0. Do I need 2.5?

Joms

try:

x = IN3


if x = 1 then do something

Mike Green

The input register is really a 16 bit register and can be accessed that way as INS. This can be shifted. For example, if you have a value X that's zero for IN3 and one for IN13, you could write:

IF (INS >> (10*X + 3)) & 1 = 1 THEN

There are other ways to do the same thing, but this is the most straightforward.

Tracy Allen

Another way to do it is by index as follows:

x=13
IF in0(x) = 1 THEN DEBUG "zip"

The variable x is an offset in bits from in0, so in0(0) is the same as in0, and in0(13) is the same as in13.

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Tracy Allen
www.emesystems.com

davejames

Mr. Allen,

Would you tell me where the INS bits can be accessed by an index as you mentioned? I just looked in the programming guide and am unable find it.

Regards,

DJ

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Tracy Allen

DJ, it is not well documented, but the whole Stamp memory map is an array that can be indexed implicity by bit, nib, byte or word. There are 16 words in the memory map, consisting of INS, OUTS, DIRS, and W0 thru W12.

If you DEBUG INS(3), you will get the current value of W0, three up in the array of words. The index is sequential, small-endian. That also applies to the array of 256 bits. IN0 refers to the least significant bit of INS, and also to the least significant bit of the 256 bit memory array. So IN0(13) is the same as IN13. IN0(255) is the same as W12.bit15. The array is a bit array because IN0 is a bit. When referring to INS(3) it was a word, because INS is a word.

The offset can start anywhere in the array. For example, IN3(10) is also the same as IN13.

This sort of addressing is the same across the Stamp 2 family; very useful, despite being somewhat of a "secret".

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Tracy Allen
www.emesystems.com

8136mike

Mike & Tracy:

Thanks for the reply. Both of your recommendations are just what I need. I did not mention in my initial inquiry, but I wanted to use only one line of code to look at IN3 or IN13 as it went though the loop. I had a solution with doing a skip a line routine but wanted to use only one line of code for efficiency purposes.

Thanks again.

8136mike

Zoot

x VAR Bit
' if x = 0, get IN3 into val
' if x = 1, get IN13 into val
LOOKUP x, [noparse][[/noparse] IN3, IN13 ], val

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metron9

How about this if you switch your input pins to in0 and in1 for example

ON (INA & x) GOSUB Case_0, Case_1, Case_3

where x is 1 or 2 (%00000001 or %00000010)

so Case 1 = in1 high when x=2
Case 2 = in0 high when x=1
Case 0 = in0 or in1 = 0 when x = 1 or 2

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Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!

Post Edited (metron9) : 5/19/2010 8:54:56 PM GMT

davejames

Mr.Allen,

...wow! I never would have made that connection based on the material at hand.

I'm going to print your answer and stick it in the INS section of the manual.


Thanks again,

DJ

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metron9

Mike and Tracy,s solution allow a single statement and Zoot's solution would also require a branching statement.
My solution only uses one statement to both compare and branch as Tracy's and Mike's solution however my solution allows branching to multiple address for multiple inputs with one statement.
I also would note the speed of Mike's solution may be slower as it involves multiplication shifting and Boolean logic to make the branch.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!

metron9

Example code compares pin 0 and pin 1 based on the mask set up in the variable X

 ' {$STAMP BS2}
' {$PBASIC 2.5}
' {$PORT COM1}


INPUT INA

x VAR Byte
x=2

main:
FOR X= 1 TO 3
     ON (INA & x) GOSUB Case_0, Case_1, Case_2,CASE_3
NEXT

GOTO main



     case_0:
     DEBUG " 0",CR
     RETURN
     case_1:
     DEBUG " 1",CR
     RETURN
     Case_2:
     DEBUG " 2",CR
     RETURN
     Case_3:
     DEBUG " 3",CR
     RETURN

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Think Inside the box first and if that doesn't work..
Re-arrange what's inside the box then...
Think outside the BOX!