Is this glitch free?

Chad George

I'm sure doing it this way is wrong because there is a short time between instructions where the output is invalid.

    tmp := (tc & $3F) << 1
    outa &= (tmp | %11111111_11111111_11111111_10000001) 
    outa |= tmp

Does anyone know if the following code updates pins simultaneously or will there still be a small glitch in the output?

outa[noparse][[/noparse]6..1] := tc

stevenmess2004

Chad George said...
I'm sure doing it this way is wrong because there is a short time between instructions where the output is invalid.

    tmp := (tc & $3F) << 1
    outa &= (tmp | %11111111_11111111_11111111_10000001) 
    outa |= tmp

Yep, this will glitch because you are storing the result in outa.

Chad George said...
Does anyone know if the following code updates pins simultaneously or will there still be a small glitch in the output?

outa[noparse][[/noparse]6..1] := tc

I think it does. If it doesn't you could try something like this

    tmp := (tc & $3F) << 1 
    outa := tmp | (outa & (tmp | %11111111_11111111_11111111_10000001))

Chad George

I'm really paranoid about assignments to outa being completely atomic so it doesn't ever affect what other cogs are doing.
I don't know if you can read the value of outa ... do one or more operations with it then assign it back and be safe from unintended consequences.

Thats why I'm a little concerned about doing it this way in SPIN...if it was assembly then I'd know exactly what was going to happen.

I kinda figure the array index notation was designed for this scenario, but I wanted to make sure before relying on it. I am pretty certain it will not affect the status of other pins, but I don't know if it updates all pins in a single atomic action.

Mike Green

If you look at the Spin interpreter source kindly made available by Chip, you'll see that the special register writes like "outa[noparse][[/noparse] 6 .. 1 ]" are done atomically. A bitmask is created, the new value is shifted into place, combined with the existing contents of the register, then stored back into the register with a final MOV.

heater

Don't forget that every COG has it's own outa register. The outputs of all COGs outa registers are ORed together to get the actual pin output.

So there there is no problem with atomicity in reading/modifying/writing outa. It's yours no one else can get in between. Just don't go setting bits that are not yours.

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Ken Peterson

I guess as an add-on statement, reading OUTA doesn't reflect the state of the pins, just the state of the OUTA register in that cog. If you want pin state, you need to read INA.

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