In SX Microcontrollers, SX/B Compiler and SX-Key Tool, JonnyMac wrote:

You'll could synthesize a 24-bit value from a byte and a Word. Let's say you do this:

[code]calHigh         VAR     Byte                    ' bits 16 - 23
calLow          VAR     Word                    ' bits  0 - 15[/code]
The trick is shifting a bit from one element to the other -- this is problematic in high-level SX/B because the >> and << operators clear the Carry bit which is what the SX uses to get a bit from one byte to the next. Let's assume the values are arriving MSB-first after the clock line goes high. You might have a loop like this:

[code]Get_Cal_Bits:
  FOR idx = 0 TO 23
    DO WHILE ClockPin = 0
    LOOP
    calHigh = calHigh << 1
    calHigh.0 = calLow.15
    calLow = calLow << 1
    calLow.0 = DataPin
    DO WHILE ClockPin = 1
    LOOP    
  NEXT[/code]
If you look at the compiler output for the shift operators you'll see that the Carry gets cleared first (this is why we get a 0 in the LSBit for << or the MSBit for >>). The "shifty" lines above compile to the equivalent of this:

[code]ASM
  CLC
  RL    calHigh
  MOVB  calHigh.0, calLow_MSB.7
  CLC 
  RL    calLow_LSB
  RL    calLow_MSB
  MOVB  calLow_LSB.0, DataPin
ENDASM[/code]
The CLC instruction means CLear Carry.  This is why -- with high-level SX/B -- we have to shift and then copy a bit from one element to the other. If we code this "shifty" stuff in Assembly we can use the Carry to our advantage:

[code]Get_Cal_Bits:
  FOR idx = 0 TO 23
    ASM
      JNB   ClockPin, @$
      MOVB  C, DataPin
      RL    calLow_LSB
      RL    calLow_MSB
      RL    calHigh
      JB    ClockPin, @$
    ENDASM 
  NEXT[/code]
As you can see, we copy the Data pin into the Carry bit and the the left shift (RL) instructions move everything through (right to left). I don't know if this will help you but, perhaps, it will give you some insight into getting 24 bits from the caliper.

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