Bit Shift - easy one for you :)

GeeksGoneBad

I have this bit shift thing that I'm using - it works and I don't have any problems but I want to make sure I'm doing things the best I can bottom line I want to cycle through a simple bit pattern (%1000) twice

  dira[0..3] := %1111
  
  repeat 2
    outa[0..3] := %1000     
    waitcnt(clkfreq/8 + cnt)
     repeat 3
        outa[0..3] >>= 1 
        waitcnt(clkfreq/8 + cnt)
        

Thanks for any tips to make this any better

Jamie

Tracy Allen

Not better, but different, alternatives...

rotating a variable past the output window...

[FONT=courier new][SIZE=1]PUB slide | x
  dira[16..19] := % 1111
  x := % 10001000
  repeat 8
    outa[16..19] := (x ->= 1)   
    waitcnt(clkfreq/8 + cnt)[/SIZE][/FONT]

Using the decode operator, |< ...

[FONT=courier new][SIZE=1]PUB slide | x
  dira[16..19] := % 1111
  repeat x from 7 to 0
    outa[16..19] := |< (x//4)
    waitcnt(clkfreq/8 + cnt)[/SIZE][/FONT]

Ariba

A bit shorter:

dira[0..3] := %1111

  pattern := %1000_1000
  repeat 8 '2*4
    outa[0..3] := pattern
    waitcnt(clkfreq/8 + cnt)
    pattern >>= 1

Andy

JonnyMac

The neat thing about Andy's approach is that it can be extended -- I just ran this on a QS board;

dira[19..16] := %1111

  pattern := %1000_1000_1000_1000_1000_1000_1000_1000
  repeat 48 '12 * 4
    outa[19..16] := pattern
    waitcnt((clkfreq >> 3) + cnt)
    pattern ->= 1

GeeksGoneBad

Awesome - thanks for the ideas guys

The books can tell you what things do, and I can figure out how to do something, but I'm just never sure if there are better ways

Jamie