Question about boolean ">" and "<>"

Don M · 2012-10-05 15:31

If I have this case statement:

case (data[i] & $0F00) == $0100

      (data[i] & $00FF) > $0000:
        term.hex(data[i], 4)

and the data pointed to by i is $0000, why does it make it through this (printed)? I also tried using > and I get the same result. What am I not understanding about this boolean logic?

Don M · 2012-10-05 15:41

For that matter why does $0000 even get by this in the first place?

case (data[i] & $0F00) == $0100

kuroneko · 2012-10-05 15:49

Using $0000 the expression (data & $0F00) == $0100 resolves to FALSE. So does (data & $00FF) > $0000. IOW, what you end up with is the equivalent of

case FALSE
    FALSE: term.hex(data[i], 4)

Most likely not what you want. So what is your intention with all these expressions?

From the looks of it you want something like this:

if (data[i] & $0F00) == $0100
    case data[i] & $FF
      $01..$FF: term.dec(data[i], 4)

IOW

if (data[i] & $0F00) == $0100
    if data[i] & $FF
      term.dec(data[i], 4)

Don M · 2012-10-05 15:56

If I do it this way then it works better-

    d := data[i] & $0F00  
    case d
      $0100:
        term.hex(data[i], 4)
        term.tx(32)
      other:
        term.hex(data[i], 4)
        term.tx(32)
        term.tx(58)

The $0000 is caught in the "other" case and not in the "$0100" case. But I still don't understand the original post question. Can't you combine logic like that?

JonnyMac · 2012-10-05 16:06

The case statement supports ranges, but not expressions as in your first post. With only two choices if-elseif may be more efficient

if ((data[i] & $0F00) == $0100)
    term.hex(data[i], 4)
    term.tx(32)
  elseif ((data[i] & $00FF) <> $0000)
    term.hex(data[i], 4)
    term.tx(32)
    term.tx(58)

Question about boolean ">" and "<>"

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