Question about boolean ">" and "<>"

Don M

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

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

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

kuroneko

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

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

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)