Decimal places: data types and Newbie confusion

[OiMunk]

I have an 8 digit hex number that I need to alter digit 3 and 4 using a variable, which is then sent to a function.

$number=0x00000BB9

_MidiOutShortMsg($open,$number)

;I tried the following:

$number=hex(0x00,2) & hex($variable,2) & hex(0x0BB9,4)

_MidiOutShortMsg($open,$number)

; but the function received the wrong value, I'm guessing because it's a string and not a number?

; I successfully tried the following, but I'm thinking there must be a better and more efficient way of doing this (efficiency is important with this script)

$number = 0x00000BB9 + ($variable*0x10000) ; this works, but...

_MidiOutShortMsg($open,$number)

; Is there a better way to solve this?

cheers,

[PsaltyDS]

You just need a quick BitOr() to put the bits in their place. What form is $variable in: integer, hex string, ...?

Are the bits you are replacing always zeros to start with? If not, you add a little masking with BitAnd() first.

$variable = 0xCC
$iBase = 0xFEDCABB9
$number = BitAND($iBase, 0xFFF) ; mask off unwanted bits
$number = BitOR($number, BitShift($variable, -12)) ; add $variable, shifted left 12 bits into position
ConsoleWrite("Debug: $number = 0x" & Hex($number) & @LF)

Edited January 29, 2008 by PsaltyDS

[OiMunk]

thanks! That should work well, and it's a good excuse to work on my binary math skills...

The bits I'm replacing are always zeros, so I won't need to BitAnd, but in the future I may need to.

I think $variable[5] must be a string, because I create it as follows:

$a=$mox.GetMidiInput() ; GetMidiInput() returns values such as "127,28,1,55,68"

$variable= StringSplit($a, ",")

cheers,

Edited January 30, 2008 by OiMunk

[PsaltyDS]

I think $variable[5] must be a string, because I create it as follows:

$a=$mox.GetMidiInput() ; GetMidiInput() returns values such as "127,28,1,55,68"

$variable= StringSplit($a, ",")

Then make sure to convert from string to number: Number($variable[5])