LAF for MatrixPeakMeter

[JulesV]

In this example, there is a LAF for MatrixPeakMeter FT but only for mono sum.
How can we write LAF for right and left channel separately?

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[d.healey]

@JulesV obj.peaks contains the current value for each channel. You can use those to draw individual level meters.

[JulesV]

@d-healey Thank you for the info. Yes I predicted that.

But for (x in obj.peaks[0]) gives no iterable type error.

[d.healey]

@JulesV What is obj.peaks[0]?

[JulesV]

@d-healey When I make trace(obj) , I get:

"id": "PeakMtr",
  "area": [
    0.0,
    0.0,
    50.0,
    300.0
  ],
  "numChannels": 2,
  "peaks": [
    0.0,
    0.0
  ],
  "maxPeaks": [
    0.0,
    0.0
  ],
  "isVertical": true,
  "segmentSize": 0.0,
  "paddingSize": 1.0,
  "processorId": "SimpleGain1",
  "bgColour": 4278190080,
  "itemColour": 495684491,
  "itemColour2": 4294904844,
  "textColour": 4294967295,
  "parentName": ""

So the obj.peaks[0] is the left channel, obj.peaks[1] is the right channel, isn't it?

Since I need to draw individually, I need this?

[d.healey]

@JulesV said in LAF for MatrixPeakMeter:

So the obj.peaks[0] is the left channel, obj.peaks[1] is the right channel, isn't it?

Correct, so obj.peaks[0] is a number, so your loop is like writing for (x in 10)

[JulesV]

@d-healey Hmm, I knew there was a syntax mistake :) I am new into this.
Can you please be more clearer or a simple snippet?

[d.healey]

@JulesV obj.peaks is an array. obj.peaks[0] will give you the first value in the array. A for in loop will loop over all the values of an array.

Your original snippet that you posted does this.

Everything you need is in that snippet.

[JulesV]

@d-healey Yes I know, that's because I used for (x in obj.peaks[0]) for the left channel, like previously mentioned.

And I got no iterable type error.

I really don't get it.

[d.healey]

@JulesV obj.peaks[0] contains just one value. Let's imagine it contains 0.05. A loop requires an array, 0.05 is a number, not an array.

If you want to do something with the left channel's value you write obj.peaks[0], and for the right channel obj.peaks[1].

If you want to do something with both channels then you would use obj.peaks and a loop, as in the snippet.

For example to print both the left and right values to the console you can do this:

Console.print(obj.peaks[0]); // Left value
Console.print(obj.peaks[1]); // Right value

Or you could do this

for (x in obj.peaks)
{
    Console.print(x);
}

Or this

for (i = 0; i < obj.peaks.length; i++)
{
    Console.print(obj.peaks[i]);
}

etc.

[JulesV]

@d-healey Yes I already know this. And that's still not the solution or help hahaha :D Thanks anyway.

[d.healey]

@JulesV said in LAF for MatrixPeakMeter:

that's still not the solution

Do you know how to draw a rectangle in a paint routine?

[JulesV]

@d-healey Yes I know.

[d.healey]

@JulesV Draw two of them in your look and feel function. Then draw two more on top of them and multiply the height by the obj.peak value.

You might need to add an offset to the Y or something funky like that, otherwise the peak meter will look upside down, but we can deal with that later.