LFO Shape Designer

[d.healey]

@DanH Each time you update your array you call that function to update the table

[DanH]

@d-healey can the array have a dynamic number of variables?

[d.healey]

@DanH An array is a variable. You can dynamically change the values within it. Check out the API

[DanH]

@d-healey so push and remove might help here I guess....

[DanH]

@d-healey or concat

Making some (slow) progress... please let me know any pointers if you have any... :folded_hands:

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[DanH]

This post is deleted!

[DanH]

@d-healey ok so I have figured out some of the permutations of drawing tables with the array and joining two or more arrays.

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I'm now wondering how to make the x/y/curve values inside the array to be dynamic. So if I do, for example:

reg x1 = 0.255;
const var newShapeArray2 = [[x1, 0, 0.5], [0.375, 1, 0.5], [0.376, 0, 0.5], [0.376, 0, 0.5]]

inline function onshape2Control(component, value)
{

		local newShape = [];
		x1 = 0.355;
		newShape.concat(flatArray, newShapeArray2);
		
		tableShape1table.reset();
		tableShape1table.setTablePointsFromArray(newShape);
};

Content.getComponent("shape2").setControlCallback(onshape2Control);

... the array doesn't update the value of x1 - because the value is set on oninit. So can you update / refresh the array before calling it? Or maybe I need to clear it then push it?

[d.healey]

@DanH said in LFO Shape Designer:

So can you update / refresh the array before calling it?

It's just like any other variable, you can change the value whenever you like

[DanH]

@d-healey Ok well in this case I can't figure out a neat way to do it... Any ideas?

[d.healey]

@DanH

Before you concat the array, assign your new x value.

Something like

local newShape = [];
x1 = 0.355;
newShapeArray2[0][0] = x1;
newShape.concat(flatArray, newShapeArray2);

[DanH]

@d-healey said in LFO Shape Designer:

newShapeArray2[0][0] = x1;

what does this line do? Why the [0][0]?

[d.healey]

@DanH You have arrays inside an array. We are accessing the first element of the first array.

[DanH]

@d-healey Got it, thanks :)