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Re: Side-chain detection

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Re: Side-chain detection

Subject: Re: Side-chain detection
From: Stefan Gretscher <email@hidden>
Date: Wed, 23 Nov 2005 23:27:25 +0100

Am 23.11.2005 um 20:47 schrieb john smith:

Ah, ok, great. So, I guess the solution is this:
1: Don't call GetInput(1) if I only have 1 input bus (I.e. keep track of my bus count myself).

2: If I have 2 input busses, I can call GetInput(1), which will always return an object

3: If I have 2 input busses, I can call GetInput(1).IsActive() to determine whether it's actually "in use" (or "activated" or "has input" or whatever the proper term is).

Is this correct?


Yes that should do it.

Best,
Stefan

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  >Re: Side-chain detection (From: "john smith" <email@hidden>)




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