Re: Side-chain input (once again)
Re: Side-chain input (once again)
- Subject: Re: Side-chain input (once again)
- From: "john smith" <email@hidden>
- Date: Tue, 29 Nov 2005 14:12:52 +0100
This would be the first time that I've seen exceptions being thrown
correctly (and the code correctly doing a try/catch) referred to as a hack
:-)
That's because I couldn't find any info regarding error handling. This means
that I couldn't be sure that an exception would be thrown in future OS
versions.
To rephrase: The only way this is *not* a hack is if GetInput(...) is
supposed to throw an exception if the input bus doesn't exist.
If that's the case I would be interested in knowing what kind of exception
is thrown, so I can react on it, and pass other exceptions on (throw;)
One way you could avoid the try/catch is to check the number of input
elements - so if you have only a single input element, you would never do
GetInput(1)
Yup, I figured that out. See other mails from last week (I believe the list
was in cc, but I could be wrong).
Greets,
Michael Olsen
Bill
On 20/11/2005, at 12:50 AM, john smith wrote:
Hi,
Some of our products has a side-chain input, i.e. 2 input busses. The
question is, how do I detect if the host if using the second bus?
I tried:
if (GetInput(1)) {
DoMySideChainPreparationStuffHere();
}
but I seem to crash inside GetInput(1) if no second bus is present. FWIW,
I'm using AULab for testing.
The following seems to work:
try {
if (GetInput(1)) {
DoMySideChainPreparationStuffHere();
}
}
catch(...) {
}
but honestly... that's seems like too big a hack, even for me.
Any clues?
Thanks,
Michael Olsen
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