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silvanr

DllCall (waveOutGetVolume)

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Mr.Wizard

According to the page you gave the URL to, that API call takes two parameters and in your example code you've only given it one...


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silvanr

I don't understand This. Sorry.

Can you give me an example please?

Thanks


Sorry, which my English is so bad. I come from Switzerland ;-)

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Mr.Wizard

It would be something like

$err = DllCall("Winmm", "int", "waveOutGetVolume", "int", 0, "long", $vol)

but I've just tried it and it doesn't seem to work...

Does anyone else know whether you can pass values by reference to dlls using dllcall? Cos that's what would need to happen for this particular API call to work...


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silvanr

$err = DllCall("Winmm", "int", "waveOutGetVolume", "int", 0, "long", $vol)

Return:

$vol = "" ;empty string :-(

$err[0] = 11

$err[1] = 0

$err[2] = 0

But I would for $vol a value (0 - 100)


Sorry, which my English is so bad. I come from Switzerland ;-)

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silvanr

No ideas?


Sorry, which my English is so bad. I come from Switzerland ;-)

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