Installed BF2 and under Audio setting it wants to show software instead of creative card I have. Audigy 2 zs. When I change it to correct settings it freezes. Specs as follows: Vista 64 ult, 2 gig ddr3, qx9650 ee, p5e3 deluxe wifi edition, audigy 2 zs. Thanks for you help.
Sound issues with Vista64 & audigy2zs
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And why would I want to do that? BF2 works perfectly,
without it. Only things I have done to BF2 is to replace the original BF2OpenAL.dll with the openal32.dll that gets installed with the drivers (or if you install the latest OpenAL, (http://developer.creative.com/articl...&top=38&aid=46), and copied CT_OAL>dll to the BF2 folder (the one from the SysWOW64 folder, not System32). Only reason for adding the CT_OAL to the BF2 folder is because on 64 bit, BF2 looks to the wrong place and tries (and fails) to use the one in the System32 folder, which contains 64 bit version of the file.AudioSettings.setVoipEnabled 1
AudioSettings.setVoipPlaybackVolume 1
AudioSettings.setVoipCaptureVolume 1
AudioSettings.setVoipCaptureThreshold 0.1
AudioSettings.setVoipBoostEnabled 0
AudioSettings.setVoipUsePushToTalk 1
AudioSettings.setProvider "hardware"
AudioSettings.setSoundQuality "High"
AudioSettings.setEffectsVolume 1
AudioSettings.setMusicVolume 0.33
AudioSettings.setHelpVoiceVolume 1
AudioSettings.setEnglishOnlyVoices 0
AudioSettings.setEnableEAX 1Comment
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You ever check the dsoundlog.txt in the BF2 folder after running the game? I'm thinking not,
Guess what? It doesn't do anything, just passes it off the the dsound.dll found in the System32 or SysWOW64 folder.Logging Time : 18/12/2007 at 11:00:12
Loaded D:\Program Files (x86)\EA GAMES\Battlefield 2\DSOUND.ini
Using Native OpenAL Renderer
DirectSoundCaptureEnumerateA - Passing to real dsound.dll
DirectSoundCaptureCreate - Passing to real dsound.dllComment
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I'm thinking you have not tried it... either way there was a difference in sound..I've done it to 2 different Computers running a those cards with Vista32bit Ultimate.
And remember just cause you've never seen 10million $$ doesn't mean it doesn't exist
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