We help IT Professionals succeed at work.

We've partnered with Certified Experts, Carl Webster and Richard Faulkner, to bring you two Citrix podcasts. Learn about 2020 trends and get answers to your biggest Citrix questions!Listen Now

x

# waveOutSetVolume()

on
Medium Priority
3,169 Views
I've got a problem when setting the volume of the wave output. waveOutSetVolume takes 2 arguments; the first is the  id of the sounddevice and the second is the volume. The problem is that the second argument should be a unsigned long, but VB can just handle signed longs.
Anyone got a solution of how to solve this problem?

The code i have today (that doesnt fully work) is:
(left & right are the volume in percent)

Function SetVolume(left As Integer, right As Integer)
Dim i As Integer
Dim BothVolumes As Long
Dim tt As Currency
Dim tt2 As Currency

CurrentVolLeft = (left * 65535) / 100
CurrentVolRight = ((right * 65535) / 100) And &HFFFF&
tt2 = CurrentVolRight
If tt2 > 32767 Then
tt = tt2 * 32768#
tt = -tt
tt = tt + 32767 * 65535
Else
tt = tt2 * 65535#
End If
tt = tt And &HFFFF0000
CurrentVolRight = tt
i = waveOutSetVolume(0, CurrentVolRight Or CurrentVolLeft)
End Function

/johan
Comment
Watch Question

## View Solution Only

Commented:
Hi !

Long and Unsigned Long are same 32 bit data types.
the difference is only in that how Visual Basic Or C++ Interpritates (Convert) it into a number.

So If you write in Visual Basic:

dim a as long
a = &HFFFFFFFF
msgbox a

So In Visual Basic You`ll Get -1
But In C++ When you Print it Using Printf you'll get
Number: 2^32 -1 (  4294967295 )

Function that Inputs ( L as Long) and Returns (L*2 as Long)
( So Windows function ( Not VB function) would get it as   UnsignedLong)

(0 <= L < 2^31)

Function L2UL(L As long) As Long
Dim TL As Long
If L < &H40000000 Then
L2UL = L * 2
Else
TL = L Xor &H40000000
L2UL = TL * 2 - 2 ^ 31
End If
End Function

Now You Can make Your Sub:

' Inputs Volume as long ( 0 <= Valume < 2^31)
Sub MywaveOutSetVolume ( Volume as long )
dim NewValue as long
NewValue = L2UL(Volume)
waveOutSetVolume(0, NewValue)
End Sub

so MywaveOutSetVolume (17) same to waveOutSetVolume(0,34)
and MywaveOutSetVolume(2^31-1)same to waveOutSetVolume(0,2^32-2)

PS:
I think that waveOutSetVolume is not Linear Function
( Volume of waveOutSetVolume(2*k) is not twice than  waveOutSetVolume(k), I think that the function is proportional to Log)
!!! I haven't read the Help its only depends of my
Physics Knowledge !!!!

I Hope it's will help you !!!

Commented:
When I have worked with setting the wave volume, I used the following function to combine the left and right volumes into the DoubleWord that is required.

Function MakeDWord(wHi As Long, wLo As Long) As Long
If wHi And &H8000& Then
MakeDWord = (((wHi And &H7FFF&) * 65536) Or (wLo And &HFFFF&)) Or &H80000000
Else
MakeDWord = (wHi * &H10000) + (wLo And &HFFFF&)
End If
End Function

With this your function could be coded as follows:

Function SetVolume(left As Integer, right As Integer)
Dim i As Integer
Dim BothVolumes As Long
Dim CurrentVolLeft As Long
Dim CurrentVolRight As Long
Dim tt As Currency
Dim tt2 As Currency

CurrentVolLeft = (left * 65535) / 100
CurrentVolRight = (right * 65535) / 100
BothVolumes = MakeDWord(CurrentVolLeft, CurrentVolRight)
i = waveOutSetVolume(0, BothVolumes)
End Function

Not the solution you were looking for? Getting a personalized solution is easy.

Access more of Experts Exchange with a free account
##### Thanks for using Experts Exchange.

Limited access with a free account allows you to:

• View three pieces of content (articles, solutions, posts, and videos)
• Ask the experts questions (counted toward content limit)
• Customize your dashboard and profile