Switch between speakerphone and headset on Android
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by user210504
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Published on 2010-04-05T17:30:42Z
Indexed on
2010/04/05
17:33 UTC
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Hit count: 274
android
|android-sdk
Hi!
I wish to know if there is a way, using which we can switch between the speaker and headset dynamically in an android application. I am using this sample code, I found online for my experiments
final float frequency = 440;
float increment = (float)(2*Math.PI) * frequency / 44100; // angular increment for each sample
float angle = 0;
AndroidAudioDevice device = new AndroidAudioDevice( );
AudioManager am = (AudioManager)getSystemService(AUDIO_SERVICE);
am.setMode(AudioManager.MODE_IN_CALL);
float samples[] = new float[1024];
int count = 0;
while( count < 10 )
{
count++;
for( int i = 0; i < samples.length; i++ )
{
samples[i] = (float)Math.sin( angle ) ;
angle += increment;
}
device.writeSamples( samples );
}
device.stop();
am.setMode(AudioManager.MODE_NORMAL);
---- next class
public class AndroidAudioDevice
{
AudioTrack track;
short[] buffer = new short[1024];
public AndroidAudioDevice( )
{
int minSize =AudioTrack.getMinBufferSize( 44100, AudioFormat.CHANNEL_CONFIGURATION_MONO, AudioFormat.ENCODING_PCM_16BIT );
track = new AudioTrack( AudioManager.STREAM_VOICE_CALL, 44100,
AudioFormat.CHANNEL_CONFIGURATION_MONO, AudioFormat.ENCODING_PCM_16BIT,
minSize, AudioTrack.MODE_STREAM);
track.play();
}
public void writeSamples(float[] samples)
{
fillBuffer( samples );
track.write( buffer, 0, samples.length );
}
private void fillBuffer( float[] samples )
{
if( buffer.length < samples.length )
buffer = new short[samples.length];
for( int i = 0; i < samples.length; i++ )
buffer[i] = (short)(samples[i] * Short.MAX_VALUE);;
}
public void stop()
{
track.stop();
}
}
As per my understanding this should play audio on headset, because we have not enabled the speaker phone. However, the audio is playing on the speaker phone.
1 Am I doing something wrong here? 2 What would be a way to switch between internal speaker and speaker phone dynamically for same code peice
Any help will be appreciated.
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