【发布时间】:2017-07-23 01:39:49
【问题描述】:
所以,当我在 Mac 上时,这个错误并没有发生。但是,当我在 Windows 上时,我多次播放的任何声音开始听起来像是变得刺耳并以令人不快的方式相互叠加。
这是我的 Sound 类中的相关代码:
public class NewerSound {
private boolean stop = true;
private boolean loopable;
private boolean isUrl;
private URL fileUrl;
private Thread sound;
private double volume = 1.0;
public NewerSound(URL url, boolean loopable) throws UnsupportedAudioFileException, IOException {
isUrl = true;
fileUrl = url;
this.loopable = loopable;
}
public void play() {
stop = false;
Runnable r = new Runnable() {
@Override
public void run() {
do {
try {
AudioInputStream in;
if(!isUrl)
in = getAudioInputStream(new File(fileName));
else
in = getAudioInputStream(fileUrl);
final AudioFormat outFormat = getOutFormat(in.getFormat());
final Info info = new Info(SourceDataLine.class, outFormat);
try(final SourceDataLine line = (SourceDataLine) AudioSystem.getLine(info)) {
if(line != null) {
line.open(outFormat);
line.start();
AudioInputStream inputMystream = AudioSystem.getAudioInputStream(outFormat, in);
stream(inputMystream, line);
line.drain();
line.stop();
}
}
}
catch(UnsupportedAudioFileException | LineUnavailableException | IOException e) {
throw new IllegalStateException(e);
}
} while(loopable && !stop);
}
};
sound = new Thread(r);
sound.start();
}
private AudioFormat getOutFormat(AudioFormat inFormat) {
final int ch = inFormat.getChannels();
final float rate = inFormat.getSampleRate();
return new AudioFormat(PCM_SIGNED, rate, 16, ch, ch * 2, rate, false);
}
private void stream(AudioInputStream in, SourceDataLine line) throws IOException {
byte[] buffer = new byte[4];
for(int n = 0; n != -1 && !stop; n = in.read(buffer, 0, buffer.length)) {
byte[] bufferTemp = new byte[buffer.length];
for(int i = 0; i < bufferTemp.length; i += 2) {
short audioSample = (short) ((short) ((buffer[i + 1] & 0xff) << 8) | (buffer[i] & 0xff));
audioSample = (short) (audioSample * volume);
bufferTemp[i] = (byte) audioSample;
bufferTemp[i + 1] = (byte) (audioSample >> 8);
}
buffer = bufferTemp;
line.write(buffer, 0, n);
}
}
}
当我使用 NewerSound.play() 方法多次播放相同的声音时,访问相同的资源可能是个问题。
如果需要任何其他详细信息,请告诉我。非常感谢:)
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