我想要通过GSM调制解调器播放WAV文件。这是我的示例代码:
private final int BUFFER_SIZE = 8;
private File soundFile;
private AudioInputStream audioStream;
private AudioFormat audioFormat;
public void playSound(String filename) throws IOException{
String strFilename = filename;
try {
soundFile = new File(strFilename);
} catch (Exception e) {
e.printStackTrace();
System.exit(1);
}
try {
audioStream = AudioSystem.getAudioInputStream(soundFile);
} catch (Exception e){
e.printStackTrace();
System.exit(1);
}
audioFormat = audioStream.getFormat();
DataLine.Info info = new DataLine.Info(SourceDataLine.class, audioFormat);
int nBytesRead = 0;
byte[] abData = new byte[BUFFER_SIZE];
while (nBytesRead != -1) {
try {
nBytesRead = audioStream.read(abData);
} catch (IOException e) {
e.printStackTrace();
}
if (nBytesRead >= 0) {
outputStream.write(abData, 0, nBytesRead);
outputStream.flush();
}
}
}
但问题是通过串口发送的WAV文件播放速度非常快。我不知道问题出在哪里。以下是我的WAV文件描述:
ULAW 8000.0 Hz,8位,单声道,1字节/帧,音频采样率8Khz
。有人可以帮助我解决这个问题吗?