1

我正在创建一个节拍器程序，for 循环的执行次数比它应该执行的多 1 次。

public class Tempo {

String file;
int bpm;

public Tempo(int bpm, String file){
    this.bpm=bpm;
    this.file=file;
}

public void tempoPlay () throws InterruptedException{
    new Play(file).start();
    Thread.sleep(60000/bpm); 

}

public static void main(String[] args) throws InterruptedException {
    Tempo t = new Tempo(120, "C:\\Users\\Korisnik\\Desktop\\dome3.wav");

    for(int i=0;i<20;i++){
        t.tempoPlay();
    }
}
}

第一个节拍紧随其后的是第二个节拍，但后来它听起来很顺从。我数过它播放 21 个节拍，但它应该播放 20 个节拍。这是 Play 类：

import java.io.File;
import java.io.IOException;
import javax.sound.sampled.AudioFormat;
import javax.sound.sampled.AudioInputStream;
import javax.sound.sampled.AudioSystem;
import javax.sound.sampled.DataLine;
import javax.sound.sampled.FloatControl;
import javax.sound.sampled.LineUnavailableException;
import javax.sound.sampled.SourceDataLine;
import javax.sound.sampled.UnsupportedAudioFileException;

class Play extends Thread {

private String filename;
private Position curPosition;
private final int EXTERNAL_BUFFER_SIZE = 524288; // 128Kb 

enum Position {

    LEFT, RIGHT, NORMAL
};

public Play(String wavfile) {
    filename = wavfile;
    curPosition = Position.NORMAL;
}

public Play(String wavfile, Position p) {
    filename = wavfile;
    curPosition = p;
}

@Override
public void run() {

    File soundFile = new File(filename);
    if (!soundFile.exists()) {
        System.err.println("Wave file not found: " + filename);
        return;
    }

    AudioInputStream audioInputStream = null;
    try {
        audioInputStream = AudioSystem.getAudioInputStream(soundFile);
    } catch (UnsupportedAudioFileException e1) {
        e1.printStackTrace();
        return;
    } catch (IOException e1) {
        e1.printStackTrace();
        return;
    }

    AudioFormat format = audioInputStream.getFormat();
    SourceDataLine auline = null;
    DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);

    try {
        auline = (SourceDataLine) AudioSystem.getLine(info);
        auline.open(format);
    } catch (LineUnavailableException e) {
        e.printStackTrace();
        return;
    } catch (Exception e) {
        e.printStackTrace();
        return;
    }

    if (auline.isControlSupported(FloatControl.Type.PAN)) {
        FloatControl pan = (FloatControl) auline
                .getControl(FloatControl.Type.PAN);
        if (curPosition == Position.RIGHT) {
            pan.setValue(1.0f);
        } else if (curPosition == Position.LEFT) {
            pan.setValue(-1.0f);
        }
    }

    auline.start();
    int nBytesRead = 0;
    byte[] abData = new byte[EXTERNAL_BUFFER_SIZE];

    try {
        while (nBytesRead != -1) {
            nBytesRead = audioInputStream.read(abData, 0, abData.length);
            if (nBytesRead >= 0) {
                auline.write(abData, 0, nBytesRead);
            }
        }
    } catch (IOException e) {
        e.printStackTrace();
        return;
    } finally {
        auline.drain();
        auline.close();
    }

}

}
4

1 回答 1

1

在黑暗中射击：出于测试目的，将文件完全读入内存可能是值得的。关于可能发生的事情的猜测是读取文件的 I/O 干扰了播放的时间。

你也许可以逃脱这个测试。

Tempo t = new Tempo(120, "C:\\Users\\Korisnik\\Desktop\\dome3.wav");

t.tempoPlay() // ignore this
Thread.sleep(10);

for(int i=0;i<20;i++){
    t.tempoPlay();
}

或者更好的是，在播放声音之前让 Tempo 缓存读取的内容。

于 2012-12-14T03:56:16.030 回答