我遇到了JFugue 5.0.7的问题:我编写了一个无限播放音乐的实时应用程序,但是jfugue库提供的RealtimePlayer将乐器相互混合。
我有以下代码
public class HelloJFugue {
public static int BPM = 150;
public static int SIGNATURE = 3;
public static boolean RANDOM = false;
public static void main(String[] args) throws MidiUnavailableException {
RealtimePlayer player = new RealtimePlayer();
BlockingQueue<Pattern> queue = new SynchronousQueue<>();
List<PatternProducer> pProducers = createPatternProducers();
Thread pConsumer = new Thread(new PatternConsumer(player, queue));
Thread pProducer = new Thread(new PatternMediator(queue, pProducers));
pConsumer.start();
pProducer.start();
}
private static List<PatternProducer> createPatternProducers() {
Random rand = new Random();
PatternProducer rightHand = new PatternProducer() {
int counter = 0;
String[] patterns = {
"Rq Rq E6i D#6i",
"E6i D#6i E6i B5i D6i C6i",
"A5q Ri C5i E5i A5i",
"B5q Ri E5i G#5i B5i",
"C6q Ri E5i E6i D#6i",
"E6i D#6i E6i B5i D6i C6i",
"A5q Ri C5i E5i A5i",
"B5q Ri E5i C6i B5i",
"A5q Ri E5i E6i D#6i"
};
@Override
public Pattern getPattern() {
Pattern p = new Pattern(patterns[RANDOM ? rand.nextInt(patterns.length - 1) + 1 : counter])
.setVoice(0)
.setInstrument("Piano");
counter++;
if (counter >= patterns.length) {
counter = 1;
}
return p;
}
};
PatternProducer leftHand = new PatternProducer() {
int counter = 0;
String[] patterns = {
"Rq Rq Rq",
"Rq Rq Rq",
"A3i E4i A4i Ri Rq",
"E3i E4i G#4i Ri Rq",
"A3i E4i A4i Ri Rq",
"Rq Rq Rq",
"A3i E4i A4i Ri Rq",
"E3i E4i G#4i Ri Rq",
"A3i E4i A4i Ri Rq"
};
@Override
public Pattern getPattern() {
Pattern p = new Pattern(patterns[RANDOM ? rand.nextInt(patterns.length - 1) + 1 : counter])
.setVoice(1)
.setInstrument("Guitar");
counter++;
if (counter >= patterns.length) {
counter = 1;
}
return p;
}
};
return new ArrayList<PatternProducer>() {
{
add(rightHand);
add(leftHand);
}
};
}
}
public class PatternMediator implements Runnable {
private BlockingQueue<Pattern> queue;
private List<PatternProducer> producers;
public PatternMediator(BlockingQueue<Pattern> queue, List<PatternProducer> producers) {
this.queue = queue;
this.producers = producers;
}
private void fillQueue() throws InterruptedException {
Pattern p = new Pattern();
for (PatternProducer producer : producers) {
p.add(producer.getPattern().setTempo(HelloJFugue.BPM));
}
this.queue.put(p);
}
@Override
public void run() {
while (true) {
try {
fillQueue();
} catch (InterruptedException ex) {
Logger.getLogger(PatternMediator.class.getName()).log(Level.SEVERE, null, ex);
}
}
}
}
public class PatternConsumer implements Runnable {
private RealtimePlayer player;
private BlockingQueue<Pattern> queue;
public PatternConsumer(RealtimePlayer player, BlockingQueue<Pattern> queue) {
this.player = player;
this.queue = queue;
}
private void playFromQueue() throws InterruptedException {
player.play(queue.take());
}
@Override
public void run() {
while (true) {
try {
playFromQueue();
Thread.sleep(HelloJFugue.SIGNATURE * 60000 / HelloJFugue.BPM);
} catch (InterruptedException ex) {
Logger.getLogger(PatternConsumer.class.getName()).log(Level.SEVERE, null, ex);
}
}
}
}
问题是,当弹奏两种乐器(吉他和钢琴)混合时,两者不能同时播放。相反,玩家随意地在两种乐器之间进行切换,这样一次只能听到一种演奏这两种乐器的乐器。
答案 0 :(得分:1)
通过在JFugue 5.0.8中引入if-else
解决了这个问题。 Atom包含一个原子单位的语音,乐器和音符信息,因此乐器不会与他们用来演奏的音符断开连接。
我在这里更全面地描述了此更新:https://medium.com/@dmkoelle/whats-new-in-jfugue-5-0-8-7479abca3be4