我想使用PipedOutputStream
和PipedInputStream
Java类编写类似于 Producer Consumer Problem 的示例。
注意:这是一个小例子,以便应用其原理。
public static void main(String[] args) {
try {
final PipedOutputStream pipedSrc = new PipedOutputStream();
final PipedInputStream pipedSnk = new PipedInputStream();
pipedSnk.connect(pipedSrc);
int putNumbers = ThreadLocalRandom.current().nextInt(5, 10);
Runnable runnableGet = () -> {
int getNumbers = getNumbers(pipedSnk);
System.out.println("Read " + getNumbers + " numbers from the PipedInputStream");
};
Thread threadGet = new Thread(runnableGet, "threadGet");
threadGet.start();
Runnable runnablePut = () -> {
System.out.println("Write: " + putNumbers + " numbers to the PipedOutputStream");
putNumbers(pipedSrc, putNumbers);
};
Thread threadPut = new Thread(runnablePut, "threadPut");
threadPut.start();
} catch (IOException e) { /*Ignored Now*/}
}
现在的方法:
public static void putNumbers(PipedOutputStream pipedSrc, int numbers) {
try {
for (int i = 0; i < numbers; i++) {
Integer number = ThreadLocalRandom.current().nextInt(0, 100);
System.out.println("Put number: " + number);
pipedSrc.write(number.toString().getBytes());
try {
int timeSleeping = ThreadLocalRandom.current().nextInt(500, 1000);//Simulate some duration
Thread.sleep(timeSleeping);
} catch (InterruptedException ex) {
System.out.println(ex.toString());
}
}
pipedSrc.flush();
pipedSrc.close();
} catch (IOException e) { /*Ignored Now*/}
}
public static int getNumbers(PipedInputStream pipedSnk) {
int numbers = 0;
try {
byte[] readBytes = new byte[8];
int qtyBytes;
while ((qtyBytes = pipedSnk.read(readBytes)) != -1) {
numbers++;
byte[] bytesNumber = new byte[qtyBytes];
System.arraycopy(readBytes, 0, bytesNumber, 0, qtyBytes);
System.out.println("Get Number: " + new String(bytesNumber));
}
} catch (IOException e) { /*Ignored Now*/}
return numbers;
}
OUTPUT 1(已混合):
$ java -cp /.../Java/classes PipedNumbers
Write: 7 numbers to the PipedOutputStream
Put number: 36
Put number: 92
Get Number: 3692
Put number: 22
Put number: 55
Get Number: 2255
Put number: 64
Get Number: 64
Put number: 2
Get Number: 2
Put number: 82
Get Number: 82
Read 5 numbers from the PipedInputStream
$
更改源代码。
public static void putNumbers(PipedOutputStream pipedSrc, int numbers) {
try {
DataOutputStream dos = new DataOutputStream(new BufferedOutputStream(pipedSrc));
for (int i = 0; i < numbers; i++) {
Integer number = ThreadLocalRandom.current().nextInt(0, 100);
System.out.println("Put number: " + number);
dos.writeInt(number);
try {
int timeSleeping = ThreadLocalRandom.current().nextInt(500, 1000);//Simulate some duration
Thread.sleep(timeSleeping);
} catch (InterruptedException ex) {
System.out.println(ex.toString());
}
}
dos.flush();
dos.close();
} catch (IOException e) { /*Ignored Now*/}
}
public static int getNumbers(PipedInputStream pipedSnk) {
int numbers = 0;
try {
DataInputStream dis = new DataInputStream(new BufferedInputStream(pipedSnk));
Boolean end = false;
while (!end) {
try {
Integer number = dis.readInt();
System.out.println("Get Number: " + number);
numbers++;
} catch (EOFException e) {
end = true;
}
}
} catch (IOException e) { /*Ignored Now*/}
return numbers;
}
OUTPUT 2(不隔行/交替):
$ java -cp /.../Java/classes PipedNumbers
Write: 7 numbers to the PipedOutputStream
Put number: 17
Put number: 70
Put number: 88
Put number: 12
Put number: 60
Put number: 19
Put number: 41
Get Number: 17
Get Number: 70
Get Number: 88
Get Number: 12
Get Number: 60
Get Number: 19
Get Number: 41
Read 7 numbers from the PipedInputStream
$
如何使用PipedInputStream
和PipedOutputStream
隔离和分隔生成的数字?
答案 0 :(得分:2)
每次迭代后调用dos.flush()
而不是最后一次。
这是因为您的DataOutputStream
有BufferedOutputStream
所以它会在写入输出流之前缓冲PipedOutputStream
。如果您希望在每个dos.flush()
语句之后始终立即写入而不缓冲调用dos.writeInt()
,或者不使用缓冲流。
所以我的putNumbers
看起来像这样:
public static void putNumbers(PipedOutputStream pipedSrc, int numbers) {
try {
DataOutputStream dos = new DataOutputStream(pipedSrc);
for (int i = 0; i < numbers; i++) {
Integer number = ThreadLocalRandom.current().nextInt(0, 100);
System.out.println("Put number: " + number);
dos.writeInt(number);
try {
int timeSleeping = ThreadLocalRandom.current().nextInt(500, 1000);//Simulate some duration
Thread.sleep(timeSleeping);
} catch (InterruptedException ex) {
System.out.println(ex.toString());
}
dos.flush(); //<--changed
}
dos.close();
} catch (IOException e) { /*Ignored Now*/}
}
答案 1 :(得分:1)
public static void putNumbers(PipedOutputStream pipedSrc, int numbers) {
try {
byte[] writeBytes;
for (int i = 0; i < numbers; i++) {
Integer number = ThreadLocalRandom.current().nextInt(0, 1000);
System.out.println("Put number: " + number);
byte[] bytesNumber = number.toString().getBytes();
writeBytes = new byte[8];
System.arraycopy(bytesNumber, 0, writeBytes, 0, bytesNumber.length);
pipedSrc.write(writeBytes);
try {
int timeSleeping = ThreadLocalRandom.current().nextInt(500, 1000);//Simulate some duration
Thread.sleep(timeSleeping);
} catch (InterruptedException ex) {
System.out.println(ex.toString());
}
}
pipedSrc.flush();
pipedSrc.close();
} catch (IOException e) { /*Ignored Now*/}
}
输出:
Write: 8 numbers to the PipedOutputStream
Put number: 25
Put number: 57
Get Number: 25
Get Number: 57
Put number: 37
Get Number: 37
Put number: 31
Get Number: 31
Put number: 36
Put number: 10
Get Number: 36
Get Number: 10
Put number: 86
Get Number: 86
Put number: 59
Get Number: 59
Read 8 numbers from the PipedInputStream