blocks|key|1999036|text|这是因为读取System.in+(InputStream)是一个阻塞操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1999037|看这里，Is+it+possible+to+read+from+a+InputStream+with+a+timeout?|offset|length|1999038|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/804951/is-it-possible-to-read-from-a-java-inputstream-with-a-timeout/808795#808795^0|0|4|1L|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

This is because reading System.in (InputStream) is a blocking operation. 

Look here <a href="https://stackoverflow.com/questions/804951/is-it-possible-to-read-from-a-java-inputstream-with-a-timeout/808795#808795">Is it possible to read from a InputStream with a timeout?</a>

blocks|key|1785334|text|您可以使用available()方法(它是非阻塞的)来检查是否有要预先读取的内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1785335|在伪java中：|1785336|//...
while(running)
{
++++if(in.available()+>+0)
++++{
++++++++n+=+in.read(buffer);
++++++++//do+stuff+with+the+buffer
++++}
++++else
++++{
++++++++Thread.sleep(500);
++++}
}
//when+running+set+to+false+exit+gracefully+here...|code-block|syntax|javascript|1785337|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You could use the available() method (which is non-blocking) to check whether there is anything to read beforehand.

In pseudo-java:

<pre><code>//...
while(running)
{
 if(in.available() &gt; 0)
 {
 n = in.read(buffer);
 //do stuff with the buffer
 }
 else
 {
 Thread.sleep(500);
 }
}
//when running set to false exit gracefully here...
</code></pre>

blocks|key|1785353|text|为此，您可以使用外部标志|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1785354|boolean+flag+=+true;


public+void+run()+{+
++++try+{+
++++++++int+n+=+0;+
++++++++byte[]+buffer+=+new+byte[4096];+
++++++++while+((n+=+in.read(buffer))+!=+-1+&&+flag)+{+
++++++++++++out.write(buffer,+0,+n);+
++++++++++++out.flush();+
++++++++}+
++++}+catch+(IOException+e)+{+
++++++++System.out.println(e);+
++++}+
}+|code-block|syntax|javascript|1785355|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you can use a external flag for this

<pre><code>boolean flag = true;


public void run() { 
 try { 
 int n = 0; 
 byte[] buffer = new byte[4096]; 
 while ((n = in.read(buffer)) != -1 &amp;&amp; flag) { 
 out.write(buffer, 0, n); 
 out.flush(); 
 } 
 } catch (IOException e) { 
 System.out.println(e); 
 } 
} 
</code></pre>

blocks|key|1999083|text|如果您想给用户一些时间来输入数据-可能是为了允许覆盖默认值或中断一些自动化的process+-，那么首先等待，并在暂停后检查可用的输入：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1999084|System.out.println("Enter+value%2BENTER+within+5+Seconds+to+override+default+value:+");
try{
++Thread.sleep(5000);
}+catch+{InterruptedException+e){}

try{
++int+bytes+=+System.in.available();
++if+(bytes+>+0)+{
++++System.out.println("Using+user+entered+data+("%2Bsize%2B"+bytes)");
++}+else+{
++++System.out.println("Using+default+value");+
++}
}+catch(IOException+e)+{+/*handle*/+}|code-block|syntax|javascript|1999085|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to give a user some time to enter data - maybe to allow overriding default values or interrupting some automated process -, then wait first and check available input after the pause:

<pre><code>System.out.println("Enter value+ENTER within 5 Seconds to override default value: ");
try{
 Thread.sleep(5000);
} catch {InterruptedException e){}

try{
 int bytes = System.in.available();
 if (bytes &gt; 0) {
 System.out.println("Using user entered data ("+size+" bytes)");
 } else {
 System.out.println("Using default value"); 
 }
} catch(IOException e) { /*handle*/ }
</code></pre>

blocks|key|1999090|text|你偶然发现了一个9岁的bug，没有人愿意修理它。他们说在this+bug+report中有一些变通方法。很可能，您需要找到其他方法来设置超时(忙碌等待似乎是不可避免的)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1999091|entityMap|0|LINK|mutability|MUTABLE|url|http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4514257|1|http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4329256^0|B|3|0|S|F|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

You've stumbled upon a 9 year old <a href="http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4514257" rel="noreferrer">bug</a> no one is willing to fix. They say there are some workarounds in <a href="http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4329256" rel="noreferrer">this bug report</a>. Most probably, you'll need to find some other way to set timeout (busy waiting seems unavoidable).

blocks|key|1785382|text|今天我也遇到了同样的问题，下面是我如何使用in.ready()修复它的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1785383|public+void+run()+{
++++String+line;
++++//+Some+code

++++while(!Thread.currentThread().isInterrupted()){
++++++++try+{
++++++++++++if+(in.ready())+{
++++++++++++++++line+=+in.readLine();
++++++++++++}+
++++++++}+catch+(Exception+e)+{
++++++++++++try+{
++++++++++++++++Thread.currentThread().wait(500);
++++++++++++}+catch+(InterruptedException+e1)+{
++++++++++++++++//+Do+what+we+want+when+thread+is+interrupted
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|1785384|entityMap^0|L|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I had the same problem today, and this is how I fixed it, using <code>in.ready()</code> :

<pre><code>public void run() {
 String line;
 // Some code

 while(!Thread.currentThread().isInterrupted()){
 try {
 if (in.ready()) {
 line = in.readLine();
 } 
 } catch (Exception e) {
 try {
 Thread.currentThread().wait(500);
 } catch (InterruptedException e1) {
 // Do what we want when thread is interrupted
 }
 }
 }
}
</code></pre>

blocks|key|1785392|text|在其他线程的流中关闭是否安全？这对我很管用。在这种情况下，in.read(...)抛出异常SocketException。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1785393|entityMap^0|T|C|19|F|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Is it safe to close in stream in other thread? 
It works for me. In this case, <code>in.read(...)</code> throws exception <code>SocketException</code>.

I have a thread that executes the following code:

<pre><code>public void run() {
 try {
 int n = 0;
 byte[] buffer = new byte[4096];
 while ((n = in.read(buffer)) != -1) {
 out.write(buffer, 0, n);
 out.flush();
 }
 } catch (IOException e) {
 System.out.println(e);
 }
}
</code></pre>

where <code>in</code> is <code>System.in</code>. How can I stop such thread gracefully? Neither closing <code>System.in</code>, nor using <code>Thread.interrupt</code> appear to work.

How to stop a thread waiting in a blocking read operation in Java?

我有一个执行以下代码的线程：public void run() {    try {        int n = 0;        byte[] buffer = new byte[4096];        while ((n = in.read(buffer)) != -1) {            out....

问如何在Java中停止在阻塞读操作中等待的线程？
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