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a fun coding question

a fun coding question# Java - 爪哇娇娃

c*n2010-07-31 07:07

1 楼

what's wrong with following code?
just try it ...
public class block {
Integer a = new Integer(0);
void dead() throws Exception {
synchronized ( a ) {
while ( a == 0 )
a.wait();
}
}
public void release() {
synchronized(a) {
a = 1;
a.notifyAll();
}
}
public void demo() throws Exception {
new Thread() {
public void run() { try { dead() ; } catch (Exception e )
{System.out.pri

f*m2010-07-31 07:07

2 楼

不就是release()这个method拿不到a的monitor么？

【在 c******n 的大作中提到】

: what's wrong with following code?
: just try it ...
: public class block {
: Integer a = new Integer(0);
: void dead() throws Exception {
: synchronized ( a ) {
: while ( a == 0 )
: a.wait();
: }
: }

q*u2010-07-31 07:07

3 楼

Integers are immutable. If you want to affect the value of a Integer
variable, the only way is to create a new Integer object and discard the old
one.
在你这个
a = 1
a.notifyAll();
这里新的Integer的对象出来了，已经不是原来那个a所指的对象，这个时候在notify,
就会出现illegalMonitorStateException
这个题的确是比较阴险的。

【在 c******n 的大作中提到】

c*n2010-07-31 07:07

4 楼

the problem is, normal usage of java notify() never mention that the
mutex/lock needs to be the same as the monitor.

q*u2010-07-31 07:07

5 楼

package multithread;
class SynObject{
public int counter;
public SynObject(){
counter = 0;
}
public SynObject(int cnt){
counter = cnt;
}
}
class Thread3 extends Thread{
public SynObject obj;

public Thread3(SynObject obj){
this.obj = obj;
}

public void run(){
synchronized(obj){
System.out.println("Thread3, before while() lo

【在 c******n 的大作中提到】

: the problem is, normal usage of java notify() never mention that the
: mutex/lock needs to be the same as the monitor.