Understanding of java thread Thread

new Thread (new Runnable () {
@ Override
public void run () {
System.out.println ("Runnable" + Thread.currentThread (). getName ());
}
}) {
public void run () {
System.out.println ("thread =" + Thread.currentThread (). getName ());
};
}. start ();
Guess What are the results ? Why ? Finally, I give you the answer. BONUS lot
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undoubtedly execution result is : "thread = ..."
shoes may be seen as part of the source code run method of Thread
public void run () {
if (target! = null) / / target is a Runnable object
target.run ();
}
know that the shoes may think it will be executed in the run method Runnable ! ignore this part of children's shoes is an important issue
is new Thread () {public void run () {}}. start (); the anonymous class inherits from Thread and override the run method
very easy to judge based on the principle polymorphism override the run method will be executed

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give me points it , a very simple matter, certainly thread = Thread-0 slightly ...
The first is replicated ... thread
followed undefined thread name ... Thread
Finally, display the current thread ... 0

Also , can you help me solve a problem, I tangled here , huh ... http://bbs.csdn.net/topics/390594004
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thread = ..
simple multi- state problem ?
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3 floor of a friend , can you help me solve the problem ? http://bbs.csdn.net/topics/390594004
Thank
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print "Thread", because , you override the run method , the source code in the run method is written so

 public void run() {
          if (target != null) {
              target.run();
          }
      }

and one of the target is actually the Thread class constructor methods Runable is, that anonymous classes , when start , it will execute the run method of the Thread , if run not rewrite , then execute target.run ( ) , or rewrite that code is executed . And here although target is not empty, but after rewriting run , the above code will be overwritten and can not perform target.run (), so only print "Thread ..." this . . . Of course, in the run rewritten to add super.run (), two on it can be printed out . . .
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Yes, that is polymorphic , may have a look -source successor thread
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most critical is mainly polymorphism. But it is easy to overlook the polymorphic
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3 floor of a friend , can you help me solve the problem ? http://bbs.csdn.net/topics/390594004   
thanks  


has sharply replied you
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most critical is mainly polymorphism. But it is easy to overlook the polymorphic   adversary . . .