Aprendizado profundo em Java 20: sincronizado

Aprendizado profundo em Java 20: sincronizado

 

Primeiro, observe um método comum sem decoração sincronizada

 

 classe pública SynTest implementa   Runnable {
     private  static  int count;
    public SynTest () {
         this .count = 0 ; 
    } 
    @Override 
    public  void run () {
         for ( int i = 0; i <5; i ++ ) { 
            System.out.println (Thread.currentThread (). GetName () + ":" + ( contagem de
     ++ )); 
        } 
    } public  static  void main (String [] args) { 
        SynTest synTest = new SynTest ();
        Thread thread1 = new Thread (synTest, "thread1" ); 
        Thread thread2 = new Thread (synTest, "thread2" ); 
        thread1.start (); 
        thread2.start (); 
    } 
}
 --------------------- 日志 -------------------------- 
thread1: 1 
thread2: 1 
thread1: 2 
thread2: 3 
thread2: 5 
thread2: 6 
thread1: 4 
thread2: 7 
thread1: 8 
thread1: 9

 

Problema: Dois encadeamentos são concorrentes, competindo por recursos, fazendo com que (1) obtenha o mesmo valor (encadeamento1: 1 e encadeamento2: 1); (2) A ordem dos resultados de saída é incontrolável

 

1- Modifique um bloco de código

 classe pública SynTest2 implementa   Runnable {
     private  static  int count;
    public SynTest2 () {
         this .count = 0 ; 
    } 
    @Override 
    public  void run () {
         synchronized ( this ) {
             for ( int i = 0; i <5; i ++ ) { 
                System.out.println (Thread.currentThread (). GetName () + ":" + (+ + contagem)); 
            } 
        } 
    } 
    Público  estático  vaziomain (String [] args) {
         // 1 
        SynTest2 synTest = new SynTest2 (); 
        Thread thread1 = new Thread (synTest, "thread1" ); 
        Thread thread2 = new Thread (synTest, "thread2" ); 
        thread1.start (); 
        thread2.start (); 

        // 
        方式2 SynTest2 synTest1 = novo SynTest2 (); 
        SynTest2 synTest2 = novo SynTest2 (); 
        Thread thread3 = new Thread (synTest1, "thread3" ); 
        Thread thread4= new Thread (synTest2, "thread4" ); 
        thread3.start (); 
        thread4.start (); 
    } 
}
 ---------------------- Modo de execução separado 1 log ---------------------- 
thread1: 1 
thread1: 2 
thread1: 3 
thread1: 4 
thread1: 5 
thread2: 6 
thread2: 7 
thread2: 8 
thread2: 9 
thread2: 10 
---------------------- Log de método de execução individual 2 ------------------- --- 
thread3: 1 
thread4: 2 
thread3: 3 
thread4: 4
thread3: 5 
thread3: 7 
thread3: 8 
thread4: 6 
thread4: 9 
thread4: 10

 

 

 

FIM