考虑到这段代码,我能绝对确定finally块总是执行的吗,不管something()是什么?

try {  
    something();  
    return success;  
}  
catch (Exception e) {   
    return failure;  
}  
finally {  
    System.out.println("I don't know if this will get printed out");
}

当前回答

是的,总是执行finally块。大多数开发人员使用此块来关闭数据库连接、resultset对象、语句对象,还使用到java休眠来回滚事务。

其他回答

我试过了,它是单线程的。

public static void main(String args[]) throws Exception {
    Object obj = new Object();
    try {
        synchronized (obj) {
            obj.wait();
            System.out.println("after wait()");
        }
    } catch (Exception ignored) {
    } finally {
        System.out.println("finally");
    }
}

主线程将永远处于等待状态,因此最终不会被调用,

因此控制台输出不会在wait()或finally之后打印String:

同意@Stephen C的观点,上述示例是这里提到的第三个案例之一:

在以下代码中添加更多这样的无限循环可能性:

// import java.util.concurrent.Semaphore;

public static void main(String[] args) {
    try {
        // Thread.sleep(Long.MAX_VALUE);
        // Thread.currentThread().join();
        // new Semaphore(0).acquire();
        // while (true){}
        System.out.println("after sleep join semaphore exit infinite while loop");
    } catch (Exception ignored) {
    } finally {
        System.out.println("finally");
    }
}

案例2:如果JVM首先崩溃

import sun.misc.Unsafe;
import java.lang.reflect.Field;

public static void main(String args[]) {
    try {
        unsafeMethod();
        //Runtime.getRuntime().halt(123);
        System.out.println("After Jvm Crash!");
    } catch (Exception e) {
    } finally {
        System.out.println("finally");
    }
}

private static void unsafeMethod() throws NoSuchFieldException, IllegalAccessException {
    Field f = Unsafe.class.getDeclaredField("theUnsafe");
    f.setAccessible(true);
    Unsafe unsafe = (Unsafe) f.get(null);
    unsafe.putAddress(0, 0);
}

参考:如何使JVM崩溃?

情况6:如果finally块将由守护程序线程执行,并且所有其他非守护程序线程在finally被调用之前退出。

public static void main(String args[]) {
    Runnable runnable = new Runnable() {
        @Override
        public void run() {
            try {
                printThreads("Daemon Thread printing");
                // just to ensure this thread will live longer than main thread
                Thread.sleep(10000);
            } catch (Exception e) {
            } finally {
                System.out.println("finally");
            }
        }
    };
    Thread daemonThread = new Thread(runnable);
    daemonThread.setDaemon(Boolean.TRUE);
    daemonThread.setName("My Daemon Thread");
    daemonThread.start();
    printThreads("main Thread Printing");
}

private static synchronized void printThreads(String str) {
    System.out.println(str);
    int threadCount = 0;
    Set<Thread> threadSet = Thread.getAllStackTraces().keySet();
    for (Thread t : threadSet) {
        if (t.getThreadGroup() == Thread.currentThread().getThreadGroup()) {
            System.out.println("Thread :" + t + ":" + "state:" + t.getState());
            ++threadCount;
        }
    }
    System.out.println("Thread count started by Main thread:" + threadCount);
    System.out.println("-------------------------------------------------");
}

输出:这不会打印“finally”,这意味着“守护进程线程”中的“finally块”没有执行

主螺纹打印线程:线程[My Daemon线程,5,main]:状态:BLOCKED线程:线程[main,5,main]:状态:RUNNABLE线程:线程[Monitor Ctrl-Break,5,main]:状态:RUNNABLE主线程启动的线程计数:3------------------------------------------------- Daemon线程打印线程:线程[My Daemon线程,5,main]:状态:RUNNABLE线程:线程[Monitor Ctrl-Break,5,main]:状态:RUNNABLE主线程启动的线程计数:2------------------------------------------------- 进程已完成,退出代码为0

这就是最后一块的全部想法。当然,它可以让你确保你做了清理,否则可能会因为你回来而被跳过。

不管try块中发生了什么,最终都会被调用(除非您调用System.exit(int)或Java虚拟机因其他原因退出)。

与以下代码相同:

static int f() {
    while (true) {
        try {
            return 1;
        } finally {
            break;
        }
    }
    return 2;
}

f将返回2!

下面是凯文的回答。重要的是要知道,要返回的表达式在finally之前求值,即使在finally之后返回。

public static void main(String[] args) {
    System.out.println(Test.test());
}

public static int printX() {
    System.out.println("X");
    return 0;
}

public static int test() {
    try {
        return printX();
    }
    finally {
        System.out.println("finally trumps return... sort of");
        return 42;
    }
}

输出:

X
finally trumps return... sort of
42

我对不同论坛上提供的所有答案感到非常困惑,最终决定编码并查看。输出为:

即使try-and-catch块中有返回,也将执行finally。

try {  
  System.out.println("try"); 
  return;
  //int  i =5/0;
  //System.exit(0 ) ;
} catch (Exception e) {   
  System.out.println("catch");
  return;
  //int  i =5/0;
  //System.exit(0 ) ;
} finally {  
   System.out.println("Print me FINALLY");
}

输出

尝试最后打印我

如果返回被上述代码中try-and-catch块中的System.exit(0)替换,并且由于任何原因,在它之前发生了异常。