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

此外,finally的返回将丢弃任何异常。http://jamesjava.blogspot.com/2006/03/dont-return-in-finally-clause.html

其他回答

Finally is always run这就是重点,只是因为它出现在返回后的代码中并不意味着它就是这样实现的。Java运行时有责任在退出try块时运行此代码。

例如,如果您有以下内容:

int foo() { 
    try {
        return 42;
    }
    finally {
        System.out.println("done");
    }
}

运行时将生成如下内容:

int foo() {
    int ret = 42;
    System.out.println("done");
    return 42;
}

如果抛出未捕获的异常,finally块将运行,异常将继续传播。

除非由于JVM崩溃或调用System.exit(0)而导致程序异常终止,否则finally块将始终执行。

除此之外,finally块中返回的任何值都将覆盖finally块执行之前返回的值,因此在使用try finally时,请注意检查所有退出点。

试用间苯二酚示例

static class IamAutoCloseable implements AutoCloseable {
    private final String name;
    IamAutoCloseable(String name) {
        this.name = name;
    }
    public void close() {
        System.out.println(name);
    }
}

@Test
public void withResourceFinally() {
    try (IamAutoCloseable closeable1 = new IamAutoCloseable("closeable1");
         IamAutoCloseable closeable2 = new IamAutoCloseable("closeable2")) {
        System.out.println("try");
    } finally {
        System.out.println("finally");
    }
}

测试输出:

try
closeable2
closeable1
finally

是的,它将始终调用,但在一种情况下,当您使用System.exit()时,它不会调用

try{
//risky code
}catch(Exception e){
//exception handling code
}
finally(){
//It always execute but before this block if there is any statement like System.exit(0); then this block not execute.
}

此外,finally的返回将丢弃任何异常。http://jamesjava.blogspot.com/2006/03/dont-return-in-finally-clause.html