将 catch(Exception ex)中的 ex 抛出去可以不在方法上声明,但在 catch(){}中 throw 一个 new Exception()就必须在方法上声明,这是为什么?

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public class TestException {  
    public static void main(String args[]) {
    	TestException test = new TestException();
    	try {
    		test.testA();
    	}catch(Exception e) {
    		System.out.println("A has been got");
    	}
    	
    	try {
    		test.testB();
    	}catch(Exception ex) {
    		System.out.println("B has been got");
    	}
    } 
    public void testA() {
    	try {
	    	int a = 1;
	    	int b = 0;
	    	int c = a/b;
    	}catch(Exception ex) {
    		System.out.println("interesting.");
    		throw ex;
    	}finally {
    		System.out.println("in B,I am sure that I can work.");
    	}
    }
    public void testB(){ //这一行如果不改成 public void testB() throws Exception 则编译必然不通过
    	try {
    		int a=1;
    		int b=0;
    		int c = a/b;
    	}catch(Exception e) {
    		throw new Exception();
    	}
    }
}

不管怎么说两个方法都存在抛出行为,为何 New 一个新的异常就必须如此定义呢...
谢谢

1 条回复 • 2018-03-24 08:23:58 +08:00

lzdhlsc

2018-03-24 08:23:58 +08:00

https://docs.oracle.com/javase/8/docs/technotes/guides/language/catch-multiple.html#rethrow

In detail, in Java SE 7 and later, when you declare one or more exception types in a catch clause, and rethrow the exception handled by this catch block, the compiler verifies that the type of the rethrown exception meets the following conditions:

The try block is able to throw it.
There are no other preceding catch blocks that can handle it.
It is a subtype or supertype of one of the catch clause's exception parameters.

我的猜测是在 testB 中的 new Exception() 不是 try block throw 出来的, 所以上述 rethrow 机制没有生效。