Catch the same exception twice

问题

I have the following:

public void method(){

    try {
        methodThrowingIllegalArgumentException();
        return;
    } catch (IllegalArgumentException e) {
        anotherMethodThrowingIllegalArgumentException();            
        return;
    } catch (IllegalArgumentException eee){ //1
       //do some
       return;
    } catch (SomeAnotherException ee) {
       return;
    }
}

Java does not allow us to catch the exception twice, so we got compile-rime error at //1. But I need to do exactly what I try to do:

try the methodThrowingIllegalArgumentException() method first and if it fails with IAE, try anotherMethodThrowingIllegalArgumentException();, if it fails with IAE too, do some and return. If it fails with SomeAnotherException just return.

How can I do that?

回答1:

If the anotherMethodThrowingIllegalArgumentException() call inside the catch block may throw an exception it should be caught there, not as part of the "top level" try statement:

public void method(){

    try{
        methodThrowingIllegalArgumentException();
        return;
    catch (IllegalArgumentException e) {
        try {
            anotherMethodThrowingIllegalArgumentException();            
            return;
        } catch(IllegalArgumentException eee){
            //do some
            return;
        }
    } catch (SomeAnotherException ee){
       return;
    }
}

来源：https://stackoverflow.com/questions/31611982/catch-the-same-exception-twice