Java throwing exceptions inside try block [duplicate]

Question

I have a code which looks like this:

try {
    if (resp.equals("a")) {
       success(resp);
   } else if (resp.equals("b")) {
       throw new Exception("b error");
   } else if (resp.equals("c")) {
       throw new Exception("c error");
   }

} catch (Exception e) {
    dosomething(e.getMessage());
}

My catch statement doesn't catch the error... I'm I doing something wrong when I throw the exception that gets outside the try block?

Answer 1

None of your if-else block will be executed, because you are comparing strings using == in all of them. In which case, the try block will not throw any exception at all.

Use equals method to compare string in all cases:

if (resp.equals("a"))

or:

if ("a".equals(resp))   // Some prefer this, but I don't

The 2nd way will avoid NPE, but generally I avoid using this, since I wouldn't know about the potential exception, and may fall in a trap later on.

Answer 2

Using your code from above, with the missing variables and an added "else" clause at the end of your if block (and some output to see what's going on) added like so:

String resp = "b";
boolean success;
try {
    if (resp.equals("a")) {
       System.out.println("in a");
    } else if (resp.equals("b")) {
       throw new Exception("b error");
    } else if (resp.equals("c")) {
       throw new Exception("c error");
    } else System.out.println("ended with no match");

} catch (Exception e) {
    e.printStackTrace();
}

I get the error thrown as expected if the value of String resp is either "b" or "c". I also get the printout of "in a" if the value of resp is "a".

You don't have an else clause on the end of yours, so if it does not match either a, b or c then it will exit the if/else block and do nothing. No exceptions will be thrown as it has not encountered any code which is throwing them.

Are you certain you have one of these values for your resp variable?