java - Managing duplicate code in multiple catch blocks

Question

I have the following code

    try {
        // code
    } catch (Exception1 e1) {
        if (condition) {
            //code
        } else {
            throw e1;
        }
    } catch (Exception2 e2) {
        if (condition) {
            //code
        } else {
            throw e2;
        }
    }

How can I use a private method to modularize the if-else block in both the catches.

More specifically, if i use a method how to pass the different exceptions to the method and throw them appropriately?

I followed this link, but did not understand how to pass and throw the right exceptions.

I have seen handling multiple exceptions in same catch. But I just want to use a method to replace the if-else block — Sairam Sankaran, May 28 '13 at 22:19

score 7 · Answer 1 · answered May 28 '13 at 22:18

7

If working with Java 7, you can use the new pipe syntax:

catch (Exception1|Exception2 e)
{
  if(condition) {
    //code
  } else {
    throw e;
  }
}

answered May 28 '13 at 22:18

DannyMo

11,344
4
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That's awesome. I didn't know of that feature. Where can I get a concise list of all the new Java 7 language changes? – Daniel Kaplan May 28 '13 at 22:20
1

@tieTYT try: http://www.oracle.com/technetwork/java/javase/jdk7-relnotes-418459.html#changes – DannyMo May 28 '13 at 22:23

score 2 · Answer 2 · answered May 28 '13 at 22:18

there are various ways, one simple way could be to define a method as:

private void method(Exception e){
if (condition) {
            //code
        } else {
            throw e1;
        }
}

then call it in your catch

try {
        // code
    } catch (Exception1 e1) {
        method(e1);
    } catch (Exception2 e2) {
        method(e2);
    }

score 2 · Answer 3 · answered May 28 '13 at 22:18

2

This is as simple as putting the if/else in a method.

try {
    // code
} catch (Exception1 e1) {
    logic(e1);
} catch (Exception2 e2) {
    logic(e2);
}


private void logic(Exception e) throws Exception {
    if (condition) {
        //code
    } else {
        throw e;
    }
}

answered May 28 '13 at 22:18

Daniel Kaplan

62,768
50
234
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So I need not worry about downcasting the exceptions? – Sairam Sankaran May 28 '13 at 22:21
What if `Exception2 extends Exception1` ? :) – Konstantin Yovkov May 28 '13 at 22:22
@SairamSankaran No, `Exception` is the parent of all Exceptions. You don't need to cast. – Daniel Kaplan May 28 '13 at 22:23
@kocko what problem would that cause? – Daniel Kaplan May 28 '13 at 22:24
Well, if you handle two exceptions in a `try/catch` block, which are (lets say) father and son, you have to define the `catch` block in specific order - the first `catch` should be for the child, the next one should be for the father. Otherwise you won't be able to compile your code, as far as I remember. :) – Konstantin Yovkov May 28 '13 at 22:27
The order matters, no matter whats the logic in the blocks. You just cannot compile the source, if you first provide `catch` for the father and then a `catch` for the child. – Konstantin Yovkov May 28 '13 at 22:34
When I try this it says unhandled exception type Exception – Sairam Sankaran May 28 '13 at 22:35

score 0 · Answer 4 · answered May 28 '13 at 22:20

The only thing you want to make sure to do in the answers people post is you need to put the Exception1 must not be superclass of Exception2 otherwise does not compile. Java 7 answer below is the best way to handle it and order matters as I explained

For list of changes in Java 7

http://www.oracle.com/technetwork/java/javase/jdk7-relnotes-418459.html

java - Managing duplicate code in multiple catch blocks

4 Answers4