How to throw again IOException in java?

Question

Code:

 catch (IOException e) {
        LOGGER.error("IOException exception happened");
        //now need throw again the same exception to be 
                       //catched in    the    upper method
    }

But when I try simply:

  catch (IOException e) {
        LOGGER.error("IOException exception happened");
        //now need throw again the same exception to be 
                       //catched in    the    upper method
               throw e;
    }

Eclipse supposes to me put "throw e" in try catch block. But this is nonsense. How fix this problem? Thanks.

I need catch - because need first make specific in bottom method, and after common in upper method — user710818, Jun 26 '12 at 14:51

score 7 · Accepted Answer · answered Jun 26 '12 at 14:46

Since IOException is a checked exception, that method needs to be declared as throwing IOException if you want it to propagate. For example:

void myMethod() throws IOException {
    try {
        //code
    }
    catch(IOException e) {
        LOGGER.error("IOException exception happened");
        throw e;
    }
}

score 5 · Answer 2 · edited Jun 26 '12 at 14:48

5

The second code fragment is just fine. Just remember that You have to declare Your method as:

public void myMethod() throws IOException {
    ...
}

edited Jun 26 '12 at 14:48

Sergey Kalinichenko

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answered Jun 26 '12 at 14:47

Tomasz Jędzierowski

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19

score 1 · Answer 3 · answered Jun 26 '12 at 14:48

1

Try adding throws IOException to your method, e.g.:

private void yourMethodName() throws IOException {
    # your method
}

Then Eclipse won't ask for the second try catch block.

answered Jun 26 '12 at 14:48

carmenism

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How to throw again IOException in java?

3 Answers3

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