How to propagate an exception in java

Question

I am a C programmer and just learning some java recently because I am developing one android application. Currently I am in a situation. Following is the one.

public Class ClassA{

public ClassA();

public void MyMethod(){

   try{
   //Some code here which can throw exceptions
   }
   catch(ExceptionType1 Excp1){
   //Here I want to show one alert Dialog box for the exception occured for the user.
   //but I am not able to show dialog in this context. So I want to propagate it
   //to the caller of this method.
   }
   catch(ExceptionType2 Excp2){
   //Here I want to show one alert Dialog box for the exception occured for the user.
   //but I am not able to show dialog in this context. So I want to propagate it
   //to the  caller of this method.
   }
   }
}

Now I wan to use call the method MyMethod() somewhere else in another class. If some one can provide me some code snippet how to propagate the exceptions to the caller of MyMethod() so that I can display them in a dialog box in the caller method.

Sorry If I am not so much clear and weird in the way of asking this question.

See also http://stackoverflow.com/questions/409563/best-practices-for-exception-management-in-java-or-c-sharp — mglauche, Feb 29 '12 at 12:34

score 27 · Accepted Answer · answered Feb 29 '12 at 12:33

Just don't catch the exception in the first place, and change your method declaration so that it can propagate them:

public void myMethod() throws ExceptionType1, ExceptionType2 {
    // Some code here which can throw exceptions
}

If you need to take some action and then propagate, you can rethrow it:

public void myMethod() throws ExceptionType1, ExceptionType2 {
    try {
        // Some code here which can throw exceptions
    } catch (ExceptionType1 e) {
        log(e);
        throw e;
    }
}

Here ExceptionType2 isn't caught at all - it'll just propagate up automatically. ExceptionType1 is caught, logged, and then rethrown.

It's not a good idea to have catch blocks which just rethrow an exception - unless there's some subtle reason (e.g. to prevent a more general catch block from handling it) you should normally just remove the catch block instead.

Don't "Log AND throw". Only do 1 or the other please. It's redundant. — djangofan, Jul 12 '22 at 19:11

score 3 · Answer 2 · answered Feb 29 '12 at 12:35

Don't catch it and rethrow again. Just do this and catch it in the place you want

public void myMethod() throws ExceptionType1, ExceptionType2 {
    // other code
}

Example

public void someMethod() {
    try {
        myMethod();
    } catch (ExceptionType1 ex) {
        // show your dialog
    } catch (ExceptionType2 ex) {
        // show your dialog
    }
}

score 0 · Answer 3 · edited May 12 '13 at 11:22

0

I always do it like this :

public void MyMethod() throws Exception
{
    //code here
    if(something is wrong)
        throw new Exception("Something wrong");
}

then when you call the function

try{
    MyMethod();
   }catch(Exception e){
    System.out.println(e.getMessage());
}

edited May 12 '13 at 11:22

laalto

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answered May 12 '13 at 10:50

MohWaylie

1

score 0 · Answer 4 · answered Feb 29 '12 at 12:31

Just rethrow the exception

throw Excp1;

You will need to add the exception type to the MyMthod() declaration like this

public void MyMethod() throws ExceptionType1, ExceptionType2  {

   try{
   //Some code here which can throw exceptions
   }
   catch(ExceptionType1 Excp1){
       throw Excp1;
   }
   catch(ExceptionType2 Excp2){
       throw Excp2;
   }
}

Or just omit the try at all since you are no longer handling the exceptions, unless you put some extra code in the catch statements doing things with the exception before rethrowing it.

How to propagate an exception in java

4 Answers4

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