Try-catch-finally order of execution appears to be random

Question

I am trying to understand how the try-catch-finally execution flow works. There are a couple of solutions from Stack Overflow users regarding their execution flow.

One such example is:

try {
    // ... some code: A
} 
catch(...) {
    // ... exception code: B
} 
finally {
    // finally code: C
}
Code A is going to be executed. If all goes well (i.e. no exceptions get thrown while A is executing), it is going to go to finally, so code C is going to be executed. If an exception is thrown while A is executed, then it will go to B and then finally to C.

However, I got different execution flows when I tried it:

try {
    int a=4;
    int b=0;
    int c=a/b;
}
catch (Exception ex)
{
    ex.printStackTrace();
}
finally {
    System.out.println("common");
}

I am getting two different outputs:

First output:

java.lang.ArithmeticException: / by zero
at substrings.main(substrings.java:15)
lication.AppMain.main(AppMain.java:140)
common

However, when I ran the same program for the second time:

Second output:

 common
    java.lang.ArithmeticException: / by zero
    at substrings.main(substrings.java:15)

What should I conclude from this? Is it going to be random?

What ever the result, but it should not Random. Cross check. — Suresh Atta, Oct 17 '15 at 17:58
different System streams produces improperly formatted output — Kumar Abhinav, Oct 17 '15 at 18:07

score 41 · Accepted Answer · answered Oct 17 '15 at 18:00

printStackTrace() outputs to standard error. System.out.println("common") outputs to standard output. Both of them are routed to the same console, but the order in which they appear on that console is not necessarily the order in which they were executed.

If you write to the same stream in both catch block and finally block (for example, try System.err.println("common")), you'll see that the catch block is always executed before the finally block when an exception is caught.

Kumar Abhinav · Answer 2 · 2015-10-17T18:22:26.570

Exception's printstacktrace() method source code(defined in Throwable class)

public void printStackTrace() {
        printStackTrace(System.err);
    }

The formatting of output occurs because your are using standard output stream through System.out.printlnand exception occurs System.err stream

Try having a variable which will check exceptions and print in same error console if exception occurs :-

boolean isException = false;
try {
        int a=4;
        int b=0;
        int c=a/b;
    }
    catch (Exception ex)
    {
       isException = true
       ex.printStackTrace();
    }
    finally {
       if(isException){
        System.err.println("common");
       }else{
        System.out.println("common");
       }
  }

Try-catch-finally order of execution appears to be random

2 Answers2

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