The result of try catch finally in java

Question

I got a try catch finally java question. The code is like this:

package com.test;

public class TestExamples {

public int testFinally(int inputNum) {
    int returnNumber = inputNum;
    try {
        returnNumber++;
        return returnNumber;

    } finally {
        returnNumber++;

    }
}

public StringBuilder testFinallyString() {
    StringBuilder builder = new StringBuilder();
    try {
        builder.append("cool");
        return builder.append("try");

    } finally {
        builder.append("finally");
    }
}


public static void main(String[] args) {
    TestExamples testExamples = new TestExamples();
    System.out.println("The result of testFinally is " + testExamples.testFinally(5));
    System.out.println("The test of testFinallyString is " + testExamples.testFinallyString());

}

    }

The results:

The result of testFinally is 6
The test of testFinallyString is cooltryfinally

If finally is executed everytime, then why is testFinally 6? I am bit puzzled that finally code block doesnt result in incrementing the number that is returned. Pls can someone throw more light on what can be the underlying reason.

Reference types and primitives. – Sotirios Delimanolis Oct 14 '13 at 19:18 — Sotirios Delimanolis, Oct 14 '13 at 19:18

score 0 · Answer 1 · answered Oct 14 '13 at 19:18

0

I think it's because return returnNumber; is computed before the finally block. At that point, returnNumber is 6. Then the finally block is executed, and returnNumber is incremented to 7, but by that time it's too late--testFinally has already decided that it's going to return 6.

answered Oct 14 '13 at 19:18

ajb

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This doesn't answer the question, just outlines the observations (poorly, because no comparison is made with the string builder) – Joost Oct 14 '13 at 19:21

score 0 · Answer 2 · answered Oct 14 '13 at 19:19

0

If finally is executed everytime, then why is testFinally 6?

The increment is happening, but it's happening after you've returned a primitive value. What you've returned isn't an object.

answered Oct 14 '13 at 19:19

Jon Lin

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I don't think this is as much to do with primitives as it is to do with reassignment. The compound operator does `x = x + y`, so if you have returned `x` already then no-one will see the reassignment. The same would be true for an `Object`. – Boris the Spider Oct 14 '13 at 19:24

The result of try catch finally in java

2 Answers2