Methods and parameters - What is the output?

Question

1.00/5 (1 vote)

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Output for the below code is:
e
0

1. I kind know the output 'e' is the 2nd character of 'Hello', but which 'System.out.println' executed that output?

2. I don't understand how could it get the 2nd output '0'? The sequence of the output seems not right.

C#

public class MyProgram
{
    public void start()
    {
        int x = 0;
        String welcome = "Hello";
        myMethod(x, welcome);

        System.out.println(x);
    }

    private void myMethod(int x, String s)
    {
        x = x + 1;

        System.out.println(s.charAt(x));

    }
}

Posted 7-Jul-14 11:27am

Member 10925581

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Paulo Zemek · Answer 1 · 2014-07-07T11:40:00

The line x = x + 1; is changing the value of a local variable x of the method myMethod, which is not changing the value of x declared in the method start().
That's why the System.out.println(x) is returning 0.

So, what's happening is this:
myMethos(x, welcome) is incrementing a local variable, which is used to read the character at index 1 of the "Hello", so, the e character.

Then, after the my method, the x from start (which is still 0) is printed.