Arrays and if Statements: Array object not passable into if statement

Question

Please consider the following code:

public JButton math_button[] = new JButton[5];
    for (int h = 0; h <math_button.length; h++) {
        if(event.getSource()==math_button[h]) {
            String button_press = math_button[h].getText();     
            if(math_button[h].getText().equals("Equals")) {
                secondn = Integer.parseInt(math_input.getText());
                System.out.println(firstn + " math operator " + secondn + " and "+ math_button[h].getText());
                System.out.println(calc.Math(button_press, firstn, secondn));
            } else {
                firstn = Integer.parseInt(math_input.getText());
                //math_input.setText("");
                //placeholder = calc.Math(math_button[h].getText(), firstn, secondn);
                //int secondn = Integer.parseInt(math_input.getText());
                //int result = calc.Math(math_button[h].getText(), firstn, secondn);
                //math_input.setText(Integer.toString(firstn));
                //math_input.setText(Integer.toString(placeholder));
            }
        }
    }

What is the reason that despite the variable button_press being set to the name of an array object outside of the second IF (nested) loop, the test condition variable math_button[h].getText()is always passed to the calc.Math method?

Is the variable of the button_press string being overridden by the nested IF statement?

Use `String#equals` to compare Strings. `==` compares object references — Reimeus, Jul 28 '13 at 18:27
Luiggi, I see you point about the possible duplication but I never would have found, or even thought of, asking about string comparisons. — obious, Jul 28 '13 at 18:41
@obious - The reason it's a dupe here is that the (ever popular) SO question about comparing a `String` in Java is the actual problem / error, and any answer to this question will simply be the same answer. — Brian Roach, Jul 28 '13 at 18:58

Ravi K Thapliyal · Accepted Answer · 2013-07-28T19:43:55.310

1

Use String#equals() as

if(math_button[h].getText().equals("Equal")) {

Equals == operator only compares the references with each other; not the actual text content.

EDIT :

The reason equals() didn't work is because the button name is Equals (notice the s)

public String[] name = {"Add", "Mulitply", "Divide", "Subtract", "Equals"};

Hence, your if condition should actually be

if(math_button[h].getText().equals("Equals")) {

edited Jul 28 '13 at 19:43

answered Jul 28 '13 at 18:31

Ravi K Thapliyal

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While your answer is correct, the problem and solution are stated in the possible dup question. – Luiggi Mendoza Jul 28 '13 at 18:34
Thanks but I've tried your suggestion and it hasn't worked. – obious Jul 28 '13 at 19:31
@obious Check update. You are missing an `s`. – Ravi K Thapliyal Jul 28 '13 at 19:44
I noticed that, and changed it accordingly but it still doesn't work – obious Jul 28 '13 at 19:48
@obious Can you add a `println()` just before the `if` and check the actual string? – Ravi K Thapliyal Jul 28 '13 at 19:50
Just done that, 'Equals', without the quotations obviously, is returned. It's interesting, if I put an 'Equals' IF statement into the calc.Math method, the code within that segment is returned. For whatever reason, only the 'Equals'variable is being passed – obious Jul 28 '13 at 19:52
It should work then. Can you update your question with the current changes in your code? – Ravi K Thapliyal Jul 28 '13 at 19:54
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/34332/discussion-between-obious-and-ravi-thapliyal) – obious Jul 28 '13 at 19:55

Arrays and if Statements: Array object not passable into if statement

1 Answers1