Android simple calculator application

Question

I'm new to android. I'm developing a simple calculator application. There is some problem in my code. There is no syntax error. When I run it, it gives some error. Written all code in a try catch but I'm still unable to catch it.

       b1.setOnClickListener(new View.OnClickListener() {

        public void onClick(View v) {
            //char x = lbldisp.getText().toString().charAt(1);

        try
        {
            String x = lbldisp.getText().toString();
            if(x.charAt(0)=='0' && x.charAt(1)!='.')
            {
                    lbldisp.setText("1");
            }
            else
            {
                lbldisp.setText(x+"1");
            }

        }
        catch(Exception e)
        {
            Toast.makeText(Calculator.this,e.getMessage().toString(), Toast.LENGTH_SHORT).show();
        }
        }
    });

try to look into the LogCat window. You can filter the messages by clicking on the type of message you want to see — rizzz86, May 11 '12 at 14:07
Actually I would like to know were the simplest place would be to start developing andriod software. I use Java and eclipse everyday which would make things very easy to start. On my old computer I downloaded a plug-in but it would seemed to work all that good for eclipse. — Doug Hauf, Dec 27 '13 at 18:31

score 0 · Answer 1 · answered May 11 '12 at 12:55

0

Try to replace this string

 if(x.charAt(0)=='0' && x.charAt(1)!='.')

By this String

 if(x.charAt(0).equals("0") && x.charAt(1).equals("."))

You have Converted the value to String so try to use .equals Method.

answered May 11 '12 at 12:55

Bhavin

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there is no method like charAt(0).equals("0") – piyush May 11 '12 at 13:02
in errorlog it just showing like : thread exiting with uncaught exception (group=0x4001d800) – piyush May 11 '12 at 13:03
You can try this then ` if(x.subSequence(0, 1).equals("0") && x.subSequence(1, 2).equals(".")) ` – Bhavin May 11 '12 at 13:06
i m completely fresher for android... can u give me a step for doing that... sry.. – piyush May 11 '12 at 13:07
Just Check this site. http://stackoverflow.com/questions/1223052/how-do-i-find-out-if-first-character-of-a-string-is-a-number – Bhavin May 11 '12 at 13:13
actuly when i remove second condition from if then its work perfectly... bt when i add it.. its giving error.. – piyush May 11 '12 at 13:18

Android simple calculator application

1 Answers1