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Android Build a Simple Android App (retired 2014) Learning the Language Objects and Random Numbers

Posted by David Esquivel
David Esquivel
977 Points

to.String

this is for the answer = Integer.toString(randomGenerator); eclipse shows an error for .toString Please help me out

David Esquivel
David Esquivel
977 Points

This is what I have:

public void onClick(View v) {
            //the button was clicked, so update the answer label with an answer
            String answer = "";

            // Randomly select one of three answers, yes, no , or maybe
            Random randomGenerator = new Random();
            int randomNumber = randomGenerator.nextInt(3);
            answer = Integer.toString(randomNumber);     <- Error
            // Update the label with our dynamic answer
            answerLabel.setText(answer);

This is the Error text: The method toString(int) in the type integer is not applicable for the arguments (Random)

2 Answers

Iskandar Fendi
Iskandar Fendi
5,700 Points
Iskandar Fendi
Iskandar Fendi
5,700 Points

Hi David,

What does the error said? I suspect: What is your randomGenerator type? Is it Random class? If it is it will give you an error. .

David Esquivel
David Esquivel
977 Points

It is Random class I think

Iskandar Fendi
Iskandar Fendi
5,700 Points
Iskandar Fendi
Iskandar Fendi
5,700 Points

Yes, it is! The error is when you put:

answer = Integer.toString(randomGenerator). In order to fix it, change it into:

answer = Integer.toString(randomNumber);

It will fix the problem.

In your Eclipse, it should tell you why (I assume "Type Don't Match").The reason for your error is:

You put in a Random Class Type instead of Integer.

In programming, you can look up toString() method in Java API

It says: static String toString(int i)

It takes integer type inside of it.

Please let me know the updates

David Esquivel
David Esquivel
977 Points

I changed it but the error is still the same. I have the error text in the comments on my question I see what you're saying though. It should work, but it won't

David Esquivel
David Esquivel
977 Points

It's working now! Thanks!

Victor Rodriguez
Victor Rodriguez
15,015 Points
Victor Rodriguez
Victor Rodriguez
15,015 Points

are you trying to take a random Number that you generated and save it as a string? if so use:

answer = randomGenerator.toString();

The toString() method should not be taking any parameters, unless you overrode the method. It is normally used: variable.toString()