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1,539 Pointsint randomNumber = randomGenerator.IntAsString + ""; Is this conversion of a random number integer to a String value?
Generating Random Numbers. The first three are correct, the last one is posing a challenge.
Random randomGenerator = new Random(); int randomNumber = randomGenerator.nextInt (); int RandomNumber = randomGenerato.nextInt (10); int randomNumber = randomGeneratot.IntAsString + '';
Random randomGenerator = new Random();
int randomNumber = randomGenerator.nextInt ();
int RandomNumber = randomGenerator.nextInt (10);
int randomNumber = randomGenerator.intAsString + "";
1 Answer
Steve Hunter
57,712 PointsHi there,
Your last line isn't quite right. It asks you to:
Declare a String variable named intAsString. Convert the randomNumber integer to a String value and store it in the intAsString variable.
So, you need to start with:
String intAsString =
That creates the variable called intAsString
which is a String. You've tried to use it as a method of the Random
class which it isn't.
What to assign to the String? It asks for the randomNumber
to be "mooshed" so it converts to a String. The final code looks like:
Random randomGenerator = new Random();
int randomNumber = randomGenerator.nextInt();
randomNumber = randomGenerator.nextInt(10);
String intAsString = randomNumber + "";
Hope that helps.
Steve.