Java Java Basics Perfecting the Prototype Parsing Integers

Manisha Gopala
PRO
Manisha Gopala
Pro Student 203 Points

Parsing Integers

Though the integer does not satisfy the question it still gives the same output. The code is below: import java.io.Console;

public class TreeStory {

public static void main(String[] args) {
    Console console = System.console();
    /*  Some terms:
        noun - Person, place or thing
        verb - An action
        adjective - A description used to modify or describe a noun
        Enter your amazing code here!
    */
 // --Happy Birthday ---- -----. God --- you and I wish you all the -------- in life"
  String ageAsString = console.readLine("How old are you?: ");
  int age = Integer.parseInt(ageAsString);
  if (age < 13); {
    // Insert exit code
    console.printf("Sorry you must be at least 13 to use this program \n");
    System.exit(0);  
  }
  String name = console.readLine("Enter a name:  ");
  String wish = console.readLine("Enter the wish:  ");
  String emotion = console.readLine("Enter an emotion:  ");

console.printf("Your TreeStory:\n----------\n");
console.printf("Happy Birthday %s  ", name);

console.printf("God %s you and I wish you all the %s in life /n", wish , emotion);

1 Answer

Calin Bogdan
Calin Bogdan
14,615 Points

Hi!

You've put a semicolon after the if statement that shouldn't be there.

if (age < 13); {  // <------ this one
    // Insert exit code
    console.printf("Sorry you must be at least 13 to use this program \n");
    System.exit(0);  
  }

Try changing that and see if it works.