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Start your free trialFabiyan Adell
365 Pointsviable input code?
In the video, craig put those curly brackets { after if (noun.equalsIgnoreCase("dork")) with } ending System.exit(0); My output worked just fine without those curly brackets. Do I still need to put those in the code and why did mine still work without them? Thanks in advance
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!
*/
// __Name__ is a __adjective__ __noun__. They are always__adverb __ __verb.
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 adjective = console.readLine("Enter an adjective: "); String noun = console.readLine("Enter an noun: "); if (noun.equalsIgnoreCase("dork")) console.printf("That langauge is not allowed. Exiting. \n\n"); System.exit(0); String adverb = console.readLine("Enter an adverb: "); String verb = console.readLine("Enter an verb: ");
console.printf("Your TreeStory:\n-------------------\n"); console.printf("%s is a %s %s. ", name, adjective, noun); console.printf("They are always %s %s. \n", adverb, verb);
1 Answer
Steven Parker
231,269 PointsThere should be a difference in performance. Without the braces, the "if" would only apply to the printf statement after it, which would cause the program to always exit at that point — even when you enter a different word.