Member Avatar for Phinocio

My code

public class BattleSystem {

    public static void main (String[] args) {
        String choice;
        
        System.out.print("What would you like to do? ");
    	Scanner input = new Scanner(System.in);
    	choice = input.nextLine();
    	System.out.println(choice);
    	
    	if(choice == "1"){
    		createChar g = new createChar();
    		g.charCreateHandler();
    		g.charStatsHandler();
    		g.closeFile();
    	}
    	else{
    		System.out.println("More choices coming soon!");
    	}
    }
}

Console

What would you like to do? 1
1
More choices coming soon!

It should not run my else statement. Help?

  • 3 Contributors
  • 3 Replies
  • 125 Views
  • 6 Hours Discussion Span
  • Latest Post Latest Post by Mattox

Dani AI

Generated

As correctly pointed out, the observed behavior comes from comparing string references instead of their contents; 's char approach works only because it compares primitive values. A few practical, defensive measures make menu-style input more reliable and easier to debug than relying on reference-equality surprises.

Trim and validate the raw input, then either compare the cleaned string or convert it to a numeric option and switch on that. Parsing to an integer and handling bad input keeps the branching clear and avoids subtle whitespace/case issues:

String raw = input.nextLine();
String s = raw.trim();
try {
    int option = Integer.parseInt(s);
    switch (option) {
        case 1:
            // create character
            break;
        default:
            System.out.println("More choices coming soon!");
    }
} catch (NumberFormatException e) {
    System.out.println("Please enter a valid number.");
}

Extra tips: prefer comparing a literal to the variable if you use content-equality to avoid null pointer problems (compare the literal first), consider equalsIgnoreCase for case-insensitive commands, and use next() instead of nextLine() if you want the first token only. If a comparison still fails, print s.length() and each character code to reveal hidden whitespace or control characters. For reference, see the Java String documentation and Scanner behavior: String.equals Javadoc and Scanner.nextLine().

Recommended Answers

All 3 Replies

Member Avatar for Mattox

Strings are not primitive types, so you can't use ==.
you have to use the .equals("String")
Ex. put this in the parenthesis of the if statement

if(choice.equals("1"))

The == operator will only work with primitive types, since a String is an object, you have to use a method to see if it equals another object (i.e. object1.equal(object2))

Member Avatar for burgercho

Not really an answer but if you use

if(choice.charAt(0)=='1')

it works, no idea why the regular comparison won't work though

Member Avatar for Mattox

choice.charAt(0) returns a char, which is a primitive type. The single quotes around 1 means that it is also a primitive type. So the == operator will work.

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