A noobs question java

[Suppersupport]

this should run the if statement right?

package turncard;

import java.util.*;

class person {// the class that sets a person
	String name;// the persons name
	int age;// that persons age
	int amount;// that persons turncard amount

	void add(int addamount) {
		amount += addamount;// add addamount to amount
	}

	void set(int setamount) {
		amount = setamount;// set amount to setamount
	}
}

public class Main {
	public static void main(String[] args) {
		person EJ = new person();
		EJ.name = "EJ";
		System.out.println("what should I do?");
		EJ.age = 14;
		EJ.amount = 4;
		Scanner input = new Scanner(System.in);
		String c = input.nextLine();
		System.out.println(c);
		if(c == "add") {
			System.out.println("how much should I add?");
		}
	}

}

 

[Mira Yurizaki]

In Java, assuming the class didn't override it, the == operator on two variables objects checks if their references are the same. That is, if they're pointing to the same object.

 

So no, that won't work. Use the String.equals() method to check if the contents are the same. See https://www.geeksforgeeks.org/difference-equals-method-java/

Edited July 30, 2019 by Mira Yurizaki

[Suppersupport]

2 minutes ago, Mira Yurizaki said:

In Java, assuming the class didn't override it, the == operator on two variables checks if their references are the same. That is, if they're pointing to the same object.

 

So no, that won't work. Use the String.equals() method to check if the contents are the same. See https://www.geeksforgeeks.org/difference-equals-method-java/

 

EDIT: The above doesn't apply to primitive data types, only to objects.

THANK YOU! I would never have found that out.

[zhick]

On 7/30/2019 at 5:49 PM, Mira Yurizaki said:

In Java, assuming the class didn't override it, the == operator on two variables objects checks if their references are the same.

Minor nitpick: Java has no operator overloading.

[AbydosOne]

On 7/30/2019 at 10:43 AM, will4623 said:

person EJ = new person();

Other minor quibble: class names always start capitalized (not primitives like int, double, etc., though); variable names always start lowercase.

[Suppersupport]

1 minute ago, AbydosOne said:

Other minor quibble: class names always start capitalized (not primitives like int, double, etc., though); variable names always start lowercase.

I know but I'm lazy.

[Mira Yurizaki]

2 hours ago, will4623 said:

I know but I'm lazy.

I'm under the belief if the compiler/interpreter/whatever doesn't strictly enforce style, then the style is left up to the coder.

 

As long as the style is consistent, it doesn't matter what you do. I might hate you for doing something though.

[Suppersupport]

package turncard;

import java.util.*;

class Person {// the class that sets a person
	String name;// the persons name
	int age;// that persons age
	int amount;// that persons turncard amount

	void add(int addamount) {
		amount += addamount;// add addamount to amount
	}

	void set(int setamount) {
		amount = setamount;// set amount to setamount
	}
}

public class Main {
	public static void main(String[] args) {
		Person EJ = new Person();
		EJ.name = "EJ";
		System.out.println("what should I do?");
		EJ.age = 14;
		EJ.amount = 4;
		String c = "";
		String errors = "";
		while (c.contentEquals("stop") != true) {
			Scanner input = new Scanner(System.in);
			c = input.next();
			if (c.contentEquals("add")) {
				System.out.println("how much should I add?");
				input = new Scanner(System.in);

				int addme = input.nextInt();
				EJ.add(addme);

			} else if (c.contentEquals("set")) {
				System.out.println("what should I set it to?");
				int setme = input.nextInt();
				EJ.set(setme);

			} else if (c.contentEquals("tell")) {
				System.out.println(EJ.amount);
			} else if (c.contentEquals("stop")) {

				System.out.println("terminating");

			} else if (c.contentEquals("errors")) {
				System.out.println(errors);
			} else {
				System.out.println("I am incapable of that at this time. Sorry!");
				errors += c + "	";

			}

		}

	}
}

fine you convensed me