Thread: Taking a square root in Java, answer appears as 0.0. Why?

OK, so I'm trying to make a program where you input a number and the program takes its square root. Yes, I realize I could do all this on a main but where is the fun in that? I know I'm close, but I'm just not seeing my error! Result shows no errors on either program, but whatever number I input the answer is 0.0
SUB

public class FunMath {
	public FunMath() {
	double number = 0;
	}
	public void enterNumber(double amount){
		factor = number + amount;
	}
	public double giveResult(){
		double result = Math.sqrt(factor);
	return result;
	}
	private double factor;
	private double number;
	}

Main

import java.util.*;
public class FunMathTester {
	public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	FunMath squareroot = new FunMath();
	System.out.print("Enter the number you wish to take a square root of: ");
		double number = in.nextDouble();
		System.out.println(squareroot.giveResult());
	}
}

Originally Posted by helloworld922

You never set number inside FunMath, but rather set a local variable which happens to have the same name as number inside your main method.

Could you elaborate? (This is my own program, not an assigned one)

Thank you for the assistance. I haven't had time to work on my programming recently, but after taking a look at your advice, I finally found my error!

Behold, the working program of a noobie!

 public class FunMath {
	public FunMath() {
	factor = 0;
	}
	public void setNumber(double initialNumber){
		factor = initialNumber;
	}
	public double getResult(){
		double newFactor = Math.sqrt(factor);
		factor = newFactor;
	return factor;
	}
	private double factor;
	}

Main

import java.util.*;
public class FunMathTester {
	public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	FunMath factor = new FunMath();
	System.out.print("Enter the number you wish to take a square root of: ");
		double initialNumber = in.nextDouble();
		factor.setNumber(initialNumber);
		System.out.println(factor.getResult());
	}
 
	}

Please criticize me if there is anything too criticize. I'd like to make more complex math programs going forward as efficient and easy to understand as possible!

Well, though how you have it is perfectly fine, at least that I'm aware of, you could possibly pass a double parameter called number to your class.

 public class FunMath {
	public FunMath(double initialNumber) {
	setNumber(initialNumber);
	}
	public void setNumber(double initialNumber){
		factor = initialNumber;
	}
	public double getResult(){
		double newFactor = Math.sqrt(factor);
		factor = newFactor;
	return factor;
	}
	private double factor;
	}

import java.util.*;
public class FunMathTester {
	public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	System.out.print("Enter the number you wish to take a square root of: ");
		double initialNumber = in.nextDouble();
		FunMath factor = new FunMath(initialNumber);
		System.out.println(factor.getResult());
	}
 
	}

you can try :

class SquareRoot
{
public static void findSqrt(double n)
{
double a=Math.sqrt(n);
System.out.print("Square Root="+a);
}
}
as a simple code..