Temperature Conversion Method Outputting Java

Question

I have to write a class called temperature and my only problem is when i try to call the conversion method. Whenever I run the main program, the conversion output is 0.0

Does it have something to do with using a private variable?

Code is below.

Many thanks!

public class Temperature {

private double temp;
private String scale;

public Temperature(double temp) {
    this.temp = temp;
}

public String toString () {
    return "your temp is " + this.temp;
}

public double getTemp() {
    return this.temp;
}

public String getScale(String scale) {
    this.scale = scale;
    return this.scale;
}

public double conversion() {
    double temp = 0;
    if (this.scale == "fahrenheit") {
        temp = (this.temp - 32) * (double)(5/9);
    }
    else if (this.scale == "celsius") {
        temp = (this.temp * (double)(9/5)) + 32;
    }
    return temp;
}

public boolean aboveBoiling(double temp) {
    if (this.scale == "fahrenheit") {
        return temp >= 212;
    }
    else if (this.scale == "celsius") {
        return temp >= 100;
    }
    return true;

}

}

below is my test driver to test function calls

import java.util.*;
public class testdriver {

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    Temperature userTemp;
    String userScale;
    //userTemp = new Temperature(1.0);
    System.out.print("Please enter your temp scale(fahrenheit or celius): ");
    userScale = sc.next();
    System.out.println();

    System.out.print("Please enter your temp in " + userScale + ": ");
    userTemp = new Temperature(sc.nextDouble());
    System.out.println();

    System.out.println(userTemp.toString());
    System.out.println();

    System.out.println("your temp is: " + userTemp.getTemp());
    System.out.println();

    System.out.println("your scale is: " + userTemp.getScale(userScale));
    System.out.println();

    System.out.println(userTemp.conversion());
    System.out.println();

    double userTemp2 = userTemp.conversion();
    System.out.println(userTemp.aboveBoiling(userTemp2));







}

}

Emerson Dallagnol · Accepted Answer · 2017-02-13T19:00:30.373

4

Replace this.scale == "fahrenheit" and this.scale == "celcius" by this.scale.equals("fahrenheit") and this.scale.equals("celcius"). In java, the == operator compare only references or primitive values (int, long, float, double...).

Edit: I found another problem in your code:

 (this.temp - 32) * (double)(5/9);

5 and 9 are integers, then the division will result in 0. You need to cast one of the operands to double, before the / operation:

 (this.temp - 32) * ((double)5/9);

or:

(this.temp - 32) * (5./9.);

edited Feb 13 '17 at 19:00

answered Feb 13 '17 at 18:10

Emerson Dallagnol

1,269
1
12
21

1

agree, and better will be enum – Jacek Cz Feb 13 '17 at 18:13
Agreed. Please see http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java. – rajah9 Feb 13 '17 at 18:16
im still getting an output of 0.0 :/ – Feb 13 '17 at 18:33
@Roach can you post the code that call the conversion function? – Emerson Dallagnol Feb 13 '17 at 18:38
@Roach I edited my answer. – Emerson Dallagnol Feb 13 '17 at 19:01
sorry it's taken me so long to get back but it still does not work :((((( i added my function calls below the class – Feb 20 '17 at 16:28
@EmersonDallagnol nevermind i got it working! thank you very much for your help! – Feb 20 '17 at 16:38

shashwatZing · Answer 2 · 2017-02-13T18:27:36.477

1

I see two problems with your code.

1) String equality : use equals and not == for String equality in java. They are completely different in java and is used only for primitive data type equality.

2) you need to create a public method to set scale. You are not setting value of scale anywhere

public void setScale (String scale){ 
this.scale=scale;
}

and your getter should be changed to

public String getScale() {
    return scale;
}

edited Feb 13 '17 at 18:27

answered Feb 13 '17 at 18:15

shashwatZing

1,550
1
17
24

Thank you very much!! This helped a ton! – Feb 20 '17 at 16:37

score 0 · Answer 3 · answered Feb 13 '17 at 18:11

0

Isnt your problem the initialization of the variable? when you use this.temp you are acessing the "zero" in the conversion method not the external value. Actualy You dont need to use THIS in this case.

Sorry if I m mistaken

answered Feb 13 '17 at 18:11

Breno Baiardi

99
1
16

also... I think that line with temp=0 isnt doing much, right? – Breno Baiardi Feb 13 '17 at 18:14

Temperature Conversion Method Outputting Java

3 Answers3