Why am I not getting the right answer

[Ian Lubelsky]

Hello to one and all.

I am starting the journey of learning java. I am following a book I have, that gives lab assignments at the end. The assignment I am having a problem with at the moment is to convert a Fahrenheit temperature to Celsius. When I run the following code, I get an answer of "0" for Celsius when the information is printed to screen. Can someone explain to me what is wrong.

    
class TempConverter
{
    public static void main(String[] args)
    {
        double Fahrenheit = Double.parseDouble(args[0]);
        System.out.println("The value of Fahrenheit you entered is " + Fahrenheit);
        double Celsius = (5/9) * (Fahrenheit - 32);
        System.out.println("Your temperature converted to Celsius is: " + Celsius);
    }
}

Thanks in advance for any help.

[Dave Trower]

The problem with your code is the 5/9 is resolving to 0.
This is because 5 and 9 are integers by default.
One way to fix the problem is to do what is known as a primitive cast to tell the computer 5 and 9 are doubles.
Change you line to this and will work as you expect

   double Celsius = (5D/9D) * (Fahrenheit - 32);     

[Ian Lubelsky]

Holy Java Batman, it worked.

Thanks for the info.

However, I thought by including "double" at the beginning of the line: "double Celsius = ......" would make the whole line a double.

Does this mean when I use "double" I have to add a "D" at the end like when I do a "Long" by adding an "L" at the end?
I tried looking up the answer in the book I'm using, but some of the examples I see when using double, have a "D" at the end and others don't. Am I better off just adding a "D" at the end regardless?

[W. Joe Smith]

Ian Lubelsky wrote:Holy Java Batman, it worked.

Thanks for the info.

However, I thought by including "double" at the beginning of the line: "double Celsius = ......" would make the whole line a double.

Does this mean when I use "double" I have to add a "D" at the end like when I do a "Long" by adding an "L" at the end?
I tried looking up the answer in the book I'm using, but some of the examples I see when using double, have a "D" at the end and others don't. Am I better off just adding a "D" at the end regardless?

No, saying "double Celsius = ..." only denotes that the variable Celsius is a double.

[Ian Lubelsky]

W. Joe Smith wrote:

Ian Lubelsky wrote:Holy Java Batman, it worked.

Thanks for the info.

However, I thought by including "double" at the beginning of the line: "double Celsius = ......" would make the whole line a double.

Does this mean when I use "double" I have to add a "D" at the end like when I do a "Long" by adding an "L" at the end?
I tried looking up the answer in the book I'm using, but some of the examples I see when using double, have a "D" at the end and others don't. Am I better off just adding a "D" at the end regardless?

No, saying "double Celsius = ..." only denotes that the variable Celsius is a double.

After re-reading the section about double, float, long, I now realize I was a double only to Celsius and not directly to the values I was calculating.

[Henry Wong]

Ian Lubelsky wrote:
I tried looking up the answer in the book I'm using, but some of the examples I see when using double, have a "D" at the end and others don't. Am I better off just adding a "D" at the end regardless?

You actually don't need "D" in many cases. Specifying a decimal will work too.... meaning 5.0 / 9.0 works. Also, if it is mixed, then it will uses the type with the most range.... meaning 5.0 / 9 will also work. Since you are dividing a double value of 5.0 with an integer of 9, it will know that the result is to be a double.

Henry

[Serap Elbeyoglu]

Ian you asked a good question. Answers are very clear to understand what is the problem in here. It is just about data type. Choosing int will solve your problem.

[Campbell Ritchie]

Serap Elbeyoglu wrote:Choosing int will solve your problem.

No it won't. That has already been demonstrated.

[Ian Lubelsky]

Henry Wong wrote:

Ian Lubelsky wrote:
I tried looking up the answer in the book I'm using, but some of the examples I see when using double, have a "D" at the end and others don't. Am I better off just adding a "D" at the end regardless?

You actually don't need "D" in many cases. Specifying a decimal will work too.... meaning 5.0 / 9.0 works. Also, if it is mixed, then it will uses the type with the most range.... meaning 5.0 / 9 will also work. Since you are dividing a double value of 5.0 with an integer of 9, it will know that the result is to be a double.

Henry

I understand now I can use a "D" at the end or a decimal, or I could just write ((double)5/9).
However for future reference, what is more widely accepted way of coding or is just up to the coder to decide?

[Jesper de Jong]

Ian Lubelsky wrote:However for future reference, what is more widely accepted way of coding or is just up to the coder to decide?

It's a question of style, I would have written 5.0 / 9.0 myself. I don't think there is one generally agreed upon way to write this.

[Rob Spoor]

I usually use .0 as well. I think you shouldn't cast literals; not only are 5D and 5.0 shorter than (double)5, it also shows you have better knowledge of the language. I only use a cast if both operands are non-double variables:

(double)x / y