Celsius is always 0.0?

Question

I'm making a simple temperature conversion program in Java that converts Fahrenheit to Celsius. My program compiles just fine, but no matter what number I input, it always says that Celsius is 0.0! What could I be doing wrong? Here is my full code:

import javax.swing.JOptionPane;

public class FahrenheitToCelsius {

public static void main(String[] args) {

    double fahrenheit, celsius;

    String input = JOptionPane.showInputDialog("Please enter the temperature in Fahrenheit:");

    fahrenheit = Double.parseDouble(input.trim());

    celsius = (5 / 9) * (fahrenheit - 32);

    JOptionPane.showMessageDialog(null, "The temperature is " + celsius + " degrees celsius.");

    }

}

What value does this produce: `(5 / 9)` ? Print out the result of that computation. — Hot Licks, Feb 07 '14 at 18:02

asaini007 · Accepted Answer · 2014-02-07T18:05:13.023

5

It's because (5 / 9) = 0.

5 and 9 are both ints, and int division here will result in 0 (5/9 = 0.555..., which can't be an int so it is truncated to 0). Use doubles (5.0 / 9.0) and you won't have this problem.

edited Feb 07 '14 at 18:05

answered Feb 07 '14 at 17:59

asaini007

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score 4 · Answer 2 · answered Feb 07 '14 at 18:00

4

In this line:

 celsius = (5 / 9) * (fahrenheit - 32);

5 and 9 are both integer constants, which means that the calculated value will be truncated to 0.

Try instead:

celsius = (5.0 / 9) * (fahrenheit - 32);

answered Feb 07 '14 at 18:00

stevevls

10,675
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score 3 · Answer 3 · answered Feb 07 '14 at 18:00

3

You are getting 0 cause 5 / 9 is always 0.

What you can do something like this

celsius = ((fahrenheit - 32.0) * 5.0) / 9.0;

answered Feb 07 '14 at 18:00

Anirban Nag 'tintinmj'

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score 1 · Answer 4 · answered Feb 07 '14 at 18:24

This will work:

celsius = ((fahrenheit - 32) * 5) / 9;

The reason this works has to do with how the operands are grouped. In fahrenheit - 32, fahrenheit is a double and 32 is an int, so the 32 gets promoted to a double, and the result is a double. Then, in:

(fahrenheit - 32) * 5

(fahrenheit - 32) is a double (see above), and 5 is an int, so the 5 gets promoted to a double and the resulting multiplication is a double. Again:

((fahrenheit - 32) * 5) / 9

Now (fahrenheit - 32) * 5 is a double, as we saw above, and 9 is an int, so the 9 gets promoted, and we end up with double-precision floating-point division.

By contrast:

celsius = (fahrenheit - 32) * (5 / 9);

The operands of /, 5 and 9, are both int. So there's no promotion, and the program performs integer division, with an int result. The result will therefore be 0. Now, since fahrenheit - 32 is a double, the 0 will be promoted to a double--but of course it's too late.

The point is that the Java compiler is not a mind-reader; even though the two expressions look the same mathematically, the compiler has a very specific method in which it does things, and programmers must be aware of that.

Thanks, I already fixed it. – SoloMael Feb 07 '14 at 18:44 — SoloMael, Feb 07 '14 at 18:44

Celsius is always 0.0?

4 Answers4