Question about an output using the (int) statement
Folks,
I am new to java and ran a simple problem that takes a user input as a double, multiplies it by 100 and then applies (int) to that double and prints the output. See code below. I ran it with a handful of number and received the expected output except 19.99. It returns 1998.0, not the expected 1999.0.
I understand precision maybe rearing its ugly head here, I just wanted to know if anyone can shed any light on why it happens with 19.99?
PS im using eclipse with JRE 1.6.
Thank you!
import java.util.Scanner;
publicclass IntegerProblemClass{
publicstaticvoid main(String[] args){
System.out.println("Choose any number");
Scanner userInput =new Scanner(System.in);
double userNum = userInput.nextDouble();
userNum = (int)(userNum * 100);
System.out.println("Your new number is: " + userNum);
}
}
[1351 byte] By [
jmmorina] at [2007-11-26 17:22:09]
If you wish to round, you can use [url=http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Math.html#round(double)]Math.round()[/url]:
public class Test {
public static void main(String[] args) {
double input = 19.99;
long rounded = Math.round(input * 100);
System.out.println(rounded);
}
}
And if it's just a matter of display, you can always use an appropriate formatter. Example:
double d = 19.99 * 100;
System.out.println(d);// 1998.9999999999998
System.out.printf("%1.0f\n", d); // 1999
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> Thanks for the reply. Can you shed anylight on why
> 19.99 * 100 = 19.9999999999998?
[url=http://en.wikipedia.org/wiki/Floating_point]Floating point - From Wikipedia, the free encyclopedia[/url]
[url=http://java.sun.com/developer/JDCTechTips/2003/tt0204.html#2]Some things you should know about floating-point arithmetic[/url]
[url=http://docs.sun.com/source/806-3568/ncg_goldberg.html]What Every Computer Scientist Should Know About Floating-Point Arithmetic[/url]
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