For some reason, I can't seem to let this one go. I've read the comments on this blog (thanks Dan for all the information). I've talked with a few other people.

I have verified that Java and C++ represents the numbers 1.03, 0.42, and the result of 1.03 - 0.42 the same way by using

• java.lang.Double.doubleToLongBits(double)
• java.lang.Long.toHexString(long)

in Java and

• double d, *p = &d;
• long long l = *(reinterpret_cast<long long*>(p))

in C++.

So the differences are in the way doubles are printed.

Jeff Brown mentioned to me at the St. Louis JUG meeting Thursday that there might be other floating point numbers that Java prints right and other languages print wrong.

But a little experiment (a very superficial one, I admit) indicates that Java is wrong more times than C++:

[weiqi@gao] $ cat Foo.java
public class Foo {
  public static void main(String[] args) {
    double ds[] = {0.1, 0.2, 0.3, 0.4};
    for (int i = 0; i < 4; i++) {
      for (int j = i + 1; j < 4; j++) {
        System.out.println("" + ds[j] + "-" + ds[i] +
          "=" + (ds[j] - ds[i]));
      }
    }
  }
}
[weiqi@gao] $ javac Foo.java
[weiqi@gao] $ java Foo
0.2-0.1=0.1
0.3-0.1=0.19999999999999998
0.4-0.1=0.30000000000000004
0.3-0.2=0.09999999999999998
0.4-0.2=0.2
0.4-0.3=0.10000000000000003
[weiqi@gao] $ cat foo.cc
#include <iostream>
int main() {
  double ds[] = {0.1, 0.2, 0.3, 0.4};
  for (int i = 0; i < 4; i++) {
    for (int j = i + 1; j < 4; j++) {
      std::cout << ds[j] << "-" << ds[i] << "="
        << (ds[j] - ds[i]) << std::endl;
    }
  }
}
[weiqi@gao] $ make foo
g++     foo.cc   -o foo
[weiqi@gao] $ ./foo
0.2-0.1=0.1
0.3-0.1=0.2
0.4-0.1=0.3
0.3-0.2=0.1
0.4-0.2=0.2
0.4-0.3=0.1

Then Jeff Grigg interjected with the sarcastic "Java is right and everything else is wrong."

Somehow, I don't feel so confident.

The Javadoc for java.lang.Double.toString(double) details how Java prints a double. It specifically mentioned this:

How many digits must be printed for the fractional part of m or a? There must be at least one digit to represent the fractional part, and beyond that as many, but only as many, more digits as are needed to uniquely distinguish the argument value from adjacent values of type double.

This might be the root of the problem. What if two adjacent doubles have decimal representations that differ only in insignificant digits? By the above rule, insignificant digits will be printed just so that the two numbers can be distinguished.

Does such pairs of adjacent doubles exist? I don't know. My suspicion is that they do exist, and 1.03 - .42 belongs to such a pair. Of course, it could also be something more complicated that I don't understand.

Responses[7]

Posted by Weiqi Gao on July 10, 2004 10:40:11 AM CDT #