Implicit conversion/promotion of float or double in expressions between C# and Java

Question

If I have the following expression:

byte A = 69;
int B = 123;
long C = 3210;
float D = 4.9f;
double E = 11.11;

double X = (B * 100) + 338.1 - (E / B) / C;
double X1 = (B * 100) + (A * D) - (E / B) / C;

// JAVA - lost precision
System.out.println(X);  // 12638.099971861307
System.out.println(X1); // 12638.099581236307

// C# - almost the same
Console.WriteLine(X);  // 12638.0999718613
Console.WriteLine(X1)  // 12638.0999784417

I noticed that Java loses precision from X where 338.1 is implicit double while C# almost doesn't. I don't understand why because 338.1 is equals in float and in double. There is only one digit after the dot.

Answer 1

In Java, (B * 100) + (A * D) will be a float; and it will be the float that is closest to 12638.1. However, 12638 requires 14 digits to express in binary, including the initial 1; which leaves 10 digits of the significand to express the fractional part. Therefore, you're going to get the closest number of 1024ths to 0.1 - which is 102/1024. This turns out to be 0.099609375 - so the float has a rounding error of 0.000390625.

That seems to be the difference between X and X1 that you're getting in your Java program.

I'm afraid I'm not a C# expert, so I can't tell you why C# is different.