CODEWITHSUNDEEP
javac#.net
Bruno Peres
Bruno Peres

Reputation: 16365

Why the expression (double.MinValue < double.MinValue + 1) returns false?

Consider the c# below code:

public class Program
{
    public static void Main(string[] args)
    {
        //Your code goes here
        Console.WriteLine(double.MinValue < double.MinValue + 1);
        Console.WriteLine(int.MinValue < int.MinValue + 1);
    }
}

The output will be:

False
True

I know that

Console.WriteLine(double.MinValue + 1);
Console.WriteLine(double.MinValue);

will print the same -1,79769313486232E+308 value.

My question is: Why this "unexpected" and curious behavior occurs? I hope the first line returns true too.

The below Java program, for example, returns the expected true:

public class MyClass {
    public static void main(String args[]) {
        System.out.println(Double.MIN_VALUE < Double.MIN_VALUE + 1);
    }
}

Why this different behavior?

Upvotes: 1

Views: 344

Answers (2)

Oli
Oli

Reputation: 840

While both Java and C# implement the IEEE 754 specification for floating point computations, the minvalue constants are not the same.

  • In Java it is the smallest positive value (Java doc)
  • In C# it is the smallest value (C# doc)

You can try what happens (or rather "won't happen") here: http://weitz.de/ieee/

Upvotes: 1

Louis Wasserman
Louis Wasserman

Reputation: 198033

The two min values are different things.

In Java, MIN_VALUE is the name for the smallest positive double value. In C#, MinValue is the name for the double value less than all other finite values; in Java, that value would be -Double.MAX_VALUE.

And that value is so massively negative that adding 1 to it gets lost in the rounding. double doesn't have enough precision to represent MinValue differently from MinValue+1. It's, roughly, -1800000....00 where there are 307 zeroes. double can only represent, roughly, seventeen decimal digits of precision. So adding one gets lost in the rounding.

Upvotes: 11

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