CODEWITHSUNDEEP
c#floating-point
Justin
Justin

Reputation: 86789

Why are some floating point numbers accurately represented in C#?

Inspired by this question, the following doesn't do what I'd expect it to:

float myFloat = 0.6;
Console.WriteLine(myFloat);
// Output: 0.6

I'd expect the above to print out 0.60000002384185791 (the floating point representation of 0.6) - clearly there is some mechanism here which is making this work when in fact it shouldn't (although as you can see from the linked question it sometimes doesn't work)

What is this mechanism and how does it work?

Upvotes: 3

Views: 448

Answers (3)

silverclaw
silverclaw

Reputation: 178

0.6 can't be represented under IEEE754 floats. It sits between 0.599999964237213134765625 (0x3f199999) and 0.600000083446502685546875 (0x3f19999b). The round-to-nearest mode yields 0.60000002384185791015625 (0x3f19999a) which is what WriteLine is printing.

You have to either use a higher precision floating point representation (double) or limit the number of decimal places the WriteLine prints:

float f = 0.6; Console.WriteLine("{0:N6}", f);

Upvotes: 0

Brian Rasmussen
Brian Rasmussen

Reputation: 116481

If you look at the implementation of Console.WriteLine, you'll see that it ends up calling ToString on the value with a default FormatProvider. I.e. the result you're seeing is how the number appears when formatted using this format provider.

While it doesn't explain the details of how the result is produced, it does show that Console.WriteLine goes through some formatting of the value before printing it.

Upvotes: 4

KristoferA
KristoferA

Reputation: 12397

I would guess the WriteLine overload that takes a float does rounding when converting it to a string...

Upvotes: 3

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