Is NaN a comparable type in golang?

Kushal Billaiya 0
Viewed 59

I came across a golang quiz which used NaN as a maps key and it ran without any error. If map's keys are supposed to be comparable is NaN a comparable type or is this a compiler bug which allowed NaN as a key.

Here's the quiz source, the go playground link and code below.

package main

var x = 0.0

func main() {
  var a = x / x // a = NaN
  var m = map[float64]int{a: 1}
  m[a] = 2
  for k := range m {
  delete(m, k)
  }
  println(len(m)) // prints 2
}
go
4 Answers

The short answer is that this is not a compiler bug.

Maps require comparable types as keys. Float (regardless of float32 or float64) is a comparable type. What's odd about NaN (and in a different way, zero, which can be either positive or negative zero) is that while you can compare two NaNs, they always compare as unequal. So some float value x is not equal to itself when x is NaN. This produces surprises.

Issue 20660 comments note that the same kind of problems occur in a number of other languages.

I particularly like Russ Cox's comment that there are only bad answers here.

torek0

According to the documentation , math.NaN returns a float64 which is a comparable type

rrxxyy10
Kushal Billaiya0

Any float64 is comparable. There is no problem comparing any two floats. You can compare a NaN to 3.141 or NaN to NaN.

The problem is not comparability if NaNs/floats. The problem is that any comparison of a NaN with anything, even an other NaN or itself will result in false.

   NaN != x   for all  x, including NaN

This makes using NaNs as map keys basically impossible because it's impossible to find a NaN.

userazj50
Related