Lua check if a number/value is nan

Question

I have written a program to print a matrix after some computations and I am getting an output of nan for all elements. I want to break a for loop as soon as the matrix's first element becomes nan to understand the problem. How can I do this? In the terminal, I have printed the matrix a containing nan as all elements and typed a[1][1]=="nan" and a[{{1},{1}}]=="nan" both of which return false. Why are they not returning false and what statement should I use instead?

`local function isNaN( v ) return type( v ) == "number" and v ~= v end` — siffiejoe, Jun 10 '16 at 17:15
`local function isNaN( v ) return tostring(v) == tostring(0/0) end` — Egor Skriptunoff, Jun 10 '16 at 17:41
@EgorSkriptunoff -(0/0) results in a different string even though it is still nan thus returning false even though true should be returned. https://ideone.com/YoOljy — Rochet2, Jun 11 '16 at 20:04

score 20 · Accepted Answer · answered Jun 11 '16 at 03:11

20

Your test fails because you are comparing a number with a string, "nan".

If you are sure it's a number, the easiest way is:

if a[1][1] ~= a[1][1] then

because according to IEEE 754, a nan value is considered not equal to any value, including itself.

answered Jun 11 '16 at 03:11

Yu Hao

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score 8 · Answer 2 · answered Mar 21 '18 at 04:22

8

Two solutions:

local n = 0/0 -- nan

-- first solution
if ( tostring(n) == "nan" ) then
   print("is nan!!")
end

-- second solution
if (n ~= n) then
   print("is nan!!")
end

answered Mar 21 '18 at 04:22

rsc

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3

Per this question on [Is there a way to detect NaN and -NaN?](https://stackoverflow.com/a/52280708/1366033), `"nan"` is "*not portable*", presumably because that's the `tostring` output on linux while [windows uses `#IND`](https://stackoverflow.com/q/19107302/1366033). So the safer equation is to repeat the `tostring` of something we know to be NaN (i.e. `0/0`) like this: `tostring(n) == tostring(0/0)` – KyleMit Dec 16 '20 at 23:44

score -5 · Answer 3 · answered Jun 14 '16 at 00:35

-5

Try this:

for x = 1, x2 do          -- x2 depends on how big you matrix is.
  for y = 1, y2 do        -- y2 the same as x2
    -- some code depending on how your program works
    if a[x][y] == nan then
      print( "X:" .. x .. ", Y:" .. y )
      break
    end
  end
end

PS: (nan == nan) is true

answered Jun 14 '16 at 00:35

BlackHack

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2

Your answer is incorrect. The variable "nan" is not defined in Lua, so it will evaluate to (nil). (nil == nil) is indeed true. However, (0/0 == 0/0) is false! – joshuahhh Oct 17 '17 at 00:54
1

PS: (nan == nan) is **false**, as it should be. – DarkWiiPlayer Jul 19 '18 at 10:49

Lua check if a number/value is nan

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