Lua replacement for the % operator

39
votes

I want to check, if a number is divisible by another number:

for i = 1, 100 do
    if i % 2 == 0 then
        print( i .. " is divisible.")
    end
end

This should work without any problems, but with the Lua in my server the script doesn't run if there is a % in the script... I dont know whats the reason, so is there any "replacement" for that? So I could check the number divsibility?

Thank you.

6
mathluamodulo
What version of Lua is the server running? - Nicol Bolas
I think its 5.0 or later :S. - Cyclone
sounds like you have some encoding problems; maybe if you find what encoding is it, you might be able to sneak a % through. try '%%' or '\%' or '%25' - Javier
I am downvoting this question because I believe it is asking the wrong question: The real question here is "Why is % not working for me in Lua"? According to the Lua Documentation Lua supports the usual arithmetic operators: the binary + (addition), - (subtraction), * (multiplication), / (division), % (modulo), and ^ (exponentiation); and unary - (negation). - Simon Forsberg
@Javier Lua 5.0 did not support the % operator. 5.1 and later however, does. See my answer. - Simon Forsberg

6 Answers

57
votes

Use math.fmod(x,y) which does what you want:

Returns the remainder of the division of x by y that rounds the quotient towards zero.

http://www.lua.org/manual/5.2/manual.html#pdf-math.fmod

29
votes

It's not ideal, but according to the Lua 5.2 Reference Manual:

a % b == a - math.floor(a/b)*b

5
votes

Lua 5.0 did not support the % operator.

Lua supports the usual arithmetic operators: the binary + (addition), - (subtraction), * (multiplication), / (division), and ^ (exponentiation); and unary - (negation).

https://www.lua.org/manual/5.0/manual.html

Lua 5.1 however, does support the % operator.

Lua supports the usual arithmetic operators: the binary + (addition), - (subtraction), * (multiplication), / (division), % (modulo), and ^ (exponentiation); and unary - (negation).

https://www.lua.org/manual/5.1/manual.html

If possible, I would recommend that you upgrade. If that is not possible, use math.mod which is listed as one of the Mathematical Functions in 5.0 (It was renamed to math.fmod in Lua 5.1)

4
votes
for i = 1, 100 do
    if (math.mod(i,2) == 0) then
        print( i .. " is divisible.")
    end
end
3
votes

Use math.fmod, accroding lua manual math.mod was renamed to math.fmod in lua 5.1.

3
votes
function mod(a, b)
    return a - (math.floor(a/b)*b)
end