Lua replacement for the % operator

Question

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.

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, Mar 14 '12 at 04:19
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](https://www.lua.org/manual/5.1/manual.html) *Lua supports the usual arithmetic operators: the binary + (addition), - (subtraction), * (multiplication), / (division), % (modulo), and ^ (exponentiation); and unary - (negation).* — Simon Forsberg, Jan 07 '18 at 20:48
@Javier Lua 5.0 did not support the `%` operator. 5.1 and later however, does. See my answer. — Simon Forsberg, Jan 07 '18 at 20:52

score 58 · Answer 1 · edited Oct 24 '15 at 01:21

58

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

edited Oct 24 '15 at 01:21

Brad Parks

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answered Mar 14 '12 at 11:47

lhf

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4

`math.fmod` did not exist in Lua 5.0, it was renamed from `math.mod` to `math.fmod` in Lua 5.1. However, Lua 5.1 also added the `%` operator, so if the OP is running Lua 5.0, `fmod` is the wrong function to use. – Simon Forsberg Jan 07 '18 at 21:01

score 30 · Accepted Answer · answered Mar 14 '12 at 04:02

30

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

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

answered Mar 14 '12 at 04:02

ninesided

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9

This seems a workaround for older versions of Lua. At least % works fine in Lua 5.2 and later. – Henrik Erlandsson Aug 27 '13 at 07:14
3

@HenrikErlandsson `%` was added as an operator in Lua 5.1. – Simon Forsberg Jan 07 '18 at 21:01

score 7 · Answer 3 · edited Jun 20 '20 at 09:12

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)

score 5 · Answer 4 · edited Aug 18 '13 at 17:30

5

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

edited Aug 18 '13 at 17:30

Macmade

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answered Aug 18 '13 at 17:12

anonymous

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Strange that it's undocumented in recent documentation. (At least with a simple test, it seems to work like fmod.) – Henrik Erlandsson Aug 27 '13 at 07:16
@HenrikErlandsson It's not undocumented in recent documentation, the function was called `mod` in 5.0 and renamed to `fmod` in 5.1. – Simon Forsberg Jan 07 '18 at 20:57

score 5 · Answer 5 · edited Jun 11 '18 at 21:30

5

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

edited Jun 11 '18 at 21:30

Sled

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answered Dec 31 '13 at 13:20

Jim Gao

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score 3 · Answer 6 · answered Dec 15 '16 at 06:51

3

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

answered Dec 15 '16 at 06:51

vr3C

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Lua replacement for the % operator

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