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


        

Answer 1

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

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

Answer 2

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

Answer 3

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

Answer 4

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)

Answer 5

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

Answer 6

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