Lua replacement for the % operator

时光怂恿深爱的人放手 提交于 2019-11-30 16:53:49

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

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

lhf

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

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

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

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

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

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