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I used modulo (%) in Lua just a moment ago and I noticed that it was for some reason very inaccurate. I used it in World of Warcraft, but I would assume that it's at least somewhat up to date.
Anyhow, using the following example, the output would be 1;
print((0.6/(0.1^(2-1)))%1)
But when I use the following, it would return 0;
print((0.6*(10^(2-1)))%1)
And as far as I have been educated, 0.6/0.1 should be equivalent to 0.6*10.
It's not really a question or a problem, but I am merely curious about the cause. Inaccuracies in maths would be quite devastating in my opinion.
576
The Lua modulo operator is used to get the remainder of a division. The modulo operator is a built-in operator in the lua programming. Modulooperatoris an athematic operator and it is denoted by the % symbol. The modulo operation return the whole number remainder, instead of the division result.
According to the Lua Documentation Lua supports the usual arithmetic operators: the binary + (addition), - (subtraction), * (multiplication), / (division), % (modulo), and ^ (exponentiation); and unary - (negation).
The modulo operator, denoted by %, is an arithmetic operator. The modulo division operator produces the remainder of an integer division. produces the remainder when x is divided by y.
If you look in the Lua manual, the % operation is defined as
a % b == a - math.floor(a/b)*b
And so the expression
6 - math.floor(6/1) * 1
will equal zero.
However, due to floating point inaccuracies (0.1 does not exist as a floating point number, the closest should be 0.1000000014...), these should be your intermediate results:
> print(math.floor(0.6*10))
6
> print(math.floor(0.6/.1))
5
So you get
6 - 6 * 1 = 0
6 - 5 * 1 = 1
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