Due to the nature of floating-point math, .4 * .4
= 0.16000000000000003
in Julia. I want to get the mathematically correct answer of 0.16
, in a CPU-efficient way. I know round() works, but that requires prior knowledge of the number of decimal places the answer occupies, so it isn't a general solution.
The standard choice for floating-point values is Float64 , which is double precision using 64 binary bits. bitstring(1.0) "0011111111110000000000000000000000000000000000000000000000000000" The first bit determines the sign of the number: [bitstring(1.0); bitstring(-1.0)]
Getting rounded value of a number in Julia – round() Method The default round process is done to the nearest integer, with ties (fractional values of 0.5) being rounded to the nearest even integer. The specified value x is rounded to an integer value and returns a value of the given type T or of the same type of x.
Literal Float32 values can be entered by writing an f in place of e : julia> x = 0.5f0 0.5f0 julia> typeof(x) Float32 julia> 2.5f-4 0.00025f0. Values can be converted to Float32 easily: julia> x = Float32(-1.5) -1.5f0 julia> typeof(x) Float32.
Some options:
Use the inbuilt Rational
type. The most accurate and fastest way would be
16//100 * 16//100
If you're using very big numbers these might overflow, in which case you can use BigInt
s instead,
big(16)//big(100) * big(16)//big(100)
(you don't actually need to wrap them all in big
s, as the rationals will promote automatically).
You can also use rationalize(0.16)
, but this may not be quite as accurate or efficient, as the literal 0.16
has already been converted to a Float64
by the time Julia sees it, so you're converting to a binary floating point and then to a Rational
.
DecFP.jl wraps the Intel implementation of IEEE-754 Decimal floating point. This should be reasonably fast (though not as efficient as binary), but has fixed precision, so you will have to round at some point.
Decimals.jl is a "big decimal" floating point library: as it uses arbitrary precision arithmetic, it is going to be slower than DecFP.
To say which is the best would require more information about your intended use.
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