问题
Do you know how to fix the following issue with math precision?
p RUBY_VERSION # => "1.9.1"
p 0.1%1 # => 0.1
p 1.1%1 # => 0.1
p 90.0%1 # => 0.0
p 90.1%1 # => 0.0999999999999943
p 900.1%1 # => 0.100000000000023
p RUBY_VERSION # => "1.9.2"
p 0.1%1 # => 0.1
p 1.1%1 # => 0.10000000000000009
p 90.0%1 # => 0.0
p 90.1%1 # => 0.09999999999999432
p 900.1%1 # => 0.10000000000002274
回答1:
Big Decimal
As the man said;
Squeezing infinitely many real numbers into a finite number of bits requires an approximate representation.
I have however had great success using the BigDecimal class. To quote its intro
Ruby provides built-in support for arbitrary precision integer arithmetic. For example:
42**13 -> 1265437718438866624512
BigDecimal provides similar support for very large or very accurate floating point numbers.
Taking one of your examples;
>> x = BigDecimal.new('900.1')
=> #<BigDecimal:101113be8,'0.9001E3',8(8)>
>> x % 1
=> #<BigDecimal:10110b498,'0.1E0',4(16)>
>> y = x % 1
=> #<BigDecimal:101104760,'0.1E0',4(16)>
>> y.to_s
=> "0.1E0"
>> y.to_f
=> 0.1
As you can see, ensuring decent precision is possible but it requires a little bit of effort.
回答2:
This is true of all computer languages, not just Ruby. It's a feature of representing floating point numbers on binary computers:
What Every Computer Scientist Should Know About Floating Point Arithmetic
回答3:
Writing 0.1 into a floating point will always result in rounding errors. If you want 'precise' decimal representation, you should use the Decimal type.
来源：https://stackoverflow.com/questions/3777119/issue-with-precision-of-ruby-math-operations