1.824999999999999.round(2) # => 1.82 1.82499999999999999.round(2) # => 1.83 I don't understand why the result in the first case is 1.82 and for the latter 1.83. My desired result is 1.82 for both cases.
- 3You should read about what every computer scientist should know about floating-point arithmetic.toro2k– toro2k2014-02-26 12:22:31 +00:00Commented Feb 26, 2014 at 12:22
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4 Answers
As is well known, a floating number in computer has errors. Notice that both numbers that you have are close to 1.825. In your first case, the number is small enough to be differentiated from 1.825.
1.824999999999999 # => 1.824999999999999 In your second case, having enough 9s makes the value close enough to be considered exactly as 1.825:
1.82499999999999999 # => 1.825 Then, when you apply round, you get 1.83.
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In addition to the many excellent explanations here, I'd like to give an example using BigDecimal
o = BigDecimal.new('1.82499999999999999', 17) 2.0.0p247 :011 > o.round(2).to_f => 1.82 Also notice that #round takes various modes for rounding, as per the docs.
Ref:
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After a certain number of (binary) digits, Ruby cannot represent differences between Float literals, even if you've entered the value directly in source code or the console
1.82499999999999999 == 1.825 # => true If you really need a high level of precision, then you could look at the BigDecimal class. Although you could still enter enough 9s to reach limits in that class.
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Floats are notoriously imprecise as sawa and Neil have pointed out.
If this is mainly an issue for testing, you can use assert_in_delta if you're using minitest, or be_within with rSpec
