Last Updated: February 25, 2016

Floats

# Who have problems for this?
5.9 - 5.8 # => 0.10000000000000053
# Too easy solution
(5.9 - 5.8).round 14 # => 0.1

Solution and benchmark from comments (thx to sheerun):

require "benchmark"
require "bigdecimal"
n = 1_000_000
Benchmark.bm do |x|
 x.report { for i in 1..n; (5.9 - 5.8).round 14; end }
 x.report { for i in 1..n; (BigDecimal.new("5.9") - BigDecimal.new("5.8")).to_f; end }
end

 user system total real
0.240000 0.000000 0.240000 ( 0.241408)
4.140000 0.000000 4.140000 ( 4.139652)

#ruby

#float

Written by Yuri Artemev

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2 Responses

Add your response

sheerun

You can use big decimals if you require arbitrary precision:
require 'bigdecimal'; (BigDecimal.new("5.9") - BigDecimal.new("5.8")).to_f

over 1 year ago ·

artemeff

require "benchmark"
require "bigdecimal"
n = 1_000_000
Benchmark.bm do |x|
 x.report { for i in 1..n; (5.9 - 5.8).round 14; end }
 x.report { for i in 1..n; (BigDecimal.new("5.9") - BigDecimal.new("5.8")).to_f; end }
end

 user system total real
0.240000 0.000000 0.240000 ( 0.241408)
4.140000 0.000000 4.140000 ( 4.139652)

But your solution is correct, thanks.

over 1 year ago ·