From: Han Holl <han@...>
Date: 2002-06-09T23:16:52+09:00
Subject: Hash#===()


Hi,

I ran the following benchmark:

require 'benchmark'
include Benchmark

class Hash
  def ===(a)
    has_key?(a)
  end
end

n = 100000
hash = { "abc"=> 1, "def"=> 1, "kli"=> 1, "lop"=> 1, "tli"=> 1, "ppp"=> 1 }
[ "ppp", "abc", "for" ].each do |a|
 bm(7) do |x|
   x.report(a){for i in 1..n;
    case a
    when "abc", "def", "kli", "lop", "tli", "ppp"
    end
   end}
   x.report(a){for i in 1..n;
    case a
    when /^(abc|def|kli|lop|tli|ppp)$/
    end
   end}
   x.report(a){for i in 1..n;
    case a
    when /^(?:abc|def|kli|lop|tli|ppp)$/
    end
   end}
   x.report(a){for i in 1..n;
    case a
    when hash
    end
   end}

 end
end

With the following result:
             user     system      total        real
ppp      0.770000   0.000000   0.770000 (  0.765261)
ppp      0.520000   0.000000   0.520000 (  0.518373)
ppp      0.480000   0.000000   0.480000 (  0.487162)
ppp      0.290000   0.000000   0.290000 (  0.286171)
             user     system      total        real
abc      0.210000   0.000000   0.210000 (  0.204143)
abc      0.440000   0.000000   0.440000 (  0.443976)
abc      0.430000   0.000000   0.430000 (  0.425147)
abc      0.290000   0.000000   0.290000 (  0.292743)
             user     system      total        real
for      0.780000   0.000000   0.780000 (  0.776409)
for      0.360000   0.000000   0.360000 (  0.366605)
for      0.340000   0.000000   0.340000 (  0.340457)
for      0.260000   0.000000   0.260000 (  0.258510)

This shows that in a case with several alternatives, using a Hash
is almost always the best solution.
Wouldn't the above definition for Hash#=== make sense as the
default ?

Cheers,

Han Holl