core_assertions.rb: Relax assert_linear_performance
* Use an `Enumerable` as factors, instead of three arguments. * Include `assert_operator` time in rehearsal time. * Round up max expected time.
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2023-03-17 17:41:24 +00:00
@ -1783,7 +1783,7 @@ class TestRegexp < Test::Unit::TestCase
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def test_linear_performance
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pre = ->(n) {[Regexp.new("a?" * n + "a" * n), "a" * n]}
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assert_linear_performance(factor: 29, first: 10, max: 1, pre: pre) do |re, s|
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assert_linear_performance([10, 29], pre: pre) do |re, s|
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re =~ s
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end
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end
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@ -738,23 +738,34 @@ eom
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end
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alias all_assertions_foreach assert_all_assertions_foreach
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def assert_linear_performance(factor: 10_000, first: factor, max: 2, rehearsal: first, pre: ->(n) {n})
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n = first
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arg = pre.call(n)
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tmax = (0..rehearsal).map do
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# Expect +seq+ to respond to +first+ and +each+ methods, e.g.,
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# Array, Range, Enumerator::ArithmeticSequence and other
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# Enumerable-s, and each elements should be size factors.
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#
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# :yield: each elements of +seq+.
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def assert_linear_performance(seq, rehearsal: nil, pre: ->(n) {n})
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first = seq.first
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*arg = pre.call(first)
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times = (0..(rehearsal || (2 * first))).filter_map do
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st = Process.clock_gettime(Process::CLOCK_MONOTONIC)
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yield arg
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(Process.clock_gettime(Process::CLOCK_MONOTONIC) - st)
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end.max
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yield(*arg)
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t = (Process.clock_gettime(Process::CLOCK_MONOTONIC) - st)
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assert_operator 0, :<=, t
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t.nonzero?
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end
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times.compact!
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tmin, tmax = times.minmax
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tmax *= tmax / tmin
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tmax = 10**Math.log10(tmax).ceil
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(first >= factor ? 2 : 1).upto(max) do |i|
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n = i * factor
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t = tmax * factor
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arg = pre.call(n)
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message = "[#{i}]: #{n} in #{t}s"
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seq.each do |i|
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next if i == first
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t = (tmax * i).to_f
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*arg = pre.call(i)
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message = "[#{i}]: in #{t}s"
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Timeout.timeout(t, nil, message) do
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st = Process.clock_gettime(Process::CLOCK_MONOTONIC)
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yield arg
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yield(*arg)
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assert_operator (Process.clock_gettime(Process::CLOCK_MONOTONIC) - st), :<=, t, message
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end
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end
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