require 'spec_helper'
require 'thread'

describe Immutable::List do
  describe '#partition' do
    it 'is lazy' do
      -> { Immutable.stream { fail }.partition }.should_not raise_error
    end

    it 'calls the passed block only once for each item' do
      count = 0
      a,b = L[1, 2, 3].partition { |item| count += 1; item.odd? }
      (a.size + b.size).should be(3) # force realization of lazy lists
      count.should be(3)
    end

    # note: Lists are not as lazy as they could be!
    # they always realize elements a bit ahead of the current one

    it 'returns a lazy list of items for which predicate is true' do
      count = 0
      a,b = L[1, 2, 3, 4].partition { |item| count += 1; item.odd? }
      a.take(1).should == [1]
      count.should be(3) # would be 1 if lists were lazier
      a.take(2).should == [1, 3]
      count.should be(4) # would be 3 if lists were lazier
    end

    it 'returns a lazy list of items for which predicate is false' do
      count = 0
      a,b = L[1, 2, 3, 4].partition { |item| count += 1; item.odd? }
      b.take(1).should == [2]
      count.should be(4) # would be 2 if lists were lazier
      b.take(2).should == [2, 4]
      count.should be(4)
    end

    it 'calls the passed block only once for each item, even with multiple threads' do
      mutex = Mutex.new
      yielded = [] # record all the numbers yielded to the block, to make sure each is yielded only once
      list = Immutable.iterate(0) do |n|
        sleep(rand / 500) # give another thread a chance to get in
        mutex.synchronize { yielded << n }
        sleep(rand / 500)
        n + 1
      end
      left, right = list.partition(&:odd?)

      10.times.collect do |i|
        Thread.new do
          # half of the threads will consume the "left" lazy list, while half consume
          # the "right" lazy list
          # make sure that only one thread will run the above "iterate" block at a
          # time, regardless
          if i % 2 == 0
            left.take(100).sum.should == 10000
          else
            right.take(100).sum.should == 9900
          end
        end
      end.each(&:join)

      # if no threads "stepped on" each other, the following should be true
      # make some allowance for "lazy" lists which actually realize a little bit ahead:
      (200..203).include?(yielded.size).should == true
      yielded.should == (0..(yielded.size-1)).to_a
    end

    [
      [[], [], []],
      [[1], [1], []],
      [[1, 2], [1], [2]],
      [[1, 2, 3], [1, 3], [2]],
      [[1, 2, 3, 4], [1, 3], [2, 4]],
      [[2, 3, 4], [3], [2, 4]],
      [[3, 4], [3], [4]],
      [[4], [], [4]],
    ].each do |values, expected_matches, expected_remainder|
      context "on #{values.inspect}" do
        let(:list) { L[*values] }

        context 'with a block' do
          let(:result)  { list.partition(&:odd?) }
          let(:matches) { result.first }
          let(:remainder) { result.last }

          it 'preserves the original' do
            list.should eql(L[*values])
          end

          it 'returns a frozen array with two items' do
            result.class.should be(Array)
            result.should be_frozen
            result.size.should be(2)
          end

          it 'correctly identifies the matches' do
            matches.should eql(L[*expected_matches])
          end

          it 'correctly identifies the remainder' do
            remainder.should eql(L[*expected_remainder])
          end
        end

        context 'without a block' do
          it 'returns an Enumerator' do
            list.partition.class.should be(Enumerator)
            list.partition.each(&:odd?).should eql([L[*expected_matches], L[*expected_remainder]])
          end
        end
      end
    end
  end
end
