Ruby挑战 - 方法链和懒惰评估

Question

阅读文章http://jeffkreeftmeijer.com/2011/method-chaining-and-lazy-evaluation-in-ruby/后,我开始寻找方法链和懒惰评估的更好解决方案.

我想我已经用以下五个规格封装了核心问题; 谁能让他们全部通过？

一切顺利:子类化,委托,元编程,但对后者不鼓励.

将依赖关系保持在最低限度是有利的:

require 'rspec'

class Foo
    # Epic code here
end

describe Foo do

  it 'should return an array corresponding to the reverse of the method chain' do
    # Why the reverse? So that we're forced to evaluate something
    Foo.bar.baz.should == ['baz', 'bar']
    Foo.baz.bar.should == ['bar', 'baz']
  end

  it 'should be able to chain a new method after initial evaluation' do
    foobar = Foo.bar
    foobar.baz.should == ['baz', 'bar']

    foobaz = Foo.baz
    foobaz.bar.should == ['bar', 'baz']
  end

  it 'should not mutate instance data on method calls' do
    foobar = Foo.bar
    foobar.baz
    foobar.baz.should == ['baz', 'bar']
  end

  it 'should behave as an array as much as possible' do
    Foo.bar.baz.map(&:upcase).should == ['BAZ', 'BAR']

    Foo.baz.bar.join.should == 'barbaz'

    Foo.bar.baz.inject do |acc, str|
      acc << acc << str
    end.should == 'bazbazbar'

    # === There will be cake! ===
    # Foo.ancestors.should include Array
    # Foo.new.should == []
    # Foo.new.methods.should_not include 'method_missing'
  end

  it "should be a general solution to the problem I'm hoping to solve" do
    Foo.bar.baz.quux.rab.zab.xuuq.should == ['xuuq', 'zab', 'rab', 'quux', 'baz', 'bar']
    Foo.xuuq.zab.rab.quux.baz.bar.should == ['bar', 'baz', 'quux', 'rab', 'zab', 'xuuq']
    foobarbaz = Foo.bar.baz
    foobarbazquux = foobarbaz.quux
    foobarbazquuxxuuq = foobarbazquux.xuuq
    foobarbazquuxzab = foobarbazquux.zab

    foobarbaz.should == ['baz', 'bar']
    foobarbazquux.should == ['quux', 'baz', 'bar']
    foobarbazquuxxuuq.should == ['xuuq', 'quux', 'baz', 'bar']
    foobarbazquuxzab.should == ['zab', 'quux', 'baz', 'bar']
  end

end

Answer 1

这是受到Amadan答案的启发,但使用的代码更少:

class Foo < Array
    def self.method_missing(message, *args)
        new 1, message.to_s
    end
    def method_missing(message, *args)
        dup.unshift message.to_s
    end
end

Answer 2

微不足道，不是吗？

class Foo < Array
  def self.bar
    other = new
    other << 'bar'
    other
  end
  def self.baz
    other = new
    other << 'baz'
    other
  end
  def bar
    other = clone
    other.unshift 'bar'
    other
  end
  def baz
    other = clone
    other.unshift 'baz'
    other
  end
end

该to_s标准失败是因为 1.9 改变了工作方式Array#to_s。更改为兼容性：

Foo.baz.bar.to_s.should == ['bar', 'baz'].to_s

我要蛋糕。

顺便说一句 - 这里的元编程将减少代码大小并极大地提高灵活性：

class Foo < Array
  def self.method_missing(message, *args)
    other = new
    other << message.to_s
    other
  end
  def method_missing(message, *args)
    other = clone
    other.unshift message.to_s
    other
  end
end