紅寶石挑戰 - 方法鏈和懶惰評估

Question

閱讀文章http://jeffkreeftmeijer.com/2011/method-chaining-and-lazy-evaluation-in-ruby/後，我開始尋找更好的方法鏈和懶惰評估的解決方案。

我想我已經封裝了以下五個規格的核心問題;任何人都可以讓他們全部通過？

任何事情都會發生：繼承，授權，元編程，但不鼓勵後者。

這將有利於保持依賴性降到最低：

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

瑣碎的，不是嗎？因爲1.9已經改變了Array#to_s作品

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 

的to_s標準失敗。改變這種兼容性：

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

我想要的蛋糕。

BTW - 這裏的元編程將削減下來的代碼大小和增加靈活性極大：

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 

Answer 2

這是由Amadan的回答啓發，但使用更少的代碼：

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