如何获得对方法的引用？

Question

是否可以在 Ruby 中获取对对象方法的引用（我想知道这是否可以在没有 procs/lambdas 的情况下完成），例如，考虑以下代码：

class X
  def initialize
    @map = {}
    setup_map
  end

  private
  def setup_map
    # @map["a"] = get reference to a method
    # @map["b"] = get reference to b method
    # @map["c"] = get referebce to c method
  end

  public
  def call(a)
    @map["a"](a) if a > 10
    @map["b"](a) if a > 20
    @map["c"](a) if a > 30
  end

  def a(arg)
     puts "a was called with #{arg}"
  end

  def b(arg)
     puts "b was called with #{arg}"
  end

  def c(arg)
    puts "c was called with #{arg}"
  end
end

有可能做这样的事情吗？我想避免使用 procs/lambdas，因为我希望能够通过子类化来改变 A、B、C 的行为。

Answer 1

你想要Object#method:

---------------------------------------------------------- Object#method
     obj.method(sym)    => method
------------------------------------------------------------------------
     Looks up the named method as a receiver in obj, returning a Method 
     object (or raising NameError). The Method object acts as a closure 
     in obj's object instance, so instance variables and the value of 
     self remain available.

        class Demo
          def initialize(n)
            @iv = n
          end
          def hello()
            "Hello, @iv = #{@iv}"
          end
        end

        k = Demo.new(99)
        m = k.method(:hello)
        m.call   #=> "Hello, @iv = 99"

        l = Demo.new('Fred')
        m = l.method("hello")
        m.call   #=> "Hello, @iv = Fred"

现在你的代码变成了：

private
def setup_map
  @map = {
    'a' => method(:a),
    'b' => method(:b),
    'c' => method(:c)
  }
  # or, more succinctly
  # @map = Hash.new { |_map,name| _map[name] = method(name.to_sym) }
end

public
def call(arg)
  @map["a"][arg] if arg > 10
  @map["b"][arg] if arg > 20
  @map["c"][arg] if arg > 30
end

Answer 2

您可以使用 lambdas 做到这一点，同时保持更改子类行为的能力：

class X
  def initialize
    @map = {}
    setup_map
  end

  private
  def setup_map
    @map["a"] = lambda { |a| a(a) }
    @map["b"] = lambda { |a| b(a) }
    @map["c"] = lambda { |a| c(a) }
  end

  public
  def call(a)
    @map["a"].call(a) if a > 10
    @map["b"].call(a) if a > 20
    @map["c"].call(a) if a > 30
  end

  def a(arg)
     puts "a was called with #{arg}"
  end

  def b(arg)
     puts "b was called with #{arg}"
  end

  def c(arg)
    puts "c was called with #{arg}"
  end
end

Answer 3

Ruby 方法不是一流的对象；它通过消息传递实现OO。

class X
  def call(a)
    self.send(:a, a) if a > 10
    self.send(:b, a) if a > 20
    self.send(:c, a) if a > 30
  end

  def a(arg)
     puts "a was called with #{arg}"
  end

  def b(arg)
     puts "b was called with #{arg}"
  end

  def c(arg)
    puts "c was called with #{arg}"
  end
end

或者直接打电话给他们：

def call(a)
  self.a(a) if a > 10
  self.b(a) if a > 20
  self.c(a) if a > 30
end

Answer 4

您可以通过object.method(:method_name)获取对该方法的引用。

例如：获取对system 方法的引用。

m = self.method(:system)
m.call('ls')