Ruby中是否可以获取对象方法的引用(我想知道这是否可以在没有procs / lambdas的情况下完成),例如,请考虑以下代码:
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。
答案 0 :(得分:46)
您想要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"
现在您的代码变为:
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
答案 1 :(得分:4)
您可以使用lambdas执行此操作,同时保持更改子类中行为的能力:
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
答案 2 :(得分:1)
Ruby方法不是第一类对象;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
答案 3 :(得分:0)
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)