Question

是否可以执行以下操作：

class A
  def a(var)
    puts "do something with #{var}"
  end
end

class B < A
  def a(var)
    var = var + "some modification"
    #this is what I want to do:
    super.a(var)
  end
end

谢谢！

Answer 1

除非我误解你的问题，否则你应该可以自己致电super; e.g：

class A
  def a(var)
    puts "do something with #{var}"
  end
end

class B < A
  def a(var)
    var = var + "some modification"
    #this is what I want to do:
    #super.a(var)
    super
  end
end

v = B.new
v.a("hey")

产生

$ ruby test.rb
do something with heysome modification

Answer 2

您不能将方法名称与super调用

一起使用

如果你自己使用super那么它将调用传递相同参数的super类'实现

class A
  def a(var)
    puts var
  end
end

class B < A
  def a(var)
    super
  end
end

B.new.a(1) #=> 1

如果要修改变量，则可以重新分配var arg或使用传递参数的super调用

class A
  def a(var)
    puts var
  end
end

class B < A
  def a(var)
    var = 2
    super
  end
end

B.new.a(1) #=> 2

或

class A
  def a(var)
    puts var
  end
end

class B < A
  def a(var)
    super(2)
  end
end

B.new.a(1) #=> 2

Answer 3

尝试以下内容：

class A
  def a(var)
    puts "do something with #{var}"
  end
end

class B < A
  def a(var)
    var = var + "some modification"
    #this is what I want to do:
    #super.a(var)
    self.class.superclass.instance_method(:a).bind(self).call 12
  end
end

v = B.new
v.a("hello")

的输出： 的

do something with 12

另一个：

class A def a(var) puts "do something with #{var}" end end class B < A alias :old_a :a def a(var) var = var + "some modification" #this is what I want to do: #super.a(var) old_a 12 end end v = B.new v.a("hello")

的输出： 的

do something with 12

Ruby - 有没有办法调用超类的方法，重载方法的名称相同

3 个答案: