Ruby - 有没有办法用重载方法的同名调用超类的方法

Question

是否可以这样做：

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

除非我误读了您的问题，否则您应该可以自己拨打super；例如：

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")

生产

$ ruby test.rb
do something with heysome modification

Answer 2

您不能在 super 调用中使用方法名称

如果您单独使用super，那么它将调用super 类的实现并传递相同的参数

class A
  def a(var)
    puts var
  end
end

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

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

如果您想修改变量，则可以重新分配 var arg 或使用 super 调用传入参数

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

试试下面的：

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")

输出：

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")

输出：

do something with 12