紅寶石 - 有沒有辦法來調用父類的方法重載的

Question

的方法同名

是否有可能做這樣的事情：

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 

謝謝！

Answer 1

除非我誤讀了您的問題，否則您應該可以自行撥打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") 

產生

$ ruby test.rb 
do something with heysome modification 

Answer 2

如果你使用它自己的super然後它會調用super類的實現通過沿着你不能與super呼叫

使用方法名相同論點

class A 
    def a(var) 
    puts var 
    end 
end 

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

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

如果要修改的變量，那麼你可以重新分配var Arg或使用super調用傳遞參數

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

嘗試以下：

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") 

輸出：

do something with 12 

另一個

之一：

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") 

輸出：

do something with 12