調用私有方法問題

Question

我已經實現了算法來計算數組倒數（它也合併了數組），但爲了使它更整潔，我希望對輸出進行排序和計數。

方法* count_and_sort *返回數組[number_of_inversions，[sorted_array]]。 方法倒置和* merge_sort *應該返回number_of_inversions和[sorted_array]分開，而是我得到NoMethodError

我錯過了什麼？

---

class Array 

    def inversions 
     count_and_sort[0] 
    end 

    def merge_sort 
     count_and_sort[1] 
    end 

    def merge_sort! 
     replace merge_sort 
    end 

private 

    def count_and_sort 
     return [0, self] if self.size <= 1 
     left, right = partition.with_index{ |value, index| index < self.size/2} 
     left = left.count_and_sort 
     right = right.count_and_sort 
     merged = merge(left[1], right[1]) 

     [left[0] + right[0] + merged[0], merged[1]] 
    end 

    def merge(left, right) 
     inversions = 0 
     merged = [] 
     until left.empty? || right.empty? 
      if left.first < right.first 
       merged << left.shift 
      else 
       merged << right.shift 
       inversions += left.size 
      end 
     end 

     merged += left += right 
     [inversions, merged] 
    end 
end 

p [*1..20].shuffle.inversions 

---

我試着用

def inversions 
    result = self.send(:count_and_sort) 
    result[0] 
end 

更換倒置但結果是一樣的。

PS任何其他批評和評論歡迎）

Answer 1

的誤差是count_and_sort方法內，作爲錯誤信息表示：

... in 'count_and_sort': private method 'count_and_sort' called for ... 

錯誤的行是：

left = left.count_and_sort 
right = right.count_and_sort 

Ruby中的私有方法不能用明確的接收方調用，它們必須在012上隱式調用。因此，您可以使count_and_sort方法受到保護，或使用left和right上的send對其進行調用。在這種情況下，我會建議前者。