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我不确定如何解释这一点，但基本上我试图参考元素 A 的列表类前面（可以来自任何列表）。但是发生的情况是，当它遍历列表的元素时，它正在比较两个不同的列表并最终不匹配。即比较包含前面 b 的原始列表与包含元素 A 的列表。现在我只是想知道如何将元素 A 的前面设置为 b，以便我可以比较它的位置。

/*front is a dummy element used to keep position.
List is a class i have made under requirements of naming for subject.
i don't want a solution. I only want to know about how to do it.

This is what is an example code of whats causing the problem USED IN DRIVER PROGRAM
DLL.concat(DLL2);
it is basically getting DLL's front and going through the loop when it should be using DLL2's.

DLL and DLL2 are both Lists
***/


    //will return the index of the Element for comparing

    private int checkElement(Element A){

        Element b = front;

            int i = 0;  
            while (b != a && i<size)
            {
                b = b.next;
                i++;
            }

            return i;
        }


//edit: add

//size is the size of the list gets increased everytime a variable is added to the list on top of the dummy element.

//Item is a private class inside the List class. it contains the values: element,next, previous in which element contains an object, next and previous contain the next element in the list and the previous one (its a double linked list) 

// this is what causes the error to turn up in the above method as im using two different lists and joining them.

    public void concat(List L){
        if  (splice(L.first(),L.last(),last())){
            size = size+L.size;
        }
    }

//this is the splice method for cutting out elements and attaching them after t
//just using the check method to assert that a<b and will later use it to assert t not inbetween a and b

public boolean splice(Element a, Element b, Element t){

        if  (checkElement(a) < checkElement(b)){

            Element A = a.previous;
            Element B = b.next;
            A.next = B;
            B.previous = A;

            Element T = t.next;

            b.next = T;
            a.previous = t;
            t.next = a;
            T.previous = b;
        return true;
        }
        else {

        System.out.println("Splicing did not occur due to b<a");        
        return false;
        }

    }
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1 回答 1

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因此，尽管我发表了评论，但我看到了一个明显的问题。您不能在引用类型上使用相等运算符。也就是说，除了原始类型（double、int 等）之外的任何东西。发生的情况是您正在比较实例的地址，除非它们实际上是相同的对象（内存中的相同地址），否则它永远不会返回 true。也许这就是你想要的，但我怀疑不是。您需要覆盖该方法

public boolean equals(Object obj);

并使用它来比较给定类的两个实例。我的假设是否正确？

编辑好的，我认为我最初的猜测是正确的。如果它们来自同一个列表，它会起作用，因为它们最终是相同的元素（存储在相同的内存位置）。您需要使用equals()or!equals()而不是==and !=。试试看，看看它是否能解决你的问题。此外，不要只使用它们，您必须重写equals以实际比较元素的内部属性。

于 2011-04-13T01:14:51.330 回答