Java:如何在类内部引用另一个元素的类?

发布于 2024-10-31 18:31:04 字数 2317 浏览 7 评论 0原文

我不确定如何解释这一点,但基本上我试图引用 Listfront ,它是 Element A(可以来自任何列表)。但发生的情况是,当它遍历列表的元素时,它正在比较两个不同的列表,但最终不匹配。 ie 将包含前面 b 的原始列表与包含元素 A 的列表进行比较。现在我只是想知道如何将元素 A 的前面设置为 b 以便我可以比较它在哪里。

/*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;
        }

    }

I'm not to sure on how to explain this but basically I am trying to refer to the List Classes front which is of Element A(can be from any list). But what happens is that when it goes through the Elements of the list it is comparing from two different lists and ends up not matching. ie compares original list which contains the front b to list containing element A. Now I'm just wondering about how i would get the front of Element A set to b so that i can compare where it is.

/*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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甜味超标? 2024-11-07 18:31:04

因此,尽管我发表了评论,但我还是看到了一个明显的问题。不能对引用类型使用相等运算符。也就是说,除了基本类型(double、int 等)之外的任何类型。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。您需要重写该方法

public boolean equals(Object obj);

并使用它来比较给定类的两个实例。我的假设正确吗?

编辑 好的,我认为我最初的猜测是正确的。如果它们来自同一列表,则它会起作用,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()!equals() 而不是 ==!=。尝试一下,看看它是否能解决您的问题。另外,不要只是使用它们,您必须重写 equals 才能实际比较元素的内部属性。

So despite my comment, I see one glaring problem with this. You can't use equality operators on reference types. That is, anything other than a primitive type (double, int, etc). What happens is you're comparing the address of the instance and unless they are literally the same object (same address in memory), it isn't going to return true, ever. Maybe that's what you want, but I suspect not. You need to override the method

public boolean equals(Object obj);

and use that to compare two instances of a given class. Am I correct in my assumptions?

Edit Ok, I think my original guess was correct. It works if they are from the same list because they end up being the same elements (stored in the same memory location). You need to use equals() or !equals() rather than == and !=. Try that, and see if it solves your problems. Also, don't just use them, you must override equals to actually compare the elements internal properties.

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