如何返回指向链表前端节点的指针？ [英] How do I return a pointer to the front node of a linked list?

问题描述

您好，我有这个小问题...



我正在编写一个行车道并创建了一个名为getFront（）的函数，假设它返回指向车道前方的汽车。



不写所有代码，所以我只写下面的要点......

Hello, I have this minor problem...

I am programming a traffic lane and made a function called getFront(), which is suppose to return the pointer to the car at the front of the lane.

Not writing all the codes so I just wrote the main points below...

const Vehicles * TrafficSingleLane::getFront() const {

const Vehicles * car;
car = head;

if (head != NULL) {
return head;
}else return 0;

return 0;
}

所以我得到的错误是：无法将'TrafficSingleLane :: myNode * const'转换为'const Vehicles *。问题出在car = head部分......我不想知道代码，但想知道解决这个问题的步骤和建议，所以我可以自己试试。



我尝试了什么：



我没有使用这些代码，而是尝试了

So the error I am getting is: Cannot convert 'TrafficSingleLane::myNode * const' to 'const Vehicles *'. The problem is with the "car = head" part... I do not want to know the codes but wanting to know the steps and advice to fix this problem so I can try it myself.

What I have tried:

Instead of having those codes, I have tried

const Vehicles * TrafficSingleLane::getFront() const { 
return head;
}

但是有类似的错误。我在getFront（）之前有其他函数创建链表

But has similar error. I do have other functions before getFront() which creates the linked list

推荐答案

head 和汽车 类型。可能你应该发布相关的声明。

There is no trivial conversion between the head and car types. Probably you should post the relevant declarations.

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