C++ 函数返回链表

Question

我被要求做的事情:Q) 一个句子由一系列以句号结尾的字符组成。编写一个返回字符链表的函数，其中字符由用户键入并添加到列表中。返回的列表应该包括句号。它应该被称为:

LinkedList<char> *sentence;
sentence = setUpSentence();

我曾尝试编写其中的一些内容，但我正在努力，因为这是我第一次使用这样的链表和函数。

主文件

#include "LinkedList.h"
    #include "ListNode.h"
    #include "Node.h"
    #include <iostream>
    #include <stdlib.h>
    using namespace std;

    LinkedList<char> *setUpSentence() {
        //allocate the linked list objet
        LinkedList<char> *sentence = new LinkedList<char>();
        char ch;
        do {
            cout << "Enter characters to add, enter full stop to finish adding." << endl;
            ch = cin.get();
            sentence->addAtEnd(ch);
        } while (ch != '.');

        return sentence;
    }

    int main() {

        //call the function, store the returned pointer in sentence variable
        LinkedList<char> *sentence = setUpSentence();
        //working with the linked list
        sentence = setUpSentence();



        cout << sentence->getAtFront() << endl;

//delete to avoid memory leak
        delete sentence;
    }

我在开始编写此函数时尝试运行此尝试时遇到的错误是:

after entering 1 character into the console and pressing enter, the loop continues and outputs "enter characters to add...." but this appears twice each time after entering a character?

有什么想法吗？

我的 main 中使用的 LinkedList.h 文件中的函数代码:

template <typename T>
void LinkedList<T>::addAtEnd(T item)
{
    if (size == 0)
        addAtFront(item);
    else
    {
        // ListNode<T>* temp = findAt(size - 1);
        ListNode<T> *l = new ListNode<T>(item, last, nullptr);
        last->next = l;
        last = l;
        size++;
    }
}

template <typename T>
T LinkedList<T>::getAtFront()
{
    if (size > 0)
    {
        current = first;
        return first->item;
    }
    else return NULL;
}

编辑:LinkedList.h 文件中的 addAtFront 方法

template <typename T>
void LinkedList<T>::addAtFront(T item)
{
    ListNode<T> *l = new ListNode<T>(item, NULL, first);
    first = l;
    if (last == NULL)
        last = l;
    size = 1;
}

Answer 1

我不知道你为什么要在返回语句中递归调用 setUpSentence()，这没有意义。我认为它应该看起来像这样:

LinkedList<char> * setUpSentence() {
    // make sure to allocate the linked list object
    LinkedList<char> *sentence = new LinkedList<char>();
    char ch;
    do {
        cout << "Enter characters to add, enter full stop to finish adding." << endl;
        ch = cin.get();
        sentence->addAtEnd(ch);
    } while (ch != '.');

    return sentence;
}

int main() {
    // calling the function, store the returned pointer in sentence variable
    LinkedList *sentence = setUpSentence();

    // ... working with the linked list ...

    delete sentence;  // don't forget to delete it to avoid memory leak!
}