如何使用此链接列表实现？ [英] How do I use this Linked List implementation?

问题描述

我正在从Goodrich学习C ++的数据结构和算法。他们已经给出了这个LinkedList实现。我了解代码，但不是，但是我不能在主类中使用它。如何构造实例并进行插入，删除？例如，我试图按如下方式创建该类的实例：

I am learning data structures and algorithms for C++ from Goodrich. They have given this LinkedList implementation. I understand the code but I am not but I am not able to use this in main class. How do I construct an instance and make insert, delete? For example I tried to create an instance of the class as follows:

StringLinkedList（）L;

但显示错误：预期为;在‘L

But it is showing the error: expected ";" before 'L

#include <iostream>
#include <string>

using namespace std;

class StringNode {
private:
    string elem;
    StringNode* next;

    friend class StringLinkedList;
};

class StringLinkedList{
public:
    StringLinkedList();
    ~StringLinkedList();
    bool empty() const;
    const string& front() const;
    void addFront(const string& e);
    void removeFront();
private:
    StringNode* head;
};

StringLinkedList::StringLinkedList()
    :head(NULL) {}
StringLinkedList::~StringLinkedList()
    {while (!empty()) removeFront();}
bool StringLinkedList::empty() const
    {return head==NULL;}
const string& StringLinkedList::front() const
    {return head->elem;}
void StringLinkedList::addFront(const string& e){
    StringNode* v = new StringNode;
    v->elem=e;
    v->next = head;
    head=v;
}
void StringLinkedList::removeFront(){
    StringNode* old=head;
    head = old->next;
    delete old;
}


int main () {
}

推荐答案

括号（）表示函数调用。如果要声明变量，语法为

Brackets () indicate a function call. If you want to declare a variable the syntax is

Typename variable_name;

可选地，您可能需要将参数传递给构造函数

Optionally, you may need to pass parameters to a constructor

Typename variable_name(param);

在C ++ 11中，统一的初始化语法使您可以使用 {} 但我离题了。无论哪种方式，它们都在变量名之后 之后出现。
在您的情况下，这可行：

In C++11 the uniform initialisation syntax allows you to use {} but I digress. Either way they come after the variable name. In your case, this works:

StringLinkedList L;

When you say

StringLinkedList() L;

编译器看到一个类型名，然后期望一个变量名，但是得到（ ）在名称 L 之前（顺便说一句，它可能应该使用更长的名称），因此决定您必须进行函数调用，该函数调用应以a结尾分号。但这不是，它以 L; 结尾，所以您得到

the compiler sees a typename, then expects a variable name, but gets () before the name L (BTW - it might deserve a longer name) so decided you must be making a function call, which should end with a semicolon. But it doesn't, it ends with L; so you get

expected ";" before 'L

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