如何允许用户确定模板类型？

Question

我编写了一个工作的链接队列，它根据它的数据类型进行模板化，但是用户可能正在以几种不同的类型之一输入数据。如何选择在运行时使用哪种数据类型？

如果我单独使用每种类型，它工作正常;我只需要覆盖所有可能性，而无需更改代码或重写每种数据类型的重载函数。

下面我提供了我的代码的相关部分。正如我所说，我的班级成员功能没有问题。

我已经尝试过创建队列的x类型版本的switch语句，但这不能用作交换机中的后续可能性与其他队列数据类型“矛盾”。我目前正在尝试if / else if语句，除了当我尝试使用x类型的输入时没有错误，它说它是未定义的。

// From Source.cpp

#include <iostream>
#include <string>
using namespace std;
#include "LQueue.h"
int mainMenu();
int main()
{
    int value;
    bool stop = false;
    Queue<int> *theQueue;
    int choice = mainMenu();

    if (choice == 1) {
        Queue<int> theQueue;
        int dataType;
    }
    else if (choice == 2) {
        Queue<double> theQueue;
        double dataType;
    }
    else if (choice == 3) {
        Queue<string> theQueue;
        string dataType;
    }
    else if (choice == 4) {
        Queue<char> theQueue;
        char dataType;
    }

    cout << "\n\nHow many items would you like to initially"
        << " populate the queue with? ";
    int howMany;
    cin >> howMany;

    for (int i = 0; i < howMany; i++)
    {
        cin >> dataType;
        theQueue.enqueue(dataType)
    }

    theQueue.display(cout);

    theQueue.dequeue();

    theQueue.display(cout);

    return 0;
}

int mainMenu()
{
    int choice;
    cout << "What type of data will you be storing in the queue?\n"
        << "1. integers\n2. decimal numbers\n3. words\n4. chars\n\n";

    cin >> choice;
    if (choice > 0 && choice < 5)
        return choice;

    cout << "\n\nInvalid choice\n\n";
    mainMenu();
}

// Guess I'll include shown functions from the Queue class file below

//--- Definition of enqueue()
template <typename QueueElement> 
void Queue<QueueElement>::enqueue(const QueueElement & value)
{
    if (empty())
    {
        myFront = myBack = new Node(value);
    }
    else
    {
        myBack->next = new Node(value);
        myBack = myBack->next;
    }
}

//--- Definition of dequeue()
template <typename QueueElement> 
void Queue<QueueElement>::dequeue()
{
    if (empty() == false)
    {
        Queue::NodePointer oldFront = myFront;
        myFront = myFront->next;
        delete oldFront;
    }
}

//--- Definition of display()
template <typename QueueElement> 
void Queue<QueueElement>::display(ostream & out) const
{
    Queue::NodePointer ptr;
    for (ptr = myFront; ptr != 0; ptr = ptr->next)
        out << ptr->data << "  ";
    out << endl;

}

//--- Definition of front()
template <typename QueueElement> 
QueueElement Queue<QueueElement>::front() const
{
    if (!empty())
        return (myFront->data);
    else
    {
        cerr << "*** Queue is empty "
            " -- returning garbage ***\n";
        QueueElement * temp = new(QueueElement);
        QueueElement garbage = *temp;     // "Garbage" value
        delete temp;
        return garbage;
    }
}

Compiler (visual studio 2017) is showing identifier "dataType" is undefined within the following loop:
```c++

    for (int i = 0; i < howMany; i++)
        {
            cin >> dataType;
            theQueue.enqueue(dataType);
        }

2个错误：E0020和C2065上的“cin >> dataType;”线，还有下一行的另一个C2065

也许有一种更有效的方式来做到这一点？我对所有建议持开放态度，谢谢！

Answer 1

问题（问题）是你写的时候

    if (choice == 1) {
        Queue<int> theQueue;
        int dataType;
    }
    else if (choice == 2) {
        Queue<double> theQueue;
        double dataType;
    }
    else if (choice == 3) {
        Queue<string> theQueue;
        string dataType;
    }
    else if (choice == 4) {
        Queue<char> theQueue;
        char dataType;
    }

你定义了四个不同的theQueue和四个不同的dataType变量，每个变量只在相应的if的相应体内有效。

所以，当你写作

    for (int i = 0; i < howMany; i++)
    {
        cin >> dataType;
        theQueue.enqueue(dataType)
    }

    theQueue.display(cout);

    theQueue.dequeue();

    theQueue.display(cout);

没有更多的dataType和theQueue可用（所有这些都超出了范围）。

我建议如下

    if (choice == 1) {
        foo<int>();
    }
    else if (choice == 2) {
        foo<double>();
    }
    else if (choice == 3) {
        foo<std::string>();
    }
    else if (choice == 4) {
        foo<char>();
    }

其中foo()是一个模板函数，几乎是这样的（警告：代码没有经过测试）

template <typename T>
void foo ()
 {
   Queue<T> theQueue;
   T        dataType;

   std::cout << "\n\nHow many items would you like to initially"
        << " populate the queue with? ";
   int howMany;
   std::cin >> howMany;

   for (int i = 0; i < howMany; i++)
    {
      std::cin >> dataType;
      theQueue.enqueue(dataType)
    }

   theQueue.display(cout);

   theQueue.dequeue();

   theQueue.display(cout);
 }

Answer 2

编写一个模板化成员函数，它可以执行您想要的操作：

template<class DataType>
void processInput(int howMany) {
    DataType value;

    for (int i = 0; i < howMany; i++)
    {
        cin >> value;
        theQueue.enqueue(value);
    }

    theQueue.display(cout);

    theQueue.dequeue();

    theQueue.display(cout);
}

Method 1 - switch statement

然后我们可以使用switch语句在main中选择它们：

int main()
{
    int choice = mainMenu();

    cout << "\n\nHow many items would you like to initially "
            "populate the queue with? ";
    int howMany;
    cin >> howMany;

    switch(choice) {
      case 1:
        processInput<int>(howMany);
        break;
      case 2:
        processInput<double>(howMany);
        break;
      case 3:
        processInput<string>(howMany);
        break;
      case 4:
        processInput<char>(howMany);
        break;
    }
}

Method 2 - array of methods

我们可以使用数组进行查找！

using func_t = void(*)(int);

int main() {

    std::vector<func_t> options = {
        processInput<int>, 
        processInput<double>, 
        processInput<string>, 
        processInput<char>
    };

    int choice = mainMenu();

    func_t selectedOption = options[choice - 1]; 

    cout << "\n\nHow many items would you like to initially "
            "populate the queue with? ";
    int howMany;
    cin >> howMany;

    selectedOption(howMany); 
}