通用类型的gcnew操作符 [英] gcnew operator for generic type

问题描述

我有以下简单类

generic<typename T> where T:IDbConnection ref class CDbConnection
{
private:    
    IDbConnection^m_db;

    ConnectionState^ m_originalConnState;
public:
    CDbConnection();
    bool Connect(String ^ connStr);
    bool Exists(int id);
    auto GetAllData(String^ tableStr);
    ~CDbConnection();   
    !CDbConnection();
};

这里是我的构造函数

generic<typename T> CDbConnection<T>::CDbConnection()
{
    m_db=gcnew T();
    m_originalConnState=m_db->State;
}

但是编译器抱怨< 1> gcnew T用于通用类型
<2> auto 键使用错误，因为函数期望返回类型

But the compiler complains <1> the gcnew T() can't be used for generic type <2> auto key in use is wrong as the function expects a trailing return type

推荐答案

为了实现通用性，必须将类定义更改为

In order to achieve genericity, you must change your class definition to

generic<typename T> where T:IDbConnection ref class CDbConnection
{
private:    

    T m_db;
    ConnectionState^ m_originalConnState;

public:

    CDbConnection();
    bool Connect(String ^ connStr);
    bool Exists(int id);
    auto GetAllData(String^ tableStr);
    ~CDbConnection();   
    !CDbConnection();
};

由于您已经约束了 T 至少 IDbConnection 它不能是其他的。
然后你的构造函数

As you are already constraining your T to be at least IDbConnection it can't be anything else. Then your constructor

generic<typename T> CDbConnection<T>::CDbConnection()
{
    m_originalConnState=m_db.State;
}

应该像您预期的那样工作。

should work like you intended.

EDIT

似乎你不能声明一个泛型的引用。如果将对象分配给堆栈，它将工作。
请参见此条目。

It seems you cannot declare a reference to a generic. If you assign the object to the stack it will work. See this entry.

// C3229.cpp
// compile with: /clr /c
generic <class T>
ref class C {
   T^ t;   // C3229
};

// OK
generic <class T>
ref class D {
   T u;
};

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