自動轉換爲const char []爲const char *的模板函數

Question

我在與模板，由此，如果你試圖給一個模板函數的字符串參數，編譯器的「Hello World」作爲爲const char [12]的問題。我希望它是const char *。

我可以通過靜態鑄造每個字符串爲「爲const char *」「解決」的問題，而是因爲我想用這個作爲一個記錄系統的一部分，使其成爲簡單是一個很大的目標。

自很難解釋我的意思，我想出了一個簡單的再現。你會看到主函數的最後一行不能編譯。

任何幫助，將不勝感激

#include <string> 

// Trivial base class so we can use polymorphism 
class StoreItemsBase 
{ 
public: 
    StoreItemsBase() {} 
}; 

// Example of a trivial Templated class to hold some 3 items. 
// Intent to have similar classes to hold 4,5..n items 
template <typename T1, typename T2, typename T3> 
class Store3Items : public StoreItemsBase 
{ 
public: 
    Store3Items(const T1& t1, const T2& t2, const T3& t3) 
    : 
    StoreItemsBase(), 
    mT1(t1), 
    mT2(t2), 
    mT3(t3) 
    {} 

private: 
    T1 mT1; 
    T2 mT2; 
    T3 mT3; 
}; 

// Function to create a pointer to our object with added id 
// There would be similar CreateHolderFunctions for 4,5..n items 
template <typename T1, typename T2, typename T3> 
StoreItemsBase* CreateHolder(const T1& t1, const T2& t2, const T3& t3) 
{ 
    return new Store3Items<T1, T2, T3>(t1, t2, t3); 
} 

int main() 
{ 
    int testInt=3; 
    double testDouble=23.4; 
    const std::string testStr("Hello World"); 

    StoreItemsBase* Ok1 = CreateHolder(testInt, testDouble, testStr); 
    StoreItemsBase* Ok2 = CreateHolder(testDouble, testStr, testInt); 
    StoreItemsBase* Ok3 = CreateHolder(testStr, static_cast<const char*>("Hello there"), testInt); 
    // If you try a standard string, it compiler complains 
    // Although I could surround all my strings with the static cast, what I am looking for is a way 
    // to for the CreateHolder function to do the work for me 
    StoreItemsBase* NotOk4 = CreateHolder(testStr, "Hello World", testInt); 

    // Free our objects not shown in the example 
} 

編譯錯誤是： example.cpp：在構造「Store3Items :: Store3Items（常量T1 &，常量T2 &，常量T3 &）[用T1 = std :: basic_string，T2 = char [12]，T3 = int]'：T1351 = std :: basic_string，T2 = char [12]，T3 = int]' exampl e.cpp：65：74：從這裏實例化 example.cpp：21：11：error：用作初始化程序的數組

Answer 1

您可以使用元函數將作爲參數傳遞的類型轉換爲模板。字符的任意陣列將被轉化爲char*：

template< typename T > struct transform 
{ 
    typedef T type; 
}; 

template< std::size_t N > struct transform<char[N]> 
{ 
    typedef char* type; 
}; 
template< std::size_t N > struct transform< const char[N] > 
{ 
    typedef const char* type; 
}; 

然後，而不是使用Tn直接您將使用typename transform<Tn>::type。

更新：如果您在C++ 11的工作，那麼std::decay已經做你想要什麼。

Answer 2

嘗試改變模板參數爲const T1 T1，T2常量T2，T3常量T3。這將是少高性能，但它編譯

決定基於模板參數函數參數是非常困難的。你可以嘗試這個類構造函數簽名。我使用類「arg_type」（非標準）的模板特化來確保所有非const指針的參數類型都是由const ref傳遞的，並且所有的const指針都是作爲const指針傳遞的。

另外，不要忘了虛析構函數的基類或不好的事情可能發生:)

#include <string> 

// Trivial base class so we can use polymorphism 
class StoreItemsBase 
{ 
public: 
    StoreItemsBase() {} 
    virtual ~StoreItemsBase() {} 
}; 

template <typename TYPE> class arg_type 
{ 
public: 
    typedef const TYPE& type; 
}; 
template <typename TYPE> class arg_type<const TYPE*> 
{ 
public: 
    typedef const TYPE* type; 
}; 

// Example of a trivial Templated class to hold some 3 items. 
// Intent to have similar classes to hold 4,5..n items 
template <typename T1, typename T2, typename T3> 
class Store3Items : public StoreItemsBase 
{ 
    typedef typename arg_type<T1>::type arg1; 
    typedef typename arg_type<T2>::type arg2; 
    typedef typename arg_type<T3>::type arg3; 
public: 
    Store3Items(arg1 t1, arg2 t2, arg3 t3) 
    : 
    StoreItemsBase(), 
    mT1(t1), 
    mT2(t2), 
    mT3(t3) 
    {} 

private: 
    T1 mT1; 
    T2 mT2; 
    T3 mT3; 
}; 

// Function to create a pointer to our object with added id 
// There would be similar CreateHolderFunctions for 4,5..n items 
template <typename T1, typename T2, typename T3> 
StoreItemsBase* CreateHolder(const T1 t1, const T2 t2, const T3 t3) 
{ 
    return new Store3Items<T1, T2, T3>(t1, t2, t3); 
} 

int main() 
{ 
    int testInt=3; 
    double testDouble=23.4; 
    const std::string testStr("Hello World"); 

    StoreItemsBase* Ok1 = CreateHolder(testInt, testDouble, testStr); 
    StoreItemsBase* Ok2 = CreateHolder(testDouble, testStr, testInt); 
    StoreItemsBase* Ok3 = CreateHolder(testStr, static_cast<const char*>("Hello there"), testInt); 
    // If you try a standard string, it compiler complains 
    // Although I could surround all my strings with the static cast, what I am looking for is a way 
    // to for the CreateHolder function to do the work for me 
    StoreItemsBase* NotOk4 = CreateHolder(testStr, "Hello World", testInt); 

    // Free our objects not shown in the example 
} 

請記住，你的類將被儲存原料爲const char *內部（未存儲std :: string），所以確保傳入的字符串的範圍比存儲的指針長。像你的例子中的常量字符串很好，因爲它們永遠活着。