在模板函数中自动将const char []转换为const char*

Question

我遇到模板问题,如果你尝试给模板化函数一个字符串参数,编译器会将"Hello World"解释为const char [12].我希望它是const char*.

我可以通过将每个字符串静态转换为'const char*'来"解决"这个问题,但由于我正在尝试将其用作日志系统的一部分,因此将其简单化是一个很大的目标.

由于很难解释我的意思,我想出了一个简单的复制器.您将看到main函数的最后一行无法编译.

任何帮助将不胜感激

#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
}

编译器错误是:example.cpp:在构造函数'Store3Items :: Store3Items(const T1&,const T2&,const T3&)[with T1 = std :: basic_string,T2 = char [12],T3 = int]':example.cpp :50:50:从'StoreItemsBase*CreateHolder实例化(const T1&,const T2&,const T3&)[与T1 = std :: basic_string,T2 = char [12],T3 = int]'example.cpp:65:74:从这里实例化example.cpp:21:11:error:用作初始化程序的数组

Answer 1

您可以使用元函数将作为参数传递的类型转换为模板.任何字符数组都将转换为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;
};

然后,而不是Tn直接使用,你会使用typename transform< Tn >::type.

更新:如果你在C++ 11中工作,那么std::decay已经做了你想要的.