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，但是因为我试图将它作为日志记录系统的一部分使用，所以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。

非常感谢任何帮助

#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＆amp;，const T2＆amp;，const T3＆amp;）[with T1 = std :: basic_string，T2 = char [12]，T3 = int]'： example.cpp：50：50：从'StoreItemsBase * CreateHolder（const T1＆amp;，const T2＆amp;，const T3＆amp;）[与T1 = std :: basic_string，T2 = char [12]，T3 = int]'实例化 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，不要直接使用Tn，而是使用typename transform< Tn >::type。

更新：如果您使用的是 C ++ 11 ，那么std::decay已经达到了您想要的效果。

Answer 2

尝试将模板参数更改为const T1 t1，const T2 t2，const T3 t3。它的性能会降低，但它会编译

基于模板参数确定函数参数通常很困难。您可以尝试使用类构造函数签名。我使用类“arg_type”（非标准）的模板特化来确保所有不是const指针的参数类型都由const ref传递，所有的const指针都作为const指针传递。

另外，不要忘记基类上的虚拟析构函数，否则可能会发生错误：）

#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
}

请记住，您的类将在内部存储原始const char *（不存储std :: string），因此请确保传入的字符串的范围将比您存储的指针更长。像你的例子中的常量字符串很好，因为它们永远存在。

在模板函数中自动将const char []转换为const char *

2 个答案: