我正在尝试定义一个模板,该模板将指定另一个类型T的存储类型。我想使用enable_if来捕获所有算术类型。以下是我对此的尝试,抱怨模板重新声明了2个参数。我尝试将第二个虚拟parm添加到主模板,但得到了不同的错误。怎么办呢?
#include <string>
#include <type_traits>
template <typename T> struct storage_type; // want compile error if no match
// template <typename T, typename T2=void> struct storage_type; // no joy
template <> struct storage_type<const char *> { typedef std::string type; };
template <> struct storage_type<std::string> { typedef std::string type; };
template <typename T, typename std::enable_if<std::is_arithmetic<T>::value>::type* = nullptr>
struct storage_type { typedef double type; };
// Use the storage_type template to allocate storage
template<typename T>
class MyStorage {
public:
typename storage_type<T>::type storage;
};
MyStorage<std::string> s; // uses std::string
MyStorage<const char *> s2; // uses std::string
MyStorage<float> f; // uses 'double'
答案 0 :(得分:9)
您可以通过向主模板添加第二个参数,然后专门匹配它来完成此操作;你是在正确的轨道上,但没有正确地做到这一点。
#include <string>
#include <type_traits>
// template <typename T> struct storage_type; // Don't use this one.
template <typename T, typename T2=void> struct storage_type; // Use this one instead.
template <> struct storage_type<const char *> { typedef std::string type; };
template <> struct storage_type<std::string> { typedef std::string type; };
// This is a partial specialisation, not a separate template.
template <typename T>
struct storage_type<T, typename std::enable_if<std::is_arithmetic<T>::value>::type> {
typedef double type;
};
// Use the storage_type template to allocate storage
template<typename T>
class MyStorage {
public:
typename storage_type<T>::type storage;
};
MyStorage<std::string> s; // uses std::string
MyStorage<const char *> s2; // uses std::string
MyStorage<float> f; // uses 'double'
// -----
struct S {};
//MyStorage<S> breaker; // Error if uncommented.