Question

我需要检查并查看给定类型是否具有成员变量。但是，如果给出的类型不是类或结构，则会出现编译错误。

struct Vector {
    int x;
};

template <typename Type>
class has_member_x
{
   class yes { char m;};
   class no { yes m[2];};

   struct BaseMixin
   {
     int x;
   };

   struct Base : public Type, public BaseMixin {};

   template <typename T, T t>  class Helper{};

   template <typename U>
   static no deduce(U*, Helper<void (BaseMixin::*)(), &U::x>* = 0);
   static yes deduce(...);

public:
   static const bool result = sizeof(yes) == sizeof(deduce((Base*)(0)));

}; 

int main() {

    BOOST_STATIC_ASSERT(has_member_x<int>::result);
    BOOST_STATIC_ASSERT(has_member_x<Vector>::result);
}

当我尝试编译它时失败并出现以下错误。

错误：基类型'int'无法是结构或类类型

有没有办法可以在c ++或c ++ 0x中完成？

Answer 1

你可以使用boost :: enable_if和boost :: type_traits，特别是boost :: is_integral或boost :: is_class。

试试这个（我没有测试过，但这应该会给你一个想法。）：

struct Vector {
    int x;
};

template <typename Type, typename Enabled = void>
class has_member_x
{
public:
   static const bool result = false;
}; 

template <typename Type, typename Enabled = boost::enable_if<boost::is_class<Type> >::type > 
class has_member_x
{
   class yes { char m;};
   class no { yes m[2];};

   struct BaseMixin
   {
     int x;
   };

   struct Base : public Type, public BaseMixin {};

   template <typename T, T t>  class Helper{};

   template <typename U>
   static no deduce(U*, Helper<void (BaseMixin::*)(), &U::x>* = 0);
   static yes deduce(...);

public:
   static const bool result = sizeof(yes) == sizeof(deduce((Base*)(0)));

}; 

int main() {

    BOOST_STATIC_ASSERT(has_member_x<int>::result);
    BOOST_STATIC_ASSERT(has_member_x<Vector>::result);
}

Answer 2

修复原始has_member_x类中的错误，并按照其他人的建议为class类型添加专门化后，最终的工作版本如下所示：

#include <type_traits>

template <typename Type, bool Enabled = std::is_class<Type>::value>
class has_member_x
{
public:
   static const bool result = false;
}; 

template <typename Type>
class has_member_x<Type, true>
{
   class yes { char m;};
   class no { yes m[2];};

   struct BaseMixin
   {
     int x;
   };

   struct Base : public Type, public BaseMixin {};

   template <typename T, T t>  class Helper{};

   template <typename U>
   static no deduce(U*, Helper<int BaseMixin::*, &U::x>* = 0);
   static yes deduce(...);

public:
   static const bool result = sizeof(yes) == sizeof(deduce((Base*)(0)));
};

使用模板检查不在int上工作的成员变量

2 个答案: