派生类中的类模板的部分专门化会影响基类 [英] Partial specialization of a class template in derived class affects base class
问题描述
我有一个元功能:
struct METAFUNCION
{
template<class T>
struct apply
{
typedef T type;
};
};
然后我定义一个助手:
template<class T1, class T2>
struct HELPER
{
};
然后我有了第二个元函数,它源于上面的METAFUNCTION并定义了apply struct的部分特化: / p>
And then I have second metafunction which derives from the METAFUNCTION above and defines partial specialization of apply struct:
struct METAFUNCION2 : METAFUNCION
{
template<class T1, class T2>
struct apply<HELPER<T1, T2> > : METAFUNCION::apply<T2>
{
};
};
到目前为止,很好 - 代码在g ++ 4.3.2下编译。所以我使用它如下:
So far, so good - the code compiles under g++ 4.3.2. So I used it like below:
#include <typeinfo>
#include <string>
#include <cstdlib>
#include <cxxabi.h>
template<typename T>
struct type_info2
{
static std::string name()
{
char *p = abi::__cxa_demangle(typeid(T).name(), 0, 0, 0);
std::string r(p);
free(p);
return(r);
}
};
#include <boost/mpl/apply.hpp>
#include <iostream>
int main()
{
std::cout <<
type_info2<boost::mpl::apply<METAFUNCION, int>::type>::name() <<
std::endl;
std::cout <<
type_info2<boost::mpl::apply<METAFUNCION, HELPER<float, double> >::type>::name() <<
std::endl;
std::cout <<
type_info2<boost::mpl::apply<METAFUNCION2, HELPER<float, double> >::type>::name() <<
std::endl;
return(0);
}
输出:
int
double
double
这让我感到惊讶,因为我预期:
That surprised me a bit as I expected:
int
HELPER<float, double>
double
$ b $ p现在,我知道上面的代码不能在Microsoft Visual C ++ 2008 (我不记住消息,但它是沿着线,我不能专门应用的结构内部METAFUNCTION2结构)。
Now, I know that code like above does not compile under Microsoft Visual C++ 2008 (I don't remeber the message but it was something along the lines that I cannot specialize apply struct inside METAFUNCTION2 struct).
所以我的问题是 - 这是g ++行为符合标准?我有一个强烈的感觉,这里有一些问题,但我不是100%肯定。
So my question is - is this g++ behaviour conformant with the standard? I have a strong feeling that there is something wrong here but I am not 100% sure.
对于好奇 - 我有我想象的行为,当我重新定义METAFUNCTION2这样:
For the curious - I have the behaviuor as I expected when I redefine METAFUNCTION2 this way:
struct METAFUNCION2 : METAFUNCION
{
template<class T>
struct apply : METAFUNCION::apply<T>
{
};
template<class T1, class T2>
struct apply<HELPER<T1, T2> > : METAFUNCION::apply<T2>
{
};
};
推荐答案
以下代码是非法的:
struct METAFUNCION2 : METAFUNCION
{
template<class T1, class T2>
struct apply<HELPER<T1, T2> > : METAFUNCION::apply<T2>
{
};
};
根据C ++标准14.7.3 / 3:
According to C++ Standard 14.7.3/3:
明确专门化的函数模板或类模板的声明应在
明确专门化声明的范围内。
A declaration of a function template or class template being explicitly specialized shall be in scope at the point of declaration of an explicit specialization.
EDIT:根据 Core Issue 727 此限制不适用于成员模板的部分特殊化。
According to Core Issue 727 this restriction does not apply to partial specializations of member templates.
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