[英]Specialization of template template parameters
This is a sequel question (on a different topic) to an earlier question . 这是对早期问题的续集问题(关于不同的主题)。 The code below incorporates Dehstil's suggestion to use specialization. 下面的代码包含了Dehstil关于使用专业化的建议。
How should a function that has template template parameters be specialized? 具有模板模板参数的函数应该如何专门化?
The code below (in which the two specialization lines do not compile) make the question concrete. 下面的代码(两个专业化行不编译)使问题具体化。
#include <cassert>
template<typename S> struct PA1 {};
template<typename S> struct PA2 {};
template<typename S> struct PB {};
template<typename S> struct PC {};
template<typename S> struct A1 { typedef PA1<S> P; };
template<typename S> struct A2 { typedef PA2<S> P; };
template<typename S> struct B { typedef PB <S> P; };
template<typename S> struct C { typedef PC <S> P; };
template<typename S, template<typename> class T> char fn(typename T<S>::P);
template<typename S, template<typename> class T> char fn(typename T<S>::P)
{
return 'a';
}
template<typename S> char fn<B<S> >(B<S>::P) { return 'b'; }
template<typename S> char fn<C<S> >(C<S>::P) { return 'c'; }
int main()
{
PA1<int> pa1;
PA2<int> pa2;
PB<int> pb;
PC<int> pc;
assert( (fn<int, A1>(pa1)) == 'a' );
assert( (fn<int, A2>(pa2)) == 'a' );
assert( (fn<int, B>(pb)) == 'b' );
assert( (fn<int, C>(pc)) == 'c' );
}
It's important for the four function calls fn<...,...>() to have identical signatures when called since they will themselves reside in a template class that applies to the four classes A1/A2/B/C. 重要的是四个函数调用fn <...,...>()在调用时具有相同的签名,因为它们本身将驻留在适用于四个类A1 / A2 / B / C的模板类中。
Partial Specialization of function template is NOT allowed by the C++ Standard! C ++标准不允许部分专业化功能模板!
Overload your functions instead of specializing them. 重载您的功能而不是专门化它们。
Read the explanation as to Why Not Specialize Function Templates? 阅读有关为何不专业化功能模板的说明? by Herb Sutter 作者:Herb Sutter
Then read why to overload than specialize : Template Specialization and Overloading by Herb Sutter 然后阅读为什么超载而不是专门化: Herb Sutter的 模板专业化和重载
Write a class template call
and specialize them as: 编写一个类模板call
并将它们专门化为:
template<class S, template<typename> class T>
struct call
{
static char fn(typename T<S>::P &p)
{
return ::fn<S,T>(p);
}
};
template<class S>
struct call<S,B>
{
static char fn(typename B<S>::P &p)
{
return ::fn<S>(p);
}
};
template<class S>
struct call<S,C>
{
static char fn(typename C<S>::P &p)
{
return ::fn<S>(p);
}
};
Then you can use this class template to call all the functions uniformly as: 然后,您可以使用此类模板统一调用所有函数:
assert( (call<int, A1>::fn(pa1)) == 'a' );
assert( (call<int, A2>::fn(pa2)) == 'a' );
assert( (call<int, B>::fn(pb)) == 'b' );
assert( (call<int, C>::fn(pc)) == 'c' );
See the online demo : http://www.ideone.com/TISIT 请参阅在线演示: http : //www.ideone.com/TISIT
Note also the overloaded function templates in the complete solution at ideone.com (above link) 另请注意ideone.com上完整解决方案中的重载功能模板(上面的链接)
Functions can be only fully specialized. 功能只能完全专业化。 Use function overloading: 使用函数重载:
template<typename S> char fn(typename B<S>::P) { return 'b'; }
template<typename S> char fn(typename C<S>::P) { return 'c'; }
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