递归结束时的模板语法错误

Question

有人能告诉我下面的递归特化结束语法有什么问题吗？我以为我遵守了所有的规则.

#include <iostream>

template <typename StreamType = std::ostream, StreamType& stream = std::cout>
class StringList {
    template <typename...> class AddStrings;
  public:
    template <typename... Args> void addStrings (Args&&... args) {AddStrings<Args...>()(args...);}
};

template <typename StreamType, StreamType& stream>
template <typename First, typename... Rest>
class StringList<StreamType, stream>::AddStrings<First, Rest...> : AddStrings<Rest...> {
public:
    void operator()(First&& first, Rest&&... rest) {
        // do whatever
        AddStrings<Rest...>::operator()(std::forward<Rest>(rest)...);
    }
};

template <typename StreamType, StreamType& stream>
template <>
class StringList<StreamType, stream>::AddStrings<> {
    friend class StringStreamList;
    void operator()() const {}  // End of recursion.
};

int main() {
    StringList<> stringList;
//  stringList.addStrings ("dog", "cat", "bird");
}

我不明白错误信息:

Test.cpp:22:11: error: invalid explicit specialization before '>' token
 template <>
           ^
Test.cpp:22:11: error: enclosing class templates are not explicitly specialized
Test.cpp:23:39: error: template parameters not used in partial specialization:
 class StringList<StreamType, stream>::AddStrings<> {
                                       ^
Test.cpp:23:39: error:         'StreamType'
Test.cpp:23:39: error:         'stream'

Answer 1

这是重要的一点:

Test.cpp:22:11: error: enclosing class templates are not explicitly specialized

这意味着您无法定义作为非特定模板成员的事物的显式特化.

class StringList<StreamType, stream>::AddStrings<> {

这AddStrings是明确的专业,但StringList不是.这是不允许的.