使用C++模板切换算法
我有一个具有pointQuery函数的C++类。目前，当我必须在线性搜索和二进制搜索之间切换时，我会注释代码的另一部分。我看到了this的帖子，并试图重构我的代码。我的代码是这样的：
namespace searchStrategy{
struct linearSearch{};
struct binarySearch{};
}
template<typename T, typename LookupStrategy> class leaf:public rtNode<T>{
public:T points[maxCapLeaf][d]; // search to be done on this array
// some other class members, most of which used in the search
template<typename searchStrategy::linearSearch> bool pointQuery(const T* f) const{
// some code
}
template<typename searchStrategy::binarySearch> bool pointQuery(const T* f) const{
// some code
}
};类的对象被创建为：
leaf<T, LookupStrategy>* temp = new leaf<T, LookupStrategy>(some_input_params);其中LookupStrategy是searchStrategy::binarySearch或searchStrategy::linearSearch。
当我编译它时，我得到了以下错误：
/file_address/template.cpp:164:39: error: non-type template parameters of class type only available with ‘-std=c++2a’ or ‘-std=gnu++2a’
164 | template<typename searchStrategy::linearSearch> bool pointQuery(const T* f) const{
| ^~~~~~~~~~~~
/file_address/template.cpp:200:39: error: non-type template parameters of class type only available with ‘-std=c++2a’ or ‘-std=gnu++2a’
200 | template<typename searchStrategy::binarySearch> bool pointQuery(const T* f) const{
| ^~~~~~~~~~~~
/file_address/template.cpp:200:58: error: ‘template<class T, class LookupStrategy> template<<typeprefixerror><anonymous> > bool leaf<T, LookupStrategy>::pointQuery(const T*) const’ cannot be overloaded with ‘template<class T, class LookupStrategy> template<<typeprefixerror><anonymous> > bool leaf<T, LookupStrategy>::pointQuery(const T*) const’
200 | template<typename searchStrategy::binarySearch> bool pointQuery(const T* f) const{
| ^~~~~~~~~~
/file_address/template.cpp:164:58: note: previous declaration ‘template<class T, class LookupStrategy> template<<typeprefixerror><anonymous> > bool leaf<T, LookupStrategy>::pointQuery(const T*) const’
164 | template<typename searchStrategy::linearSearch> bool pointQuery(const T* f) const{
|有人能解释一下我做错了什么吗?我该如何解决这个问题？(请注意，函数需要保留在类本身中，我不能将其转移到名称空间，因为在搜索函数中使用了许多类成员)谢谢……
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