请问如何编写一个元函数来测量类中方法的大小？
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本文共计773个文字，预计阅读时间需要4分钟。
我正在尝试编写函数getSize()，它接受一些模板参数。尝试在此参数中查找方法或字段并返回size()或size.+。我的代码是：
我正在尝试编写函数getSize(),它接受一些模板参数,尝试在此参数中查找方法或字段并返回size()或size.我的代码是：
#include <iostream> #include <vector> #include <utility> #include <string> #include <type_traits> template <typename T> class has_size { private: typedef char Yes; typedef Yes No[2]; template <typename U, U> struct really_has; template<typename C> static Yes& Test(really_has <size_t (C::*)() const, &C::size>*); template<typename C> static Yes& Test(really_has <size_t (C::*)(), &C::size>*); template<typename> static No& Test(...); public: static bool const value = sizeof(Test<T>(0)) == sizeof(Yes); }; template <class T> size_t get_size(T t){ size_t res = 0; if(has_size<T>::value){ res = t.size(); }else{ res = t.size; } return res; } int main() { std::vector<float> v(10); std::cout << std::boolalpha << has_size<std::vector<float>>::value << std::endl; std::cout << std::boolalpha << has_size<std::string>::value << std::endl; size_t res = get_size(v); std::cout<< res; return 0; }
函数has_size在我的示例中正确执行,但是当我尝试调用getSize时出现错误：
prog.cpp: In function ‘int main()’: prog.cpp:47:24: error: the value of ‘v’ is not usable in a constant expression size_t res = get_size<v>; ^ prog.cpp:43:21: note: ‘v’ was not declared ‘constexpr’ std::vector<float> v(10); ^ prog.cpp:47:15: error: cannot resolve overloaded function ‘get_size’ based on conversion to type ‘size_t {aka long unsigned int}’ size_t res = get_size<v>; ^~~~~~~~~~~ 所以稍微升级你的代码:(对于 c++11)
struct MyStruct{ int size = 12; }; // This function will compile only if has_size is true template <class T, typename std::enable_if<has_size<T>::value, int>::type = 0> size_t get_size(const T& t){ return t.size(); } // This function will compile only if has_size is FALSE (check negation !has_size) template <class T, typename std::enable_if<!has_size<T>::value, int>::type = 0> size_t get_size(const T& t){ return t.size; } int main(){ std::vector<float> v(10); std::cout << get_size(v) << std::endl; MyStruct my; std::cout << get_size(my) << std::endl; return 0; }
关于std::enable_if的文件
所以我使用#4案例,通过模板参数启用.
因此,每个get_size函数的情况都将存在于最终程序中,具体取决于enable_if结果.所以编译器会忽略不满足我们的条件函数来编译.
所以稍微升级你的代码:(从c++17)
template <class T> size_t get_size(const T& t){ size_t res = 0; if constexpr(has_size<T>::value){ res = t.size(); }else{ res = t.size; } return res; } int main(){ std::vector<float> v(10); std::cout<< get_size(v) << std::endl; return 0; }
所以更少的代码和更可读:)
该解决方案使用C 17 if constexpr的特征
为什么您的解决方案不起作用：
if(has_size<T>::value){ // <--- this is compile time result (has_size<T>::value) so always true or always false depends on template argument which is deduced from argument type res = t.size(); // this need to compile always, so if it is vector then ok if something else that doesn't have such method will fail to compile }else{ res = t.size; // this need to compile always, again as above }
从较小的错误/改进：
>通过const&
本文共计773个文字，预计阅读时间需要4分钟。
我正在尝试编写函数getSize()，它接受一些模板参数。尝试在此参数中查找方法或字段并返回size()或size.+。我的代码是：
我正在尝试编写函数getSize(),它接受一些模板参数,尝试在此参数中查找方法或字段并返回size()或size.我的代码是：
#include <iostream> #include <vector> #include <utility> #include <string> #include <type_traits> template <typename T> class has_size { private: typedef char Yes; typedef Yes No[2]; template <typename U, U> struct really_has; template<typename C> static Yes& Test(really_has <size_t (C::*)() const, &C::size>*); template<typename C> static Yes& Test(really_has <size_t (C::*)(), &C::size>*); template<typename> static No& Test(...); public: static bool const value = sizeof(Test<T>(0)) == sizeof(Yes); }; template <class T> size_t get_size(T t){ size_t res = 0; if(has_size<T>::value){ res = t.size(); }else{ res = t.size; } return res; } int main() { std::vector<float> v(10); std::cout << std::boolalpha << has_size<std::vector<float>>::value << std::endl; std::cout << std::boolalpha << has_size<std::string>::value << std::endl; size_t res = get_size(v); std::cout<< res; return 0; }
函数has_size在我的示例中正确执行,但是当我尝试调用getSize时出现错误：
prog.cpp: In function ‘int main()’: prog.cpp:47:24: error: the value of ‘v’ is not usable in a constant expression size_t res = get_size<v>; ^ prog.cpp:43:21: note: ‘v’ was not declared ‘constexpr’ std::vector<float> v(10); ^ prog.cpp:47:15: error: cannot resolve overloaded function ‘get_size’ based on conversion to type ‘size_t {aka long unsigned int}’ size_t res = get_size<v>; ^~~~~~~~~~~ 所以稍微升级你的代码:(对于 c++11)
struct MyStruct{ int size = 12; }; // This function will compile only if has_size is true template <class T, typename std::enable_if<has_size<T>::value, int>::type = 0> size_t get_size(const T& t){ return t.size(); } // This function will compile only if has_size is FALSE (check negation !has_size) template <class T, typename std::enable_if<!has_size<T>::value, int>::type = 0> size_t get_size(const T& t){ return t.size; } int main(){ std::vector<float> v(10); std::cout << get_size(v) << std::endl; MyStruct my; std::cout << get_size(my) << std::endl; return 0; }
关于std::enable_if的文件
所以我使用#4案例,通过模板参数启用.
因此,每个get_size函数的情况都将存在于最终程序中,具体取决于enable_if结果.所以编译器会忽略不满足我们的条件函数来编译.
所以稍微升级你的代码:(从c++17)
template <class T> size_t get_size(const T& t){ size_t res = 0; if constexpr(has_size<T>::value){ res = t.size(); }else{ res = t.size; } return res; } int main(){ std::vector<float> v(10); std::cout<< get_size(v) << std::endl; return 0; }
所以更少的代码和更可读:)
该解决方案使用C 17 if constexpr的特征
为什么您的解决方案不起作用：
if(has_size<T>::value){ // <--- this is compile time result (has_size<T>::value) so always true or always false depends on template argument which is deduced from argument type res = t.size(); // this need to compile always, so if it is vector then ok if something else that doesn't have such method will fail to compile }else{ res = t.size; // this need to compile always, again as above }
从较小的错误/改进：
>通过const&