enable-if

问题 I have something this: template<typename T> class Image { Image(int w, int h, T defaultVal){ for(int i=0; i<h; i++) for(int j=0; j<w; j++) pixel(j, i) = defaultVal; } template<typename F> Image(int w, int h, F initializer){ for(int i=0; i<h; i++) for(int j=0; j<w; j++) pixel(j, i) = initializer(j, i); } // ... }; My intention is to be able to instantiate an Image like this: Image<int> img0(w, h, 0); // image of zeroes Image<int> imgF(w, h, [](int j, int i){ // checkerboard image return (j/10

问题 Why char* pstr="hello"; pushArg(pstr); still invoke such template? You see there is &&(!std::is_same<char, typename std::remove_cv_t<std::remove_pointer_t<T>>>::value) already. template <typename T, typename std::enable_if<(!std::is_same<lua_CFunction, T*>::value) && std::is_pointer<T>::value && (!std::is_same<std::string*, T>::value) && (!std::is_same<char, typename std::remove_cv<std::remove_pointer<T>>>::value) && (!std::is_same<unsigned char, typename std::remove_cv<std::remove_pointer<T>

来源： https://stackoverflow.com/questions/63094844/using-enable-if-to-prevent-declaration

来源： https://stackoverflow.com/questions/63094844/using-enable-if-to-prevent-declaration

来源： https://stackoverflow.com/questions/63094844/using-enable-if-to-prevent-declaration

问题 I've been trying to understand the way C++ selects templates. Namely, consider the following code sample: template <typename R> class Curious { public: template <typename T, typename std::enable_if<std::is_const<T>::value, int>::type = 33> void test1() {} template <typename T, typename std::enable_if<!std::is_const<T>::value, int>::type = 33> void test1() {} template <typename T, typename = typename std::enable_if<std::is_const<T>::value>::type> void test2() {} template <typename T, typename

问题 I'm trying to make a simple template function that given some parameters, it outputs them with a space in between. Some of those can be elements of an enum, and in that case I want to output its integer value. This is the code I have: #include <iostream> #include <type_traits> using std::cerr; using std::endl; // Output integer value if parameter is an element of an enum template<typename T, typename = typename std::enable_if_t<std::is_enum<T>::value>> constexpr std::ostream& debug(const T& a

问题 MSVC produces error ("function template has already been defined") for the following code: template<typename T, typename = std::enable_if_t<std::is_default_constructible<T>::value>> auto foo(T&& val) { return 0; } // note difference from above ---> ! template<typename T, typename = std::enable_if_t<!std::is_default_constructible<T>::value>> auto foo(T&& val) { return 0; } I thought it would work because there are mutually exclusive sfinae conditions. Can you help me with the hole in my

问题 MSVC produces error ("function template has already been defined") for the following code: template<typename T, typename = std::enable_if_t<std::is_default_constructible<T>::value>> auto foo(T&& val) { return 0; } // note difference from above ---> ! template<typename T, typename = std::enable_if_t<!std::is_default_constructible<T>::value>> auto foo(T&& val) { return 0; } I thought it would work because there are mutually exclusive sfinae conditions. Can you help me with the hole in my

问题 I can not get my head around the following problem. I don't even really know how I could approach it. Consider this code: struct fragment_shader { std::string mPath; }; struct vertex_shader { std::string mPath; }; template <typename T> T shader(std::string path) { return T{ path }; } To create the different structs, I can write the following: auto fragmentShader = shader<vertex_shader>("some_shader.frag"); auto vertexShader = shader<fragment_shader>("some_shader.vert"); I am wondering, if it