template-specialization

问题 I only have an hpp file for a school assignment in C++ (I am not allowed to add a cpp file, declaration and implementation should be both written in the file). I wrote this code inside it: template<class T> class Matrix { void foo() { //do something for a T variable. } }; I would like to add another foo method, but this foo() will be specialized for only an <int> . I have read in some places that I need to declare a new specialization class for this to work. But what I want is that the

问题 I only have an hpp file for a school assignment in C++ (I am not allowed to add a cpp file, declaration and implementation should be both written in the file). I wrote this code inside it: template<class T> class Matrix { void foo() { //do something for a T variable. } }; I would like to add another foo method, but this foo() will be specialized for only an <int> . I have read in some places that I need to declare a new specialization class for this to work. But what I want is that the

来源： https://stackoverflow.com/questions/30518804/find-template-type-of-a-template-type-c

来源： https://stackoverflow.com/questions/64275733/access-checking-rules-for-template-argument-list-in-particularly-explicit-spec

问题 Here's what I mean: // test.h class cls { public: template< typename T > void f( T t ); }; - // test.cpp template<> void cls::f( const char* ) { } - // main.cpp int main() { cls c; double x = .0; c.f( x ); // gives EXPECTED undefined reference (linker error) const char* asd = "ads"; c.f( asd ); // works as expected, NO errors return 0; } This is completely fine, right? I started doubting this, because I just ran over the specialization of '...' after instantiation error, which was new to me.

问题 Here's what I mean: // test.h class cls { public: template< typename T > void f( T t ); }; - // test.cpp template<> void cls::f( const char* ) { } - // main.cpp int main() { cls c; double x = .0; c.f( x ); // gives EXPECTED undefined reference (linker error) const char* asd = "ads"; c.f( asd ); // works as expected, NO errors return 0; } This is completely fine, right? I started doubting this, because I just ran over the specialization of '...' after instantiation error, which was new to me.

问题 Please check out the following code snippet (pseudo code, didn't compile it): A.h template <typename T> void SubTest(T t) = delete; template <> void SubTest(int i) { cout << i; } template <typename T> class MyClass { public: void Test(T t) { SubTest(t); } } B.h class X{}; template <> void SubTest(X x) { cout << x; } As you can see I want a class template method to call a function template. That function is specialized in various ways. Some of the specializations are located in A.h, some in

问题 If I have a header foo.h which I include all over my project, it seems to work fine when all it contains is: template<typename T> void foo(const T param) { cout << param << endl; } But I get one definition rule (ODR) errors when I add a specalization to foo.h: template<> void foo(const bool param) { cout << param << endl; } Obviously I can solve this by inline 'ing the specialization. My question is, why do I need to? If the template doesn't violate ODR, why does a specialization? 回答1: An

问题 If I have a header foo.h which I include all over my project, it seems to work fine when all it contains is: template<typename T> void foo(const T param) { cout << param << endl; } But I get one definition rule (ODR) errors when I add a specalization to foo.h: template<> void foo(const bool param) { cout << param << endl; } Obviously I can solve this by inline 'ing the specialization. My question is, why do I need to? If the template doesn't violate ODR, why does a specialization? 回答1: An

问题 If I have a header foo.h which I include all over my project, it seems to work fine when all it contains is: template<typename T> void foo(const T param) { cout << param << endl; } But I get one definition rule (ODR) errors when I add a specalization to foo.h: template<> void foo(const bool param) { cout << param << endl; } Obviously I can solve this by inline 'ing the specialization. My question is, why do I need to? If the template doesn't violate ODR, why does a specialization? 回答1: An