I know you already mentioned it but I would suggest using haml-js with Jammit. Simply include haml.js in your javascripts and in your assets.yml add template_function: Haml as well as including your template files in to a package. e.g. javascript_templates: – app/views/**/*.jst.haml Then in your views you can include this package (= include_javascripts :javascript_templates) and … Read more
I think it is referring to the following case: template <typename T> struct base { void foo() { std::cout << “generic” << std::endl; } void bar() { std::cout << “bar” << std::endl; } }; template <> void base<int>::foo() // specialize only one member { std::cout << “int” << std::endl; } int main() { base<int> i; … Read more
It’s actually quite straight forward. For t[1], overload resolution has these candidates: Candidate 1 (builtin: 13.6/13) (T being some arbitrary object type): Parameter list: (T*, ptrdiff_t) Candidate 2 (your operator) Parameter list: (TData<float[100][100]>&, something unsigned) The argument list is given by 13.3.1.2/6: The set of candidate functions for overload resolution is the union of the … Read more
There is a trick to get this to work with gcc. The feature isn’t fully implemented yet, but you can structure the code to avoid the unimplemented sections. Manually expanding a variadic template into a parameter list won’t work. But template specialization can do that for you. template< char head, char … rest > struct … Read more
Define the function in the .h file. Works fine for me a.h #include <vector> #include <iostream> using namespace std; class A { public: template< typename T> static double foo( vector<T> arr ); }; template< typename T> double A::foo( vector<T> arr ){ cout << arr[0]; } main.cpp #include “a.h” int main(int argc, char *argv[]) { A … Read more
Welcome to the world of Abominable Function Types. void() & is not a reference to void(). The way to spell that would be void(&)() (which if you remove_reference_t, you would get back void() — that is remove_reference_t does work on references to functions, if what you provide it is actually a reference to function type). … Read more
The type of pointer-to-member-function is different from pointer-to-function. The type of a function is different depending on whether it is an ordinary function or a non-static member function of some class: int f(int x); the type is “int (*)(int)” // since it is an ordinary function And int Mat::f2(int x); the type is “int (Mat::*)(int)” … Read more
It’s apparently this overload of operator<< that’s stepping in your way and making the expression in traling return type valid: template< class CharT, class Traits, class T > basic_ostream< CharT, Traits >& operator<<( basic_ostream<CharT,Traits>&& os, const T& value ); See (3) on this reference page. It’s a simple forwarder (calling os << value) that was … Read more
It’s to allow you to accrue state in your function and then return it to your calling code. For instance, your function (as a functor class) could have a member int for counting the number of times it had been called. Here is a page with some examples : https://web.archive.org/web/20171127171924/http://xenon.arcticus.com/c-morsels-std-for-each-functors-member-variables
Template parameters can be inferred for function templates when the parameter type can be deduced from the template parameters So it can be inferred here: template <typename T> void f(T t); template <typename T> void f(std::vector<T> v); but not here: template <typename T> T f() { return T(); } And not in class templates. So … Read more