c++ class template can be instantiated but a function template instantiation with the same template parameters fails -


i have wrapper class binded function calls (a helper class fight legacy code issues):

template <class result, class... args> class functionwrapper {     std::function<result()> func_; public:     functionwrapper(std::function<result(args...)> f, args&&... args) :         func_(std::bind(f, std::forward<args>(args)...))     {     }     //...some methods using func_ };

i can write following code, compiles , works fine:

double f(int i, double d) {     return i*d; } //... functionwrapper<double, int, double> w(f, 2, 4.5); //calling methods of w ...

now want save me typing when defining wrapper instance, i've introduced make_wrapper function:

template <class result, class... args> functionwrapper<result, args...> make_wrapper(std::function<result(args...)> f, args&&... args) {     return functionwrapper<result, args...>(f, std::forward<args>(args)...); }

although template parameter list function identical 1 of the wrapper class can not compiled (adding template parameters "help" compiler doesn't either):

auto w1=make_wrapper(f, 2, 4.5); //error: no matching function call 'make_wrapper', candidate template ignored: not match 'function<type-parameter-0-0 (type-parameter-0-1...)>' against 'double (*)(int, double)'  auto w2=make_wrapper<double, int, double>(f, 2, 4.5); //error: no matching function call 'make_wrapper', candidate template ignored: not match 'function<double (int, double, type-parameter-0-1...)>' against 'double (*)(int, double)'

the compiler llvm 6.1 (the current xcode one). so, going on here? possible fix make function?

the problem first argument make_wrapper() doesn't have type claim has. although function pointer convertible corresponding std::function<...> compiler won't use std::function<...> deduce template arguments. if you'd make nested type haveargs...be deduce other argument, theresult` type cannot deduced conversion.

if want bind function pointers, should work expect @ function pointer argument:

template <class result, class... args> functionwrapper<result, args...> make_wrapper(result (*f)(args...), args&&... args) {     return functionwrapper<result, args...>(f, std::forward<args>(args)...); }

when arguments of function pointer , arguments passed disagree, may necessary have separate template argument lists function arguments , parameters bound:

template <class result, class... fargs, class... args> functionwrapper<result, fargs...> make_wrapper(result (*f)(fargs...), args&&... args) {     return functionwrapper<result, fargs...>(f, std::forward<args>(args)...); }

i go alternative doesn't care actual type of function object argument , deduces whatever function type produced:

template <class fun, class... args> auto make_wrapper(fun fun, args&&... args)     -> functionwrapper<decltype(fun(std::forward<args>(args)...)), args...> {     return functionwrapper<decltype(fun(std::forward<args>(args)...)), args...>(f, std::forward<args>(args)...); }

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