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考虑以下几点:如何在类模板之外定义类模板的构造函数模板?
template<typename R>
struct call {
template<typename F, typename... Args>
explicit call(F&& f, Args&&... args);
R result;
};
template<typename R, typename F, typename... Args>
call<R>::call(F&& f, Args&&... args)
: result(std::forward<F>(f)(std::forward<Args>(args)...)) { }
铛骂我:
utility.tpp:40:1: error: too many template parameters in template redeclaration template<typename R, typename F, typename... Args> ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ utility.hpp:36:5: note: previous template declaration is here template<typename R> ^~~~~~~~~~~~~~~~~~~~
我完全摸不着头脑。它有可能吗?