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Home/ Questions/Q 8109959
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T01:35:04+00:00 2026-06-06T01:35:04+00:00

Consider: template <typename Function, typename …Args> auto wrapper(Function&& f, Args&&… args) -> decltype(f(args…)) {

  • 0

Consider:

template <typename Function, typename ...Args>
auto wrapper(Function&& f, Args&&... args) -> decltype(f(args...)) {
//...
}

Is there a way to partially specialize the above template for all the cases where decltype(f(args...)) is a pointer?

EDIT:
I think it can be done with an template helper class which takes decltype(f(args...)) as template argument, and specialize the helper class. If you know better solutions let me know.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-06T01:35:05+00:00Added an answer on June 6, 2026 at 1:35 am

    An SFINAE-based solution:

    #include <type_traits>
    
    template<
        typename Functor
        , typename... Args
        , typename Result = decltype(std::declval<Functor&>()(std::declval<Args>()...))
        , typename std::enable_if<
            std::is_pointer<Result>::value
            , int
        >::type = 0
    >
    Result wrapper(Functor&& functor, Args&&... args)
    { /* ... */ }
    
    template<
        typename Functor
        , typename... Args
        , typename Result = decltype(std::declval<Functor&>()(std::declval<Args>()...))
        , typename std::enable_if<
            !std::is_pointer<Result>::value
            , int
        >::type = 0
    >
    Result wrapper(Functor&& functor, Args&&... args)
    { /* ... */ }
    

    You can adapt the test (here, std::is_pointer<Result>) to your needs.

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