Return of indicator to function



  • How can the index be returned to function from function?

    void f0(char) {}
    

    void (*)(char) f1() {return f0;} // В возвращаемом типе ошибка

    May I announce auto I'm not interested.



  • The following ways can be used to re-enter the function.

    (1) Clear indication of type.

    void (*f1())(char) { return f0; }
    

    (2) Synonym announcement through typedef

    typedef void (*TFunc)(char);
    TFunc f1() { return f0; }
    

    (3) Synonym announcement through using (sighs)c++11)

    using TFunc = void (*)(char);
    TFunc f1() { return f0; }
    

    (4) Semi-automatic definition of re-entry typec++11)

    auto f1() -> decltype(&f0) { return f0; }
    

    (5) Automatic definition of re-entry type (c)c++14)

    auto f1() { return f0; }
    

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