Is template enable_if function implementation possible?

Question

Using c++14, I have some function declarations that resemble the following.

template <class... Args>
struct potato {
template <class T, class = std::enable_if_t<!std::is_same<T, int>::value>>
const T& blee(size_t blou) const;

template <class T, class = std::enable_if_t<std::is_same<T, int>::value>>
const T& blee(size_t blou) const;
};

Is it possible to implement the functions separately? From what I can tell, there is no way the compiler can figure out what is implementing what. For example :

template <class... Args>
template <class T, class>
const T& potato<Args...>::blee(size_t blou) const {
    // do something
}

template <class... Args>    
template <class T, class>
const T& potato<Args...>::blee(size_t blou) const {
    // do something
}

The enable_if information is lost at that point. Am I missing a trick in my toolbag to make this work? Note that I'd rather not use return type enable_if or argument enable_if as they are ungodly.

edit : Updated to represent my use-case better.

Whomever posted the question has already been answered, that is not true. The linked question doesn't even use declaration and implementation seperately. — scx, Nov 08 '18 at 15:50
So, your actual question is how to define an out of line function that uses SFINAE, right? — NathanOliver, Nov 08 '18 at 15:55
Then this is basically a dupe of [this](https://stackoverflow.com/questions/48480469/function-implementation-with-enable-if-outside-of-class-definition) or [this](https://stackoverflow.com/questions/43818137/is-out-of-line-sfinae-on-template-member-functions-possible) — NathanOliver, Nov 08 '18 at 16:04
@NathanOliver Ah, yes absolutely. Thanks for pointing out the correct dupes! — scx, Nov 08 '18 at 16:05

Nelfeal · Accepted Answer · 2018-11-08T16:09:50.020

2

You don't really need enable_if for that:

template<class T>
const T& blee(size_t blou) const {
    // do something
}

template<>
const int& blee<int>(size_t blou) const {
    // do something
}

Edit: since your functions are inside a class template, you will have to use tag dispatching:

template<class... Args>
struct potato {
    template<class T>
    void blee() const;

private:
    void realBlee(std::true_type) const;
    void realBlee(std::false_type) const;
};

template<class... Args>
template<class T>
void potato<Args...>::blee() const {
    realBlee(std::is_same<T, int>());
}

template<class... Args>
void potato<Args...>::realBlee(std::true_type) const {
    std::cout << "int\n";
}
template<class... Args>
void potato<Args...>::realBlee(std::false_type) const {
    std::cout << "generic\n";
}

Live on Coliru

or something similar, like a constexpr if:

template<class... Args>
struct potato {
    template<class T>
    void blee() const;

private:
    void intBlee() const;
};

template<class... Args>
template<class T>
void potato<Args...>::blee() const {
    if constexpr (std::is_same_v<T, int>) {
        intBlee();
    } else {
        std::cout << "generic\n";
    }
}

template<class... Args>
void potato<Args...>::intBlee() const {
    std::cout << "int\n";
}

Live on Coliru

edited Nov 08 '18 at 16:09

answered Nov 08 '18 at 15:34

Nelfeal

12,593
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Partial function specialization?.. – bipll Nov 08 '18 at 15:38
There is no partial specialization, @bipll – SergeyA Nov 08 '18 at 15:43
I'm trying to make this work, but it isn't. My functions are member functions, which I don't think makes them eligible. – scx Nov 08 '18 at 15:44
@scx They probably are, just show the real declaration. – Nelfeal Nov 08 '18 at 15:45
Its the same, in a variadic struct. Updating the question. – scx Nov 08 '18 at 15:46
@SergeyA I edited it out a few seconds after his comment – Nelfeal Nov 08 '18 at 15:46
Can you elaborate on tag dispatch? I've never used it. Unfortunately I don't have `if constexpr` since I'm using c++14 :/ – scx Nov 08 '18 at 16:03
@scx Here you go. Note that I take advantage of the fact that `std::is_same` is either a `std::true_type` or a `std::false_type` (instead of creating a redondant tag). – Nelfeal Nov 08 '18 at 16:11
@SergeyA Yes I'm pretty aware of that, thank you. (Though a similar effect is achieved via `if constexpr`.) – bipll Nov 09 '18 at 07:06

score 0 · Answer 2 · answered Nov 08 '18 at 15:35

0

The enable_if information is lost at that point.

It is not lost, it is int in both cases. Just one template is not being instantiated.

answered Nov 08 '18 at 15:35

Maxim Egorushkin

131,725
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Is template enable_if function implementation possible?

2 Answers2