variadic template only using type parameter

Question

I would like to do something like this:

#include <iostream>

class a {

public:
    a() : i(2) {}

    template <typename ...ts>
    void exec() {
        f<ts...>();
        std::cout << "a::()" << std::endl;
    }

    int i;
private:

    template <typename t>
    void f() {
        i += t::i;
    }

    template <typename t, typename ...ts>
    void f() {
        f<t>();
        f<t, ts...>();
    }
};

struct b {
    static const int i = -9;
};

struct c {
    static const int i = 4;
};


int main()
{
    a _a;

    _a.exec<b,c>();

    std::cout << _a.i << std::endl;
}

The idea is to get the same information from a group of classes, without the need of an object of each class.

Does anyone know if it is possible?

Thanks!

Are you looking for std::integral_constant? Or maybe make exec have template <int ...ts> template arguments? — François Andrieux
– François Andrieux, Commented May 29, 2018 at 20:08
f<t, ts...>(); would do infinite recursive call, you probably mean f<ts...>(); — Jarod42
– Jarod42, Commented May 29, 2018 at 20:09
Jarod42, you are right. I posted the code without this correction. But, even them, it does not compile, reporting ambiguous call from two calls to the same 'f'. — canellas
– canellas, Commented May 29, 2018 at 20:11
Variant of your code, using tag parameter to select correct overload Demo. — Jarod42
– Jarod42, Commented May 29, 2018 at 21:00

bartop · Accepted Answer · 2018-05-29 20:14:58Z

5

In case Your compiler does not support C++17:

template <typename ...ts>
void f() {
    for ( const auto &j : { ts::i... } )
        i += j;
}

answered May 29, 2018 at 20:14

bartop

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4 Comments

Jarod42 Over a year ago

Which looks even better than folding expression IMO.

Piotr Siupa Over a year ago

@Jarod42 It is shorter, not more readable. Can someone explain to me how this works, btw?

bartop Over a year ago

@NO_NAME Long story short - it creates an initializer list from is from all the classes and iterates over them. It works ONLY because the is are all the same type

Piotr Siupa Over a year ago

I retract what I said. It only seem shorter because you didn't write code of the class. Although now, as I understand this, it strikes me that this loop is an exact equivalent to the @Jarod42 's code. C++17 just provides a shorter way (maybe also more flexible) to write such loops.

Jarod42 · Accepted Answer · 2018-05-29 20:12:52Z

3

In C++17, your class would simply be

class a {

public:
    a() : i(2) {}

    template <typename ...ts>
    void exec() {
        ((i += ts::i), ...); // Folding expression // C++17
        std::cout << "a::()" << std::endl;
    }

    int i;
};

Possible in C++11 too, but more verbose.

answered May 29, 2018 at 20:12

Jarod42

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Comments

Piotr Siupa · Accepted Answer · 2018-05-29 20:53:02Z

1

Reasons why your code is not compiling:

Syntax of specializing templates is a little different.
You need to put the most general case first.
You can't partially specialize functions, only classes.
Partial specialization is not allowed within classes, only in namespaces.

Here is an example for C++11.

#include <iostream>

template<typename t, typename ...ts>
class a {
public:
    static constexpr int x = t::i + a<ts...>::x;
};

template<typename t>
class a<t> {
public:
    static constexpr int x = 2 + t::i;
};

struct b {
    static constexpr int i = -9;
};

struct c {
    static constexpr int i = 4;
};

int main()
{
    constexpr int result = a<b,c>::x;
    std::cout << result << std::endl;
}

Remember that templates are calculated during compilation so, for optimization sake, it is a good idea to write them in a way that allows them to be constexpr.

answered May 29, 2018 at 20:53

Piotr Siupa

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2 Comments

Jarod42 Over a year ago

"Syntax of specializing templates is a little different": It is actually overloads and not specialization.

Piotr Siupa Over a year ago

@Jarod42 I think that overloading applies only if functions have different parameters.

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