CODEWITHSUNDEEP
c++templates
user1715664
user1715664

Reputation: 153

C++ using default values in template declaration

I have the following code...

#include <iostream>

using namespace std;

template<typename R, R V = R()> R X() { return V; }

int main()
{
    cout << boolalpha << X<bool>() << endl;    
    cout << boolalpha << X<bool, true>() << endl;

    cout << X<int>() << endl;
    cout << X<int, 5>() << endl;

    cout << X<void>() << endl;   // compiler error

    return 0;
}

...which works for the bool and int cases, but does not compile in the void case. Is there a way to handle this?

I know code like this is acceptable...

void F()
{
    return void();
}

...so need to get that behavior out of the template somehow.

Upvotes: 1

Views: 64

Answers (2)

user2249683
user2249683

Reputation:

You can create an invalid type (None) and specify the return type with type traits:

#include <iostream>

struct None {};
// It may not be reasonable o provide the operator:
inline std::ostream& operator << (std::ostream& stream, None) {
    return stream;
}

template<typename R>
struct Traits {
    typedef R return_type;
};

template<>
struct Traits<void> {
    typedef None return_type;
};

template<typename R>
typename Traits<R>::return_type X() { return typename Traits<R>::return_type(); }

template<typename R, typename Traits<R>::return_type V>
typename Traits<R>::return_type X() { return V; }

int main()
{
    std::cout << std::boolalpha << X<bool>() << std::endl;
    std::cout << std::boolalpha << X<bool, true>() << std::endl;

    std::cout << X<int>() << std::endl;
    std::cout << X<int, 5>() << std::endl;

    std::cout << X<void>() << std::endl;

    return 0;
}

Also, the function X is split into two, to avoid the problem with the default template parameter.

Upvotes: 0

Mankarse
Mankarse

Reputation: 40623

Use std::enable_if to choose between two function templates. Live Example:

#include <iostream>
#include <type_traits>
using namespace std;

template<typename R, R V = R()>
typename std::enable_if<!is_same<R, void>::value, R>::type X() { return V; }

template<typename R>
typename std::enable_if<is_same<R, void>::value, R>::type X() { return; }

int main()
{
    cout << boolalpha << X<bool>() << endl;    
    cout << boolalpha << X<bool, true>() << endl;

    cout << X<int>() << endl;
    cout << X<int, 5>() << endl;

    X<void>(); // You can't print `void` with standard iostreams...

    return 0;
}

Upvotes: 1

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