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c++c++11constexprsfinae
Luka Mikec
Luka Mikec

Reputation: 435

SFINAE and sizeof vs constexpr

Sometimes people write something like the following (source):

template<typename T>
class is_class {
    typedef char yes[1];
    typedef char no [2];
    template<typename C> static yes& test(int C::*); // selected if C is a class type
    template<typename C> static no&  test(...);      // selected otherwise
  public:
    static bool const value = sizeof(test<T>(0)) == sizeof(yes);
};

Is there some reason not to replace such constructs (sizeof-based) with constexpr-based code, such as the following?

template<typename T>
class is_class {
    template<typename C> static constexpr bool test(int C::*) { return true; } // selected if C is a class type
    template<typename C> static constexpr bool test(...) { return false; }     // selected otherwise
public:
    static bool constexpr value = test<T>(0);
};

I know constexpr is a relatively new addition to the language, but is there any reason to prefer the first version other than having to use an old standard (pre-C++11)?

Upvotes: 3

Views: 252

Answers (1)

Dmitry Gordon
Dmitry Gordon

Reputation: 2324

Both options would work. But the difference between them is that the first one doesn't require C++11 while the second one does. And if you are free to use at least C++11 - there is no need to use any of them, there is already std::is_class in the standard library.

So if you see such code in some project then either this project is supposed to compile without C++11 support or it's some legacy.

Upvotes: 2

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