Proper use of std::enable_if or how to replace it

c++, c++11, type-traits

Solution

Your usage is fine, and very much idiomatic.

The equivalent to Boost.MPL's `if_` would be `std::conditional`:

typedef typename std::conditional<
        std::is_floating_point<T>::value,
        std::uniform_real_distribution<T>,
        std::uniform_int_distribution<T>>::type dist_type;

Problem

Is this proper use of std::enable_if? It works, but is it correct? ``` //*.h file template <typename T> static typename std::enable_if<std::is_integral<T>::value, T>::type randomFrom(const T min, const T max); template <typename T> static typename std::enable_if<std::is_floating_point<T>::value, T>::type randomFrom(const T min, const T max); ``` . ``` //*.inl file template <typename T> inline typename std::enable_if<std::is_integral<T>::value, T>::type Math::randomFrom(const T min, const T max) { static std::default_random_engine re((unsigned long)time(0)); std::uniform_int_distribution<T> uni(min, max); return static_cast<T>(uni(re)); } template <typename T> inline typename std::enable_if<std::is_floating_point<T>::value, T>::type Math::randomFrom(const T min, const T max) { static std::default_random_engine re((unsigned long)time(0)); std::uniform_real_distribution<T> uni(min, max); return static_cast<T>(uni(re)); } ``` How can I rewrite it, to achieve cleaner interface? Like: ``` template <typename T> static T randomFrom(const T min, const T max); ``` BTW with boost I had something like: (i do not want to use boost) ``` typedef typename boost::mpl::if_< boost::is_floating_point<T>, boost::uniform_real<>, boost::uniform_int<>>::type dist_type; ``` and whole behavior was solved in single function. But there is nothing like `std::if` right?

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