SFINAE: detect existence of a template function that requires explicit specialization

问题

As a follow-up to my previous question, I am trying to detect the existence of a template function that requires explicit specialization.

My current working code detects non-template functions (thanks to DyP's help), provided they take at least one parameter so that dependent name lookup can be used:

// switch to 0 to test the other case
#define ENABLE_FOO_BAR 1

namespace foo {
  #if ENABLE_FOO_BAR
    int bar(int);
  #endif
}

namespace feature_test {
  namespace detail {
    using namespace foo;
    template<typename T> decltype(bar(std::declval<T>())) test(int);
    template<typename> void test(...);
  }
  static constexpr bool has_foo_bar = std::is_same<decltype(detail::test<int>(0)), int>::value;
  static_assert(has_foo_bar == ENABLE_FOO_BAR, "something went wrong");
}

^{(the ENABLE_FOO_BAR macro is just for testing purpose, in my real code I don't have such a macro available otherwise I wouldn't be using SFINAE)}

This also works perfectly with template functions when their template arguments can automatically be deduced by the compiler:

namespace foo {
  #if ENABLE_FOO_BAR
    template<typename T> int bar(T);
  #endif
}

However when I try to detect a template function that requires explicit specialization, the static_assert kicks in when foo::bar() exists:

namespace foo {
  #if ENABLE_FOO_BAR
    template<typename T, typename U> T bar(U);
  #endif
}

//...
// error: static assertion failed: something went wrong

Obviously the compiler can't deduce the template arguments of bar() so the detection fails. I tried to fix it by explicitly specializing the call:

template<typename T> decltype(bar<int, T>(std::declval<T>())) test(int);
//      explicit specialization  ^^^^^^^^

This works fine when foo::bar() exists (the function is correctly detected) but now all hell breaks loose when foo::bar() doesn't exist:

error: ‘bar’ was not declared in this scope
     template<typename T> decltype(bar<int, T>(std::declval<T>())) test(int);
                                   ^
error: expected primary-expression before ‘int’
     template<typename T> decltype(bar<int, T>(std::declval<T>())) test(int);
                                       ^
// lots of meaningless errors that derive from the first two

It seems my attempt at explicit specialization failed because the compiler doesn't know that bar is a template.

I'll spare you everything I tried to fix this and get straight to the point: how can I detect the existence of a function such as template<typename T, typename U> T bar(U); that requires explicit specialization in order to be instantiated?

回答1:

Following may help you:

// Helper macro to create traits to check if function exist.
// Note: template funcName should exist, see below for a work around.
#define HAS_TEMPLATED_FUNC(traitsName, funcName, Prototype)                          \
    template<typename U>                                                             \
    class traitsName                                                                 \
    {                                                                                \
        typedef std::uint8_t yes;                                                    \
        typedef std::uint16_t no;                                                    \
        template <typename T, T> struct type_check;                                  \
        template <typename T = U> static yes &chk(type_check<Prototype, &funcName>*); \
        template <typename > static no &chk(...);                                    \
    public:                                                                          \
        static bool const value = sizeof(chk<U>(0)) == sizeof(yes);                  \
    }

So with provided namespace with bar and without bar2

// namespace to test
namespace foo {
    template<typename T, typename U> T bar(U);
    // bar2 not present
}

Code which check the presence of bar<int, int> and bar2<int, int>.

// dummy class which should be never used
namespace detail {
    struct dummy;
}

// Trick, so the names exist.
// we use a specialization which should never happen
namespace foo {
    template <typename T, typename U>
    std::enable_if<std::is_same<detail::dummy, T>::value, T> bar(U);

    template <typename T, typename U>
    std::enable_if<std::is_same<detail::dummy, T>::value, T> bar2(U);
}

#define COMMA_ , // trick to be able to use ',' in macro

// Create the traits
HAS_TEMPLATED_FUNC(has_foo_bar, foo::bar<T COMMA_ int>, int(*)(int));
HAS_TEMPLATED_FUNC(has_foo_bar2, foo::bar2<T COMMA_ int>, int(*)(int));

// test them
static_assert(has_foo_bar<int>::value, "something went wrong");
static_assert(!has_foo_bar2<int>::value, "something went wrong");

来源：https://stackoverflow.com/questions/18936259/sfinae-detect-existence-of-a-template-function-that-requires-explicit-specializ

标签

c++

templates

c++11

sfinae

explicit-specialization