如何区分功能类型与SFINAE

Question

I'm reading book C++ Templates . It mentions SFINAE(substitution failure is not an error) principal can be used to detect function type. Code example:

template <typename T>
class IsFunctionT {
private:
   typedef char One;
   typedef struct { char a[2]; } Two;
   template<typename U> static One test(...);
   template<typename U> static Two test(U (*)[1]); // This test overloading I cannot understand 
public:
   enum { Yes = sizeof(IsFunctionT<T>::test<T>(0) == 1};
   enum { No = !Yes };
};

I understand its intent is to find functions which cannot be categorized as arrays, but how does it work with U (*)[1] . I have never seen this before.

Answer 1

The U( )[1] is an unnamed pointer to an array of 1 element. And the 0 is either treated as an int or a null pointer to an U ( )[1];

The sizeof is testing the function return type, the actual result of the test function isn't used as test() is never actually called, only its return type tested.

Answer 2

In a nutshell, SFINAE works by substituting a type and not giving a hard failure if the expression is ill-formed. For example you cannot have an array of functions, so if U is a function, the substitution will fail and that overload will be discarded. Now as a comment pointed out, the code you posted alone does not cover all cases. You need additional specializations, ie:

template<typename T>
class IsFunctionT<T&> {
  public:
    enum { Yes = 0 };
    enum { No = !Yes };
};

template<>
class IsFunctionT<void> {
  public:
    enum { Yes = 0 };
    enum { No = !Yes };
};

template<>
class IsFunctionT<void const> {
  public:
    enum { Yes = 0 };
    enum { No = !Yes };
};

Which I pulled from the freely available source code from the book .