问题
Is there a way to use Substitution failure is not an error (SFINAE) for enum?
template <typename T>
struct Traits
{
}
template <>
struct Traits<A>
{
};
template <>
struct Traits<B>
{
enum
{
iOption = 1
};
};
template <T>
void Do()
{
// use Traits<T>::iOption
};
Then, Do<B>();
works and Do<A>();
fails. However, I can supply a default behavior when iOption does not exist.
So I separate out some part of Do to DoOption.
template <typename T, bool bOptionExist>
void DoOption()
{
// can't use Traits<T>::iOption. do some default behavior
};
template <typename T>
void DoOption<T, true>()
{
// use Traits<T>::iOption
};
template <T>
void Do()
{
// 'Do' does not use Traits<T>::iOption. Such codes are delegated to DoOption.
DoOption<T, DoesOptionExist<T> >();
};
Now, the missing piece is DoesOptionExist<T>
- a way to check whether iOption exists in the struct.
Certainly SFINAE works for function name or function signature, but not sure
it works for enum value.
回答1:
If you can use C++11, this is completely trivial:
template<class T>
struct has_nested_option{
typedef char yes;
typedef yes (&no)[2];
template<class U>
static yes test(decltype(U::option)*);
template<class U>
static no test(...);
static bool const value = sizeof(test<T>(0)) == sizeof(yes);
};
The C++03 version is (surprisingly) similar:
template<class T>
struct has_nested_option{
typedef char yes;
typedef yes (&no)[2];
template<int>
struct test2;
template<class U>
static yes test(test2<U::option>*);
template<class U>
static no test(...);
static bool const value = sizeof(test<T>(0)) == sizeof(yes);
};
Usage:
struct foo{
enum { option = 1 };
};
struct bar{};
#include <type_traits>
template<class T>
typename std::enable_if<
has_nested_option<T>::value
>::type Do(){
}
int main(){
Do<foo>();
Do<bar>(); // error here, since you provided no other viable overload
}
来源:https://stackoverflow.com/questions/8597394/substitution-failure-is-not-an-error-sfinae-for-enum