问题
Last time I'm founding many answers regarding SFINAE which suggest using void_t helper.
But I don't seem to understand what's so special about it in regard to:
decltype (..., void()).
Consider the example:
template <typename...>
using void_t = void;
template <typename T, typename = void>
struct has_foo : std::false_type {};
template <typename T>
struct has_foo <T, decltype (T().foo(), void())> : std::true_type {};
template <typename T, typename = void>
struct has_bar : std::false_type {};
template <typename T>
struct has_bar <T, void_t <decltype (T().bar())> > : std::true_type {};
class MyClass1
{
public:
int foo() { return 3; }
};
class MyClass2
{
public:
double bar() { return 5.4; }
};
int main() {
std::cout << has_foo<MyClass1>::value << std::endl;
std::cout << has_foo<MyClass2>::value << std::endl;
std::cout << has_bar<MyClass1>::value << std::endl;
std::cout << has_bar<MyClass2>::value << std::endl;
return 0;
}
The output is as expected for both traits, which makes me think that both implementations are the same. Am I missing something?
回答1:
It's a more expressive, less cumbersome way of saying the same thing.
That's it.
来源：https://stackoverflow.com/questions/35986886/difference-between-decltype-void-and-void-t