问题
According to the reference, the name of a non-type template parameter is optional, even when assigning a default value (see (1) and (2)). Therefore these template structs are valid:
template <int> struct Foo {};
template <unsigned long = 42> struct Bar {};
I haven't seen a possibility of accessing the values of the non-type parameters. My question is: What's the point of unnamed/anonymous non-type template parameters? Why are the names optional?
回答1:
What's the point of unnamed/anonymous non-type template parameters?
I can think of specialization:
template<int = 42>
struct Foo{
char x;
};
template<>
struct Foo<0> {
int x;
};
template<>
struct Foo<1> {
long x;
};
Then:
Foo<0> a; // x data member is int
Foo<1> b; // x data member is long
Foo<7> c; // x data member is char
Foo<> d; // x data member is char
回答2:
First, we can split declaration from definition. So name in declaration is not really helpful. and name might be used in definition
template <int> struct Foo;
template <unsigned long = 42> struct Bar;
template <int N> struct Foo {/*..*/};
template <unsigned long N> struct Bar {/*..*/};
Specialization is a special case of definition.
Then name can be unused, so we might omit it:
template <std::size_t, typename T>
using always_t = T;
template <std::size_t ... Is, typename T>
struct MyArray<std::index_sequence<Is...>, T>
{
MyArray(always_t<Is, const T&>... v) : /*..*/
};
or used for SFINAE
template <typename T, std::size_t = T::size()>
struct some_sized_type;
回答3:
Oh, you can access them!
template <int> struct Foo {};
template <int N>
int get(Foo<N>) {
return N;
}
int main() {
Foo<3> foo;
return get(foo);
}
This might be a bit contrived. But in general for some templates you don't want to name them and then it is convenient that you don't have to.
来源:https://stackoverflow.com/questions/59824884/whats-the-point-of-unnamed-non-type-template-parameters