Since C++20 concepts aren't standardized yet, I'm using static_assert
as a makeshift concept check, to provide helpful error messages if a type requirement isn't met. In this particular case, I have a function which requires that a type is callable before getting its result type:
template <typename F, typename... Args>
void example() {
static_assert(std::is_invocable_v<F, Args...>, "Function must be callable");
using R = std::invoke_result_t<F, Args...>;
// ...
}
In addition, I require that the callable's result must be some kind of std::optional
, but I don't know what type the optional will hold, so I need to get that type from it:
using R = // ...
using T = typename R::value_type; // std::optional defines a value_type
However, this will fail if type R
doesn't have a value_type
, e.g. if it's not a std::optional
as expected. I'd like to have a static_assert
to check for that first, with another nice error message if the assertion fails.
I could check for an exact type with something like std::is_same_v
, but in this case I don't know the exact type. I want to check that R
is some instance of std::optional
, without specifying which instance it must be.
One way to do that is with a helper trait:
template <typename T>
struct is_optional { static constexpr bool value = false; };
template <typename T>
struct is_optional<std::optional<T>> { static constexpr bool value = true; };
template <typename T>
constexpr bool is_optional_v = is_optional<T>::value;
…and then I can write:
static_assert(is_optional_v<R>, "Function's result must be an optional");
That works, but it seems a little awkward to pollute my namespace with a helper trait just for a one-off check like this. I don't expect to need is_optional
anywhere else, though I can imagine possibly ending up with other one-off traits like is_variant
or is_pair
too.
So I'm wondering: is there a more concise way to do this? Can I do the pattern matching on instances of std::optional
without having to define the is_optional
trait and its partial specialization?
Following the suggestion by several respondents, I made a re-usable trait:
template <typename T, template <typename...> typename Tpl>
struct is_template_instance : std::false_type { };
template <template <typename...> typename Tpl, typename... Args>
struct is_template_instance<Tpl<Args...>, Tpl> : std::true_type { };
template <typename T, template <typename...> typename Tpl>
constexpr bool is_template_instance_v = is_template_instance<T, Tpl>::value;
…so that I can write:
static_assert(is_template_instance_v<R, std::optional>, "Function's result must be an optional");
This is just as many lines and declarations as the is_optional
trait, but it's no longer a one-off; I can use the same trait for checking other kinds of templates (like variants and pairs). So now it feels like a useful addition to my project instead of a kluge.
Can I do the pattern matching on instances of
std::optional
without having to define theis_optional
trait and its partial specialization?
Maybe using implicit deduction guides for std::optional
?
I mean... something as
using S = decltype(std::optional{std::declval<R>()});
static_assert( std::is_same_v<R, S>, "R isn't a std::optional" );
Explanation.
When R
is std::optional<T>
for some T
type, std::optional{r}
(for an r
value of type R
) should call the copy constructor and the resulting value should be of the same type R
.
Otherwise, the type should be different (std::optional<R>
).
The following is a full compiling example.
#include <iostream>
#include <optional>
template <typename T>
bool isOptional ()
{
using U = decltype(std::optional{std::declval<T>()});
return std::is_same_v<T, U>;
}
int main ()
{
std::cout << isOptional<int>() << std::endl; // print 0
std::cout << isOptional<std::optional<int>>() << std::endl; // print 1
}
Anyway, I support the suggestion by super: create a more generic type-traits that receive std::option
as template-template argument.
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