Reputation: 3574
I want to write a concept that checks if the type has a static method called foo
. That method will have a templated parameter (the function will be called multiple times later with different parameter types).
Because of that templated parameter, it's quite difficult to check it. For the start, I thought I only check if there is a member at all with that name.
The following code compiles with Clang, but doesn't compile with GCC, because it cannot resolve the address of the overloaded function T::foo
.
template <typename T>
concept HasFoo = requires { T::foo; };
class Bar {
public:
template <typename T>
static void foo(T t);
};
static_assert(HasFoo<Bar>);
How do you correctly check for the existence of a templated static method (working in Clang and GCC)?
And ideally, can you even check more than this? Like checking if the return type is void
, or if it is callable.
One way would be to include the templated type into the concept, but as I want to use the method with multiple different types. So checking with only one type, like in the following code, is not enough.
template <typename T, typename T2>
concept HasFoo = requires { T::template foo<T2>; };
static_assert(HasFoo<Bar, int>);
Upvotes: 0
Views: 1164
Reputation: 66230
How do you correctly check for the existence of a templated static method (working in Clang and GCC)? And ideally, can you even check more than this? Like checking if the return type is void, or if it is callable.
I do have some constraints on template arguments, for the sake of the simplified example in the question we can just assume it's an integer type.
To check if the class support a static template method foo()
that is callable with an integer and return void
, you can simply check
template <typename T>
concept HasFoo = std::is_same_v<decltype(T::foo(0)), void>;
If you also want to be sure that the foo()
method is a template one, I suppose you can also check that converting &T::foo
to different function pointer types you get different values, so (for example)
( (void*)(&T::template foo<int>)
!= (void*)(&T::template foo<long>))
Combining the two requirements,
template <typename T>
concept HasFoo = ( (void*)(&T::template foo<int>)
!= (void*)(&T::template foo<long>))
&& std::is_same_v<decltype(T::foo(0)), void>;
With
struct Bar1
{ template <typename T> static void foo (T) {} };
struct Bar2
{ static void foo (int) {} };
struct Bar3
{ template <typename T> static T foo; };
template <typename T>
T Bar3::foo;
struct Bar4
{ template <typename T> static int foo (T) { return 0; } };
you have
static_assert(HasFoo<Bar1>);
static_assert(not HasFoo<Bar2>); // not because foo() isn't template
static_assert(not HasFoo<Bar3>); // not because foo isn't callable
static_assert(not HasFoo<Bar4>); // not becasue foo() return int
Upvotes: 2