问题
In short:
Is there a way I can feed a General
templated class with something that only represent an enum
type? Something like:
template <typename T> struct General {};
struct EnumSpecific : General<any_enum_type> {};
<int>
is too much / does not work in my case.
My specific case:
- A templated
Holder
class handles any kind of data in a generic way. - An abstract
General
class implements specific algorithms relying on theHolder
s' behaviour. - Template specifications of
General
(likeIntSpecific
,DoubleSpecific
,StringSpecific
,MoreSophisticatedTypeSpecific
..) define how to deal with some concreteHolder
types. - How can I properly define an
EnumSpecific
specification?
Here is boilt down code making my problem occur:
// A templated value holder:
template <typename T>
class Holder {
public:
Holder(T const& t) : _value(t) {};
// generic methods
void generics() {};
// methods concerning the value:
void set(T const& t /*, setInfo */) {
// .. check for an actual change, notify buddies of the change..
_value = t;
};
T value(/*readInfo*/) {
// .. do stuff depending on how / why the value is read..
return _value;
};
private:
T _value;
};
// (in reality, all `generics` methods come from a parent, untemplated class)
// A generic process involving such `Holder`s:
template <typename T>
class General {
public:
typedef bool /* or anything */ KnownReturnTypes;
General(Holder<T>* const a
, Holder<T>* const b)
: _a(a)
, _b(b)
{};
void methods() {
// Use common behavior of all `Holder`'s
_a->generics();
// .. or methods that rely on the actual values:
KnownReturnTypes knr( valuedMethods() );
if (knr) {} else {}
// ...
};
// Use polymorphism to adapt to each situation..
virtual KnownReturnTypes valuedMethods() = 0;
protected:
Holder<T>* _a;
Holder<T>* _b;
};
// Example of specialization for integral types (there might be others)
class IntSpecific : General<int> {
public:
IntSpecific(Holder<int>* const a
, Holder<int>* const b)
: General<int>(a, b)
{};
// implement the valuedMethods:
virtual KnownReturnTypes valuedMethods() {
return _a->value() > _b->value(); // dummy
}
};
// Specialization for enum types:
// * * * class EnumSpecific : General<any_enum_type> { // does not exist * *
class EnumSpecific : General<int> {
public:
EnumSpecific( Holder<int>* const a
, Holder<int>* const b)
: General<int>(a, b)
{};
// only use properties and methods offered by an enum type:
virtual KnownReturnTypes valuedMethods() {
return _a->value() == _b->value(); // dummy
}
};
// One particular case
typedef enum {One, Two, Three} Enum;
typedef Holder<Enum> EnumHolder;
int main() {
// Check that `IntSpecific` works fine.
Holder<int>* i( new Holder<int>(3) );
Holder<int>* j( new Holder<int>(5) );
IntSpecific is(i, j); // ok.
// Try the `EnumSpecific`
EnumHolder* a( new EnumHolder { One } );
EnumHolder* b( new EnumHolder { Two } );
EnumSpecific es(static_cast<Holder<int>*>(a) // invalid cast
, static_cast<Holder<Enum>*>(b)); // unexpected type
// This is because the compiler doesn't know enough about what
// EnumSpecific actually *is*. How to tell him more about it?
return EXIT_SUCCESS;
}
What should I feed the template argument with in EnumSpecific : General<??>
to make things clear for the compiler?
Do I need to use some kind of enum_type
concept and more sophisticated tools from generic programming?
回答1:
We can accomplish this with std::enable_if and std::is_enum. As a sample this is a class that will take an enum type as a template parameter.
#include <type_traits>
enum Enum { FOO, BAR};
template<typename T, typename std::enable_if<std::is_enum<T>::value>::type* = nullptr>
class Test {};
int main()
{
Test<Enum> a; // ok
Test<double> b; // error
}
来源:https://stackoverflow.com/questions/32316875/give-a-generic-enum-type-as-template-argument