How to fix error refactoring decltype inside template

Question

edit Possibly can\'t be done, see Clean implementation of function template taking function pointer although answer 1 there has a C macro work-around https://stacko

Answer 1

I think for c++11 it is impossible to achieve what you want.

template<typename T, typename D>
class unique_gptr : public std::unique_ptr<T, decltype(&D)> {
    public: unique_gptr(T* t) : std::unique_ptr<T, decltype(&D)>(t, D) { };
};

using int_gptr = unique_gptr<int, ::free_int>;
int_gptr ig(new int(2));

Notice that decltype is applied on D, which is declared as a typename. So D is a type. But decltype can't be used on a type.

Firstly the code tries to get the address of a type (&). Secondly, the argument of decltype is expected to be an expression, but not a type or "the address of a type". To make it easier to understand, we can say that decltype expects its argument to be a “variable”, like the following example.

int main()
{
    int i = 10;
    decltype(i) j;
    decltype(int) k; /* Compiler error. */
    decltype(&int) l; /* Compiler error. */
    return 0;
}

You also want the compiler to replace D with ::free_int. And ::free_int is passed in as a non-type template argument. However D is a type template parameter. A non-type template parameter starts with an actual type (e.g. int, struct a* or a type template name). While a type template parameter starts with typename or class. An easier example for non-type template parameter,

template<int INIT>
void func2(void)
{
    decltype(INIT) j = INIT;
}

int main()
{
    func2<10>();
    return 0;
}

When you pass in a function pointer like ::free_int, you need a non-type template parameter, which must be preceded by a type. And you want the type of the function pointer to be "replaceable". So the type of the function pointer has to be a type template parameter. These make them two template parameters.

That's the reason you need three template parameters in template<typename T, typename D, D F>, which is the best result you have.

Answer 2

You can use some macro magic:

template<class T, class D, D d>
struct UniqueDeleter : public std::unique_ptr<T, D> {
    UniqueDeleter(T* t) : std::unique_ptr<T, D>(t, d) { };
};

template<class R, class T>
T decl_unique_deleter_ptr(R (*d)(T*));

#define DECL_UNIQUE_DELETER1(f) UniqueDeleter<decltype(decl_unique_deleter_ptr(&f)), decltype(&f), &f>
#define DECL_UNIQUE_DELETER2(T, f) UniqueDeleter<T, decltype(&f), &f>
#define DECL_UNIQUE_DELETER_GET_MACRO(_1, _2, NAME, ...) NAME
#define DECL_UNIQUE_DELETER_EXPAND(x) x
#define decl_unique_deleter(...) DECL_UNIQUE_DELETER_EXPAND(DECL_UNIQUE_DELETER_GET_MACRO( \
    __VA_ARGS__, DECL_UNIQUE_DELETER2, DECL_UNIQUE_DELETER1, _)(__VA_ARGS__))

Now you can use it like such:

decl_unique_deleter(int, free_int) ig(new int(2));

Or with using and some more magic:

using int_gptr = decl_unique_deleter(free_int); // If accepted pointer type matches.
int_gptr ig(new int(2));

---

I might also recommend an alternative solution:

auto i = new int(2);
BOOST_SCOPE_EXIT_ALL(&) { delete i; };

Answer 3

You cannot use decltype with a type as the goal is to obtain the type of a variable. But in decltype(&D), D is a type.

I would rather pass throught a template function:

template<typename T, typename D, D F>
class unique_gptr : public std::unique_ptr<T, D> {
    public: unique_gptr(T* t) : std::unique_ptr<T, D>(t, F) { };
};

template <typename T, typename D>
unique_gptr<T, D F> make_unique_gptr(T pointer, D deleter)
{
    return unique_gptr<T, D, deleter>(pointer);
}

auto ptr = make_unique(new int(2), ::free_int);