问题
Given the following classes:
template <typename T>
class Child : public T {};
I also have a templatized function:
template <typename T>
void foo(const T& bar)
After doing some template gymnastics I have a section of code that has determined that bar is a Child of some sort. But I need to find of what sort.
I want to be able to make a call on bar that yields the type inherited. So fake syntax to find the type of Parent would be:
decltype(foo.parent) bar;
Is there any actual syntax to accomplish this?
回答1:
As requested by the OP, here's an expanded version of my comment above:
One common way of handling this if Child and its siblings are under your control is to add a nested type to them:
template<typename T> class Child : public T
{
public:
using parent = T;
};
template<typename T> void foo(const T& bar)
{
typename T::parent goo;
}
If T is not directly available (in a lambda, for example), then this should work:
auto foo2 = [](const auto& bar)
{
typename std::remove_reference_t<decltype(bar)>::parent goo;
};
To make it slightly nicer and shorter,
template<typename T> using parent = typename T::parent;
template<typename T> void foo(const T& bar)
{
parent<T> goo;
}
auto foo2 = [](const auto& bar)
{
parent<std::remove_reference_t<decltype(bar)>> goo;
};
(Although parent is a relatively common name; beware of clashes. parent_t, maybe?)
For types that aren't under your control, if the set of types is known, parent<T> could also be an external type trait.
回答2:
Just take the type as your argument:
template <typename T>
void foo(const Child<T>& bar) {
}
bar is a Child that inherits from T. You can even add a separate overload for non-Child:
template <typename NonChild>
void foo(const NonChild& baz) { }
来源：https://stackoverflow.com/questions/31429363/get-the-type-of-a-parent