[英]Template function to handle nested std::any_of
考慮以下代碼:
#include <iostream>
#include <vector>
#include <algorithm>
struct Animal { virtual ~Animal() = default; };
struct Dog : Animal {};
struct Person {
std::vector<struct Toddler*> children;
const std::vector<Toddler*>& getChildren() const {return children;}
};
struct Toddler : Person {
std::vector<Animal*> pets;
const std::vector<Animal*>& getPets() const {return pets;}
};
int main() {
Person* bob = new Person;
Toddler* tom = new Toddler;
tom->pets.push_back(new Dog);
bob->children.push_back(tom);
std::vector<Person*> people = {tom, bob};
// Output: Does anybody in 'people' have a child that has a pet that is a dog?
std::cout << std::boolalpha << std::any_of(people.begin(), people.end(),
[](const Person* p) {return std::any_of(p->getChildren().begin(), p->getChildren().end(),
[](const Toddler* t) {return
std::any_of(t->getPets().begin(), t->getPets().end(),
[](const Animal* a) {return dynamic_cast<const Dog*>(a) != nullptr;});
});
}) << std::endl; // true
}
我的目標是使用模板函數重寫上面的輸出,如下所示:
std::cout << anyOf (people, &Person::getChildren, &Toddler::getPets,
[](const Animal* a) {return dynamic_cast<const Dog*>(a) != nullptr;}) << std::endl;
因此,僅需要指定一個lambda函數,其他所有內容也更易於讀寫。 這是我到目前為止的內容,但是它出了可怕的錯誤:
#include <type_traits>
template <typename...> struct AnyOf;
template <typename Predicate, typename... A>
bool anyOf (Predicate pred, const A&... a) {
return AnyOf<Predicate, A...>::execute (pred, a...);
}
template <typename Predicate, typename Container>
struct AnyOf<Predicate, Container> {
static bool execute (Predicate pred, const Container& c) {
return std::any_of (c.begin(), c.end(), [pred](const typename Container::value_type& x) {return pred(x);});
};
};
template <typename Predicate, typename Container, typename First, typename... Rest>
struct AnyOf<Predicate, Container, First, Rest...> : AnyOf<Predicate, typename std::result_of<First(void)>::type, Rest...> {
using Base = AnyOf<Predicate, typename std::result_of<First(void)>::type, Rest...>; // ???
static bool execute (Predicate pred, const Container& c, First first, Rest... rest) {
return std::any_of (c.begin(), c.end(), [=](const typename Container::value_type& x) {
return Base::execute (pred, (x->*first)(), rest...);}); // ???
};
};
有人可以幫我完成這個嗎? 還是完全想出一個新設計?
// Base case: only a predicate.
// Call std::any_of and we are done.
template<class Cont, class Pred>
bool anyOf(Cont&& c, Pred p) {
using std::begin; using std::end; // enable ADL
return std::any_of(begin(c), end(c), p);
}
// Recursive case: at least one Callable object to be applied to
// each element in the range.
// rest... contains both any remaining Callables and the predicate
template<class Cont, class F1, class... Rest>
bool anyOf(Cont&& c, F1 f, Rest... rest) {
return anyOf(std::forward<Cont>(c), [&](auto&& a){
return anyOf(std::ref(f)(std::forward<decltype(a)>(a)), rest...);
});
}
std::ref(f)(...)
利用std::reference_wrapper
的operator()
,它使用INVOKE
並處理指向成員的指針和其他Callable對象。
template <typename Predicate, typename Container, typename... MemberFunctions> struct AnyOf;
template <typename Predicate, typename... A>
bool anyOf (Predicate pred, const A&... a) {
return AnyOf<Predicate, A...>::execute (pred, a...);
}
template <typename Predicate, typename Container, typename First, typename... Rest>
struct AnyOf<Predicate, Container, First, Rest...> {
static bool execute (Predicate pred, const Container& container, First first, Rest... rest) {
const auto element = (typename std::remove_pointer<typename Container::value_type>::type().*first)();
using NextContainer = std::vector<typename decltype(element)::value_type>;
NextContainer nextContainer;
for (const typename Container::value_type& x : container)
std::copy ((x->*first)().begin(), (x->*first)().end(), std::back_inserter(nextContainer));
return AnyOf<Predicate, NextContainer, Rest...>::execute (pred, nextContainer, rest...);
};
};
template <typename Predicate, typename Container>
struct AnyOf<Predicate, Container> {
static bool execute (Predicate pred, const Container& c) {
return std::any_of (c.begin(), c.end(), [pred](const typename Container::value_type& x) {return pred(x);});
};
};
我歡迎更好的解決方案。
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