我最近发现使用std :: tuple<>的问题只有一个元素.我创建了一个类型擦除类,并保留了N个引用计数对象.但是,如果引用计数对象是std :: tuple<>中唯一的对象,则不会保留它.
难道我做错了什么?
class token {
public:
template<typename... Types>
token(Types... types) : _self(std::make_shared<const std::tuple<Types...>>(std::make_tuple(std::move(types)...))) {}
// Why do I need this special version of the constructor?
// Uncomment and the code will work!
//template<typename T>
//token(T t) : _self(std::make_shared<const T>(std::move(t))) {}
private:
std::shared_ptr<const void> _self;
};
示例(使用Xcode 8.0测试):
token make_token() {
std::shared_ptr<int> shared(new int(), [](int* i) {
// Called immediately if using only tuple constructor!
});
return token(shared);
}
token my_token = make_token(); // std::shared_ptr<> is already gone!
最佳答案 从我的角度来看,你的代码应该可以正常运行,msvc和gcc似乎在
this snippet中同意我的观点.来自T.C.评论,这看起来像clang的真正问题,并在clang trunk中修复
作为现在的解决方法,我建议采用这种方法,(special_decay_t取自cppreference):
#include <iostream>
#include <tuple>
#include <memory>
template <class T>
struct unwrap_refwrapper
{
using type = T;
};
template <class T>
struct unwrap_refwrapper<std::reference_wrapper<T>>
{
using type = T&;
};
template <class T>
using special_decay_t = typename unwrap_refwrapper<typename std::decay<T>::type>::type;
class token {
public:
template<typename... Types>
token(Types&&... types) : _self(std::make_shared<std::tuple<special_decay_t<Types>...>>(std::forward<Types>(types)...)) {}
private:
std::shared_ptr<void> _self;
};
token make_token() {
return token(std::shared_ptr<int>(new int(), [](int* i) {
std::cout << "freed\n";
delete i;
}));
}
int main()
{
token my_token = make_token();
std::cout << __LINE__ << '\n';
}
见demo