英文:
std::unordered_map using a template class pointer as a key
问题
我是cpp的初学者，我有一些类似这样的代码：```cpp#include <iostream>#include <utility>#include <map>#include <unordered_map>template<typename T,typename EX>class Attachable{static std::unordered_map<T*,Attachable<T,EX>> attachCaches;std::unordered_map<std::string,EX> datas;Attachable(T* sender):m_attachedObj(sender){}bool operator==(const Attachable& other)const{return m_attachedObj == other.m_attachedObj;}public:static Attachable<T,EX> attach(T* sender){if(attachCaches.find(sender)==attachCaches.end()){attachCaches.insert(std::make_pair(sender,Attachable<T,EX>(sender)));}return attachCaches[sender];}void set(std::string key,EX properties){datas.insert(key,properties);}EX get(std::string key){return datas.at(key);}private:T* m_attachedObj;};class Tester{public :Tester(){}};int main(){Tester* a =new Tester();Attachable<Tester,std::string>::attach(a).set("test","typeOfInt");std::cout<< Attachable<Tester,std::string>::attach(a).get("test");}
当我尝试编译这段代码时，我会在emplace(*_First);的函数处：
template <class _Iter, class _Sent>void _Insert_range_unchecked(_Iter _First, const _Sent _Last) {for (; _First != _Last; ++_First) {emplace(*_First);}}
我尝试使用一些基本的指针，比如int，但仍然不起作用：
int* a = new int(4);Attachable<int,std::string>::attach(a).set("test","typeOfInt");std::cout<< Attachable<int,std::string>::attach(a).get("test");
<details><summary>英文:</summary>**I'm a beginner for cpp,and I have some codes like this:**```cpp#include <iostream>#include <utility>#include <map>#include <unordered_map>template<typename T,typename EX>class Attachable{static std::unordered_map<T*,Attachable<T,EX>> attachCaches;std::unordered_map<std::string,EX> datas;Attachable(T* sender):m_attachedObj(sender){}bool operator==(const Attachable& other)const{return m_attachedObj == other.m_attachedObj;}public:static Attachable<T,EX> attach(T* sender){if(attachCaches.find(sender)==attachCaches.end()){attachCaches.insert(std::make_pair((sender),Attachable<T,EX>(sender)));}return attachCaches[(sender)];}void set(std::string key,EX properties){datas.insert(key,properties);}EX get(std::string key){return datas.at(key);}private:T* m_attachedObj;};class Tester{public :Tester(){}};int main(){Tester* a =new Tester();Attachable<Tester,std::string>::attach(a).set("test","typeOfInt");std::cout<< Attachable<Tester,std::string>::attach(a).get("test");}
When I try to compile this code , I will receive an error at <xhash>,line 952,at the function called
emplace(*_First);
template <class _Iter, class _Sent>void _Insert_range_unchecked(_Iter _First, const _Sent _Last) {for (; _First != _Last; ++_First) {emplace(*_First);}}
I am trying to using some basic pointer like int,but this still not worked:
int* a =new int(4);Attachable<int,std::string>::attach(a).set("test","typeOfInt");std::cout<< Attachable<int,std::string>::attach(a).get("test");
答案1
得分: 3
attachCaches[(sender)] 可能需要默认构造一个 Attachable 对象，如果发现 sender 不在映射中。但是 Attachable 并没有提供默认构造函数。由于您实际上知道键始终在映射中，您可以这样实现 attach（也更有效，因为它消除了冗余的查找）：
static Attachable<T, EX>& attach(T* sender){auto it = attachCaches.find(sender);if (it == attachCaches.end()){it = attachCaches.emplace(sender, Attachable<T, EX>(sender)).first;}return it->second;}
请注意，函数现在通过引用返回 Attachable。您的原始代码返回映射元素的临时副本；然后在该副本上调用 set；然后销毁了该副本。映射中的原始值从未更新，因此随后的 get 没有找到传递给 set 的值。
英文:
attachCaches[(sender)] may need to default-construct an Attachabe object, if it turns out that sender is not in the map. But Attachable doesn't provide a default constructor. Since you actually know the key is always in the map, you can implement attach this way (also more efficient, as it eliminates redundant lookups):
static Attachable<T,EX>& attach(T* sender){auto it = attachCaches.find(sender);if (it == attachCaches.end()){it = attachCaches.emplace(sender, Attachable<T,EX>(sender)).first;}return it->second;}
Note also how the function now returns Attachable by reference. Your original code returned a temporary copy of the map element; then called set on that copy; then that copy was destroyed. The original in the map was never updated, so a subsequent get didn't find the value passed to set.
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