c++ - 这里为什么使用引用接收返回值

Question

hash_map<string, list<uint32_t> > g_hmLimits;
...
list<uint32_t>& lsErrorTime = g_hmLimits[strDomain];
lsErrorTime.clear();
为什么返回值是：list<uint32_t>& 类型呢？
hash_map中的value存储的是一个list对象，并不是对象的指针，这个时候返回的Value是原来Value的副本吗？

Answer 1

First of all, I would like to emphasize that hash_map is not part of the C++ standard library, and there is no universal standard. What I said below are all based on C++11's unordered_map.
Overloading the unordered_map operator in [] is defined as

mapped_type& operator[] ( const key_type& k );
mapped_type& operator[] ( key_type&& k );

As you can see, what is returned here is a reference, so there is no problem in accepting it by reference.