在应该返回引用的函数内调用 exit()

In a library I have a function which searches for a key in a database and return a non-const reference to an object. I want to handle the case in which the key is not found, which is typically caused by a mistake when calling the function. This situation is so bad that the program cannot continue, so I print a message to help to spot the bug and call exit(1). The problem is with the return statement which will never be executed in this case, but have to be there anyway. If it was a pointer I could just return nullptr; but with a reference? Should I do something like this pseudo code?

 Type & get(const Key & k) {
     if (my_db.key_exists(k)) {
       return my_db.at(k);
     }
     std::cerr << k << " not found
";
     exit(1); 
     return *(new Type(some_dummy_parameters));
 }

看起来好可怕！也许我应该避免这样的功能.请告诉我你的意见！