c++ - 如何获取 map 值的引用？

Question

通常 std::map::operator[] 返回对象的引用。

https://en.cppreference.com/w/cpp/container/map/operator_at

std::map<char*, char*>* myMap1 = new std::map<char*, char*>() ;
char* c1 = new char[100];
strcpy(c1,"fggf");
char* c2 = new char[100];
strcpy(c2 ,"gdgfd");
myMap1->insert(std::pair<char*,char*>(c1,c2));
char* v1=  (*myMap1)[c1];
char** v2=  &(*myMap1)[c1];

在这个程序中 (*myMap1)[c1] 没有返回引用，这个语法只有 &(*myMap1)[c1] 返回引用，为什么？

Answer 1

C++和C最好不要混用!当你可以使用字符串时，不要使用 char*(如果你有 C++17，则不要使用 string_view)

map<string, string> myMap; //C++ prefers value sementatics, don't go for pointers 
                            //unless absolutely necessary, and if it is...
auto myMapPtr = make_unique<map<string, string>>(); //Use smart pointers! 
myMap["fggf"] = "gdgfd"; //Lots of people don't know that [] also inserts, no need to construct a pair...

string c2 = MyMap.at("fggf"); //that's it!

就是这样，您只需要三行。不要混淆。 :)