How to dereference an std::map member pointer inside a class?

Question

I'm currently learning C++ and ran across this that has me stumped. I have this class:

class MyClass {
  public:      
    std::map<int,int> *myMaps;
}

How do I dereference myMaps? This does not work.

int main() {
  MyClass *test = new MyClass();
  std::map<int,int> *testMap = new std::map<int,int>();
  (*testMap)[1] = 1;

  test->myMaps = testMap;
  std::cout << *test->myMaps[1] << std::endl;

As a follow up, what if I had a std map inside a map and how to dereference that?

class MyClass {
  public:      
    std::map<int,std::maps<int,int>> *myMaps;
}

Need brackets to enforce precedence. Just like you did when dereferencing the testMap — ChrisMM
– ChrisMM, Commented Dec 6, 2021 at 22:46
Is there a good reason why myMaps is a pointer? I can't think of any. This looks like something that Java programmers do when they start writing C++. — paddy
– paddy, Commented Dec 6, 2021 at 22:48
(*test->myMaps)[1], exactly the same pattern you used above. BTW, make sure to run it with valgrind and fix all the errors. You are leaking memory all over the place. new without delete is a recipe for tears and, in a way, so are both of them combined. Either nest myMaps directly into MyClass, no need for explicit dynamic allocation, or, if you insist on dynamically allocating an STL container (which is an antipattern), please use std::unique_ptr<std::map<int, int>>. Your future self will thank you and valgrind will praise you. — Andrej Podzimek
– Andrej Podzimek, Commented Dec 6, 2021 at 22:51
Another side note: Why not std::map<int,int> myMaps; in MyClass and then, in main(), std::map<int,int> testMap; testMap[1] = 1; test->myMaps = std::move(testMap);? Avoid raw pointers; you don’t need them in this case. — Andrej Podzimek
– Andrej Podzimek, Commented Dec 6, 2021 at 22:54

Remy Lebeau · Accepted Answer · 2021-12-06 23:00:45Z

How do I dereference myMaps?

The same way you dereference any other pointer. Exactly as you are doing in (*testMap)[1] = 1;

This does not work.

No, it doesn't, but this does:

test->myMaps = testMap;

std::cout << (*(test->myMaps))[1] << std::endl;
// or cleaner:
// std::cout << (*test->myMaps)[1] << std::endl;
// due to operator precedence...

Note the extra parenthesis around the pointer before invoking map::operator[] on the dereferenced std::map object.

As a follow up, what if I had a std map inside a map and how to dereference that?

First you dereference myMaps, as shown above, then you access the inner map by key, exactly the same way you access any value by key, eg:

(*(test->myMaps))[1][2] = ...;
//(*test->myMaps)[1][2] = ...;

cout << (*(test->myMaps))[1][2];`
//cout << (*test->myMaps)[1][2];`

That being said, there is no good reason to use a pointer to a std::map at all. Use this instead:

class MyClass {
  public:      
    std::map<int,int> myMaps;
};

int main() {
  MyClass test;
  std::map<int,int> testMap;
  testMap[1] = 1;

  test.myMaps = std::move(testMap);
  // or, prior to C++11:
  // test.myMaps.swap(testMap);

  std::cout << test.myMaps[1] << std::endl;
}

class MyClass {
  public:      
    std::map<int,std::map<int,int>> myMaps;
};

int main() {
  MyClass test;
  std::map<int,std::map<int,int>> testMap;
  testMap[1][2] = 1;

  test.myMaps = std::move(testMap);
  // test.myMaps.swap(testMap);

  std::cout << test.myMaps[1][2] << std::endl;
}

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