what is the meaning of this line of code? [closed]

Here is the code.

std::shared_ptr<MyType> f() const { return f_; }

I understand that std::share_ptr is a smart pointer. MyType is a template parameter. f() is a function, right? const here means this function will be read-only. And then what is the relation between the function definition/body and this smart pointer? Many thanks for your comments. EDIT: I just realized that std::shared_ptr<MyType> is the function return type from all your comments/answers. Now, my question is, f_ is an object of std::shared_ptr<MyType>, is it correct? How do I know that? Do I need to declare the variable f_ anywhere? This is a piece of code from my working_on project and I did not find it very clear about the variable f_ or is there anything else I need to know about the relation between method f and variable f_?


The above code snippet can be read as:

f is a const member function that returns a std::shared_ptr<MyType> that is it returns a shared_ptr<> to MyType object.

As you already mentioned that the function is read-only, is there something else you want to ask, if there is you can edit your question. Also,note that the variable f_ that you’re returning from inside the body of function f is of type std::shared_ptr<MyType> .

Answer to your Edit

Do I need to declare the variable f_ anywhere?

The below example shows one possible scenario:

#include <iostream>
#include <memory>
class MyType
  int value = 0;
  //some other things here 
          std::cout<<"MyType Default constructor"<<std::endl;

class Name 
    std::shared_ptr<MyType> f_ = std::make_shared<MyType>(); //note that you can also write std::shared_ptr<MyType> f_{new MyType};
        std::shared_ptr<MyType> f() const //f is a const member function returning std::shared_ptr<MyType>
            std::cout<<"inside const member function f"<<std::endl;
            return f_; 

int main()
  const Name n;
    return 0;

The output of the program can be seen here.