C++ Correct way to pass a std::unique_ptr object as reference on a function argument

Question

I have a std::unique_ptr<T> object and a library function that takes T& as a parameter. This function will change T object data.

What is the better way to pass the std::unique_ptr<T> to that function ? Is the above code correct ? Is there a better approach ?

#include <iostream>
#include <string>
#include <memory>

class Test {

   public: 
       std::string message = "Hi";
};

void doSomething(Test& item)
{
    item.message = "Bye";
}

int main()
{
     std::unique_ptr<Test> unique = std::unique_ptr<Test>(new Test());

     std::cout << unique->message << " John." << std::endl;

     doSomething(*unique.get());

     std::cout << unique->message << " John." << std::endl;
}

*unique is enough.

Jarod42
– Jarod42

2015-12-03 10:37:02 +00:00
Commented Dec 3, 2015 at 10:37 — Jarod42
– Jarod42, Commented Dec 3, 2015 at 10:37

TartanLlama · Accepted Answer · 2015-12-03 10:36:59Z

11

You don't need to call get, just dereference:

doSomething(*unique);

It is best to pass around references to the stored object rather than passing the std::unique_ptr around by reference, as the lifetime and storage of an object shouldn't need to be communicated to every function using that object.

answered Dec 3, 2015 at 10:36

TartanLlama

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Comments

Dietmar Kühl · Accepted Answer · 2015-12-03 10:39:37Z

6

The standard library smart pointers are supposed to be used like raw pointers when it comes to dereferencing the pointer. That is, you'd simply use

std::unique_ptr<Test> unique(new Test());
std::cout << unique->message << " John.\n";
doSomething(*unique);

I included the declaration to show the simplified use and one output to emphasize not to use std::endl.

answered Dec 3, 2015 at 10:39

Dietmar Kühl

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