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15.5 — std::unique_ptr

At the beginning of the chapter, we discussed how use of pointers can lead to bugs and memory leaks in some situations. For example, this can happen when a function early returns, or throws an exception, and the pointer is not properly deleted.

Now that we’ve covered the fundamentals of move semantics, we can return to the topic of smart pointer classes. As a reminder, a smart pointer is a class that manages a dynamically allocated object. Although smart pointers can offer other features, the defining characteristic of a smart pointer is that it manages a dynamically allocated resource, and ensures the dynamically allocated object is properly cleaned up at the appropriate time (usually when the smart pointer goes out of scope).

Because of this, smart pointers should never be dynamically allocated themselves (otherwise, there is the risk that the smart pointer may not be properly deallocated, which means the object it owns would not be deallocated, causing a memory leak). By always allocating smart pointers statically (as local variables or composition members of a class), we’re guaranteed that the smart pointer will properly go out of scope when the function or object it is contained within ends, ensuring the object the smart pointer owns is properly deallocated.

C++11 standard library ships with 4 smart pointer classes: std::auto_ptr (which you shouldn’t use -- it’s being deprecated in C++17), std::unique_ptr, std::shared_ptr, and std::weak_ptr. std::unique_ptr is by far the most used smart pointer class, so we’ll cover that one first. In the next lesson, we’ll cover std::shared_ptr and std::weak_ptr.

std::unique_ptr

std::unique_ptr is the C++11 replacement for std::auto_ptr. It should be used to manage any dynamically allocated object that is not shared by multiple objects. That is, std::unique_ptr should completely own the object it manages, not share that ownership with other classes. std::unique_ptr lives in the <memory> header.

Let’s take a look at a simple example smart pointer example:

Because the std::unique_ptr is allocated on the stack here, it’s guaranteed to eventually go out of scope, and when it does, it will delete the Resource it is managing.

Unlike std::auto_ptr, std::unique_ptr properly implements move semantics.

This prints:

Resource acquired
res1 is not null
res2 is null
Ownership transferred
res1 is null
res2 is not null
Resource destroyed

Because std::unique_ptr is designed with move semantics in mind, copy initialization and copy assignment are disabled. If you want to transfer the contents managed by std::unique_ptr, you must use move semantics. In the program above, we accomplish this via std::move (which converts res1 into an r-value, which triggers a move assignment instead of a copy assignment).

Accessing the managed object

std::unique_ptr has an overloaded operator* and operator-> that can be used to return the resource being managed. Operator* returns a reference to the managed resource, and operator-> returns a pointer.

Remember that std::unique_ptr may not always be managing an object -- either because it was created empty (using the default constructor or passing in a nullptr as the parameter), or because the resource it was managing got moved to another std::unique_ptr. So before we use either of these operators, we should check whether the std::unique_ptr actually has a resource. Fortunately, this is easy: std::unique_ptr has a cast to bool that returns true if the std::unique_ptr is managing a resource.

Here’s an example of this:

This prints:

Resource acquired
I am a resource
Resource destroyed

In the above program, we use the overloaded operator* to get the Resource object owned by std::unique_ptr res, which we then send to std::cout for printing.

std::unique_ptr and arrays

Unlike std::auto_ptr, std::unique_ptr is smart enough to know whether to use scalar delete or array delete, so std::unique_ptr is okay to use with both scalar objects and arrays.

However, std::array or std::vector (or std::string) are almost always better choices than using std::unique_ptr with a fixed array, dynamic array, or C-style string.

Rule: Favor std:array, std::vector, or std::string over a smart pointer managing a fixed array, dynamic array, or C-style string

std::make_unique

C++14 comes with an additional function named std::make_unique(). This templated function constructs an object of the template type and initializes it with the arguments passed into the function.

The code above prints:

3/5
0/1

Use of std::make_unique() is optional, but is recommended over creating std::unique_ptr yourself. This is because code using std::make_unique is simpler, requires less typing (when used with automatic type deduction), and it resolves an exception safety issue that can result from C++ leaving the order of evaluation for function arguments unspecified.

Rule: use std::make_unique() instead of creating std::unique_ptr and using new yourself

Returning std::unique_ptr from a function

std::unique_ptr can be safely returned from a function by value:

In the above code, createResource() returns a std::unique_ptr by value. If this value is not assigned to anything, the temporary return value will go out of scope and the Resource will be cleaned up. If it is assigned (as shown in main()), move semantics will be employed to transfer the Resource from the return value to the object assigned to (in the above example, ptr). This makes returning a resource by std::unique_ptr much safer than returning raw pointers!

In general, you should not return std::unique_ptr by pointer (ever) or reference (unless you have a specific compelling reason to).

Passing std::unique_ptr to a function

If you want a function to use a std::unique_ptr without taking ownership of the resource, pass it by (const) reference:

The above program prints:

Resource acquired
I am a resource
Ending program
Resource destroyed

If you want the function to take ownership of the contents of the pointer, pass it by value. Note that because copy semantics have been disabled, you’ll need to use std::move to actually pass the variable in.

The above program prints:

Resource acquired
I am a resource
Resource destroyed
Ending program

Note that in this case, ownership of the Resource was transferred to takeOwnership(), so the Resource was destroyed at the end of takeOwnership() rather than the end of main().

std::unique_ptr and classes

You can, of course, use std::unique_ptr as a composition member of your class. This way, you don’t have to worry about ensuring your class destructor deletes the dynamic memory, as the std::unique_ptr will be automatically destroyed when the class object is destroyed. However, do note that if your class object is dynamically allocated, it is at risk for not being properly deallocated, in which case even a smart pointer won’t help.

Misusing std::unique_ptr

There are two easy ways to misuse std::unique_ptrs, both of which are easily avoided. First, don’t let multiple classes manage the same resource. For example:

While this is legal syntactically, the end result will be that both res1 and res2 will try to delete the Resource, which will lead to undefined behavior.

Second, don’t manually delete the resource out from underneath the std::unique_ptr.

If you do, the std::unique_ptr will try to delete an already deleted resource, again leading to undefined behavior.

Note that std::make_unique() prevents both of the above cases from happening inadvertently.

Quiz time

1) If your class has a smart pointer member, why should you try to avoid allocating objects of that class dynamically?

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2) Convert the following program from using a normal pointer to using std::unique_ptr:

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15.6 -- std::shared_ptr
Index
15.4 -- std::move

3 comments to 15.5 — std::unique_ptr

  • Bobix Louis

    Hi Alex,
      I was wondering why it is not allowed to pass a unique_ptr to a function by value(without using std::move) even though it allows to return a unique_ptr by value from a function.

    Since the copy constructors are not implemented for unique_ptr, ideally it should not allow somebody to return a unique_ptr by value from a function right?

    • Alex

      Great question. Function parameters are l-values (because they have an identifier). If you were to pass such a function an l-value std::unique_ptr, you’d be invoking copy semantics, which is disallowed by std::unique_ptr. You _can_ pass a unique_ptr to a function by value without using std::move as long as it’s an r-value unique_ptr. If your unique_ptr is an l-value, you have to convert it to an r-value using std::move to ensure move semantics are invoked.

      However, return values are treated as r-values, because they have no name. So when we return a unique_ptr from a function, there’s an implicit conversion to r-value that happens as part of the return process.

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