C++17: std::any

When trying to implement something that will store a value of an unknown data type (to be as generic as possible, for example), we had these possibilities before C++17:

  • Having a void* pointer to something that will be assigned at runtime. The problem with this approach is that it leaves all responsibility for managing the lifetime of the data pointed to by this void pointer to the programmer. Very error prone.
  • Having a union with a limited set of data types available. We can use still use this approach using C++17 variant.
  • Having a base class (e.g. Object) and store pointers to instances derived of that class (à la Java).
  • Having an instance of template typename T (for example). Nice approach, but to make it useful and generic, we need to propagate the typename T throughout the generic code that will use ours. Probably verbose.

So, let’s welcome to std::any.

std::any, as you already guess it, is a class shipped in C++17 and implemented in header <any> that can store a value of any type, so, these lines are completely valid:

std::any a = 123;
std::any b = "Hello";
std::any c = std::vector<int>{10, 20, 30};

Obviously, this is C++ and you as user need to know the data type of what you stored in an instance of std::any, so, to retrieve the stored value you have to use std::any_cast<T> as in this code:

#include <any>
#include <iostream>

int main()
    std::any number = 150;
    std::cout << std::any_cast<int>(number) << "\n";

If you try to cast the value stored in an instance of std::any to anything but the actual type, a std::bad_any_cast exception is thrown. For example, if you try to cast that number to a string, you will get this runtime error:

terminate called after throwing an instance of 'std::bad_any_cast'
  what():  bad any_cast

If the value stored in an instance of std::any is an instance of a class or struct, the compiler will ensure that the destructor for that value will be invoked when the instance of std::any goes of scope.

Another really nice thing about std::any is that you can replace the existing value stored in an instance of it, with another value of any other type, for example:

std::any content = 125;
std::cout << std::any_cast<int>(content) << "\n";

content = std::string{"Hello world"};
std::cout << std::any_cast<std::string>(content) << "\n";

About lifetimes

Let’s consider this class:

struct A
  int n;
  A(int n) : n{n} { std::cout << "Constructor\n"; }
  ~A() { std::cout << "Destructor\n"; }
  A(A&& a) : n{a.n} { std::cout << "Move constructor\n"; }
  A(const A& a) : n{a.n} { std::cout << "Copy constructor\n"; }
  void print() const { std::cout << n << "\n"; }

This class stores an int, and prints it out with “print”. I wrote constructor, copy constructor, move constructor and destructor with logs telling me when the object will be created, copied, moved or destroyed.

So, let’s create a std::any instance with an instance of this class:

std::any some = A{4516};

This will be the output of such code:

Move constructor

Why two constructors and two destructors are invoked if I only created one instance?

Because the instance of std::any will store a copy (ok, in this case a “moved version”) of the original object I created, and while in my example it may be trivial, in a complex object it cannot be.

How to avoid this problem?

Using std::make_any.

std::make_any is very similar to std::make_shared in the way it will take care of creating the object instead of copying/moving ours. The parameters passed to std::make_any are the ones you would pass to the object’s constructor.

So, I can modify my code to this:

auto some = std::make_any<A>(4517);

And the output will be:


Now, I want to invoke to the method “print”:

auto some = std::make_any<A>(4517);

And when I do that, the output is:

Copy constructor

Why such extra copy was created?

Because std::any_cast<A> returns a copy of the given object. If I want to avoid a copy and use a reference, I need to explicit a reference in std::any_cast, something like:

auto some = std::make_any<A>(4517);

And the output will be:


It is also possible to use std::any_cast<T> passing a pointer to an instance of std::any instead of a reference.

In such case, if the cast is possible, will return a valid pointer to a T* object, otherwise it will return a nullptr. For example:

auto some = std::make_any(4517);
std::cout << std::any_cast<int>(&some) << "\n";

In this case, notice that I am passing a pointer to “some” instead of a reference. When this occurs, the implementation returns a pointer to the target type if the stored object is of the same data type (as in the second line) or a null pointer if not (as in the third line, where I am trying to cast my object from type A to int). Using this version overloaded version with pointers avoids throwing an exception and allows you to check if the returned pointer is null.

std::any is a very good tool for storing things that we, as implementers of something reusable, do not know a priori; it could be used to store, for example, additional parameters passed to threads, objects of any type stored as extra information in UI widgets (similar to the Tag property in Windows.Forms.Control in .NET, for example), etc.

Performance wise, std::any needs to store stuff in the heap (this assert is not completely correct: Where the stuff is actually stored depends on the actual library implementation and some of them [gcc’s standard library] store locally elements whose sizeof is small [thanks TheFlameFire]) and also needs to do some extra verification to return the values only if the cast is valid, so, it is not as fast as having a generic object known at compile time.