Skip to main content

Binding std::function to member functions

I realized that std::function can be bound to member functions without requiring the *this object. Consider the following examples.
// std::string::empty is a const function. All variables from e1 to e5 are fine.
std::function<bool(std::string)> e1 = &std::string::empty;
std::function<bool(std::string &)> e2 = &std::string::empty;
std::function<bool(const std::string &)> e3 = &std::string::empty;
std::function<bool(std::string *)> e4 = &std::string::empty;
std::function<bool(const std::string *)> e5 = &std::string::empty;

// std::string::push_back is not a const function. p4 and p5 don't compile.
std::function<void(std::string, char)> p1 = &std::string::push_back;
std::function<void(std::string &, char)> p2 = &std::string::push_back;
std::function<void(std::string *, char)> p3 = &std::string::push_back;

// These two don't compile because push_back is a non-const function 
std::function<void(const std::string &, char)> p4 = &std::string::push_back;
std::function<void(const std::string *, char)> p5 = &std::string::push_back;
I thought I knew how to do that but this time I found that the syntax is a little different than what I had in mind.

I used tho think (incorrectly) that just like function types for free-standing functions, one would create a member-function type. While, it's straight-forward to create function types for free-standing functions, I think it's not possible to create member function types. Don't get me wrong. One can create pointer-to-member-function type just like pointer-to-a-function type. Here's what I mean.
using F = int(const char *); // a function type of a free standing function.
F f = std::atoi; // error because a there are no instances of a function type.
F* fnew = new F(); //error because new can't be applied to a function type.
F* fptr = &std::atoi; // OK. A pointer to a function type is initialized.

// However, there's no function type for member functions
using GPTR = bool (std::string::*)() const; // OK. This is a pointer-to-member-function-type. 
GPTR gptr = &std::string::empty; // OK. This is a pointer to a member function.
string s;
std::cout << (s.*gptr)(); // OK. prints 1
using H = decltype(*gptr); // error. It makes no sense without a std:string object. Illformed.

bool (*x)(const std::string &) = gptr; // error. Incompatible types.
std::function<bool (const std::string &)> fobj = gptr; // OK! Neat!
Therefore, std::function uses the syntax of free-standing function types to bind to pointer to member functions. That's neat.

Comments

Popular posts from this blog

Multi-dimensional arrays in C++11

What new can be said about multi-dimensional arrays in C++? As it turns out, quite a bit! With the advent of C++11, we get new standard library class std::array. We also get new language features, such as template aliases and variadic templates. So I'll talk about interesting ways in which they come together.

It all started with a simple question of how to define a multi-dimensional std::array. It is a great example of deceptively simple things. Are the following the two arrays identical except that one is native and the other one is std::array?

int native[3][4];
std::array<std::array<int, 3>, 4> arr;

No! They are not. In fact, arr is more like an int[4][3]. Note the difference in the array subscripts. The native array is an array of 3 elements where every element is itself an array of 4 integers. 3 rows and 4 columns. If you want a std::array with the same layout, what you really need is:

std::array<std::array<int, 4>, 3> arr;

That's quite annoying for two r…

Folding Monadic Functions

In the previous two blog posts (Understanding Fold Expressions and Folding Functions) we looked at the basic usage of C++17 fold expressions and how simple functions can be folded to create a composite one. We’ll continue our stride and see how "embellished" functions may be composed in fold expressions.

First, let me define what I mean by embellished functions. Instead of just returning a simple value, these functions are going to return a generic container of the desired value. The choice of container is very broad but not arbitrary. There are some constraints on the container and once you select a generic container, all functions must return values of the same container. Let's begin with std::vector.
// Hide the allocator template argument of std::vector. // It causes problems and is irrelevant here. template <class T> struct Vector : std::vector<T> {}; struct Continent { }; struct Country { }; struct State { }; struct City { }; auto get_countries…

Covariance and Contravariance in C++ Standard Library

Covariance and Contravariance are concepts that come up often as you go deeper into generic programming. While designing a language that supports parametric polymorphism (e.g., templates in C++, generics in Java, C#), the language designer has a choice between Invariance, Covariance, and Contravariance when dealing with generic types. C++'s choice is "invariance". Let's look at an example.
struct Vehicle {}; struct Car : Vehicle {}; std::vector<Vehicle *> vehicles; std::vector<Car *> cars; vehicles = cars; // Does not compile The above program does not compile because C++ templates are invariant. Of course, each time a C++ template is instantiated, the compiler creates a brand new type that uniquely represents that instantiation. Any other type to the same template creates another unique type that has nothing to do with the earlier one. Any two unrelated user-defined types in C++ can't be assigned to each-other by default. You have to provide a c…