Reputation: 1757
I was reading about closures on the net. I was wondering if C++ has a built-in facility for closures or if there is any way we can implement closures in C++?
Upvotes: 65
Views: 64782
Reputation: 2043
Yes, This shows how you could implement a function with a state without using a functor.
#include <iostream>
#include <functional>
std::function<int()> make_my_closure(int x) {
return [x]() mutable {
++x;
return x;
};
}
int main() {
auto my_f = make_my_closure(10);
std::cout << my_f() << std::endl; // 11
std::cout << my_f() << std::endl; // 12
std::cout << my_f() << std::endl; // 13
auto my_f1 = make_my_closure(1);
std::cout << my_f1() << std::endl; // 2
std::cout << my_f1() << std::endl; // 3
std::cout << my_f1() << std::endl; // 4
std::cout << my_f() << std::endl; // 14
}
Upvotes: 24
Reputation: 176
You can achive similar functionality using static variables and lambdas.
#include <iostream>
#include<functional>
int main()
{
std::function<std::function<int()>()> generator_function=[]()->std::function<int()>{
static int i=0;
return [&]()->int{
return i++;
};
};
std::function<int()> iterator_function=generator_function();
std::cout<<iterator_function()<<std::endl; //1
std::cout<<iterator_function()<<std::endl; //2
std::cout<<iterator_function()<<std::endl; //3
std::cout<<iterator_function()<<std::endl; //4
return 0;
}
Upvotes: 0
Reputation: 991
If you understand closure as a reference to a function that has an embedded, persistent, hidden and unseparable context (memory, state), then yes:
class add_offset {
private:
int offset;
public:
add_offset(int _offset) : offset(_offset) {}
int operator () (int x) { return x + offset; }
}
// make a closure
add_offset my_add_3_closure(3);
// use closure
int x = 4;
int y = my_add_3_closure(x);
std::cout << y << std::endl;
The next one modifies its state:
class summer
{
private:
int sum;
public:
summer() : sum(0) {}
int operator () (int x) { return sum += x; }
}
// make a closure
summer adder;
// use closure
adder(3);
adder(4);
std::cout << adder(0) << std::endl;
The inner state can not be referenced (accessed) from outside.
Depending on how you define it, a closure can contain a reference to more than one function or, two closures can share the same context, i.e. two functions can share the same persistent state.
Closure means not containing free variables - it is comparable to a class with only private attributes and only public method(s).
Upvotes: 26
Reputation: 344
Strictly speaking. 'Closure' is LISP only. Use Let returns lambda as last commands. 'Let Over Lambda'. This is possible only for LISP because of infinite scope with lexical scoping. I don't know any other language support this natively until know.
(defun my-closure ()
(let ((cnt 0))
(lambda ()
(format t "called : ~A times" (incf cnt)))))
Upvotes: 0
Reputation: 1238
The latest C++ standard, C++11, has closures.
http://en.wikipedia.org/wiki/C%2B%2B11#Lambda_functions_and_expressions
http://www.cprogramming.com/c++11/c++11-lambda-closures.html
Upvotes: 44
Reputation: 153919
I suspect that it depends on what you mean by closure. The meaning I've always used implies garbage collection of some sort (although I think it could be implemented using reference counting); unlike lambdas in other languages, which capture references and keep the referenced object alive, C++ lambdas either capture a value, or the object refered to is not kept alive (and the reference can easily dangle).
Upvotes: 9
Reputation: 9089
Yes, C++11 has closures named lambdas.
In C++03 there is no built-in support for lambdas, but there is Boost.Lambda implementation.
Upvotes: 7