Do we have closures in C++?

问题

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++?

回答1:

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

回答2:

If you understand closure as a reference to a function that has an embedded, persistent, hidden and unseparable context (memory, state), than 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 cloure
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).

回答3:

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
}

I forgot the mutable keyword which introduced an undefined behaviour (clang version was returning a garbage value). As implemented, the closure works fine (on GCC and clang)

回答4:

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).

回答5:

Yes, C++11 has closures named lambdas.

In C++03 there is no built-in support for lambdas, but there is Boost.Lambda implementation.

来源：https://stackoverflow.com/questions/12635184/do-we-have-closures-in-c

标签

c++

closures

lexical-closures