In C++11 lambda syntax, heap-allocated closures?

Question

C++11 lambdas are great!

But one thing is missing, which is how to safely deal with mutable data.

The following will give bad counts after the first count:

#include <cstdio>
#include <functional>
#include <memory>

std::function<int(void)> f1()
{
    int k = 121;
    return std::function<int(void)>([&]{return k++;});
}

int main()
{
    int j = 50;
    auto g = f1();
    printf("%d\n", g());
    printf("%d\n", g());
    printf("%d\n", g());
    printf("%d\n", g());
}

gives,

$ g++-4.5 -std=c++0x -o test test.cpp && ./test
121
8365280
8365280
8365280

The reason is that after f1() returns, k is out of scope but still on the stack. So the first time g() is executed k is fine, but after that the stack is corrupted and k loses its value.

So, the only way I've managed to make safely returnable closures in C++11 is to allocate closed variables explicitly on the heap:

std::function<int(void)> f2()
{
    int k = 121;
    std::shared_ptr<int> o = std::shared_ptr<int>(new int(k));
    return std::function<int(void)>([=]{return (*o)++;});
}

int main()
{
    int j = 50;
auto g = f2();
    printf("%d\n", g());
    printf("%d\n", g());
    printf("%d\n", g());
    printf("%d\n", g());
}

Here, [=] is used to ensure the shared pointer is copied, not referenced, so that memory handling is done correctly: the heap-allocated copy of k should be freed when the generated function g goes out of scope. The result is as desired,

$ g++-4.5 -std=c++0x -o test test.cpp && ./test
121
122
123
124

It's pretty ugly to refer to variables by dereferencing them, but it should be possible to use references instead:

std::function<int(void)> f3()
{
    int k = 121;
    std::shared_ptr<int> o = std::shared_ptr<int>(new int(k));
    int &p = *o;
    return std::function<int(void)>([&]{return p++;});
}

Actually, this oddly gives me,

$ g++-4.5 -std=c++0x -o test test.cpp && ./test
0
1
2
3

Any idea why? Maybe it's not polite to take a reference of a shared pointer, now that I think about it, since it's not a tracked reference. I found that moving the reference to inside the lambda causes a crash,

std::function<int(void)> f4()
{
    int k = 121;
std::shared_ptr<int> o = std::shared_ptr<int>(new int(k));
    return std::function<int(void)>([&]{int &p = *o; return p++;});
}

giving,

g++-4.5 -std=c++0x -o test test.cpp && ./test
156565552
/bin/bash: line 1: 25219 Segmentation fault      ./test

In any case, it would be nice if there was a way to automatically make safely returnable closures via heap allocation. For example, if there was an alternative to [=] and [&] that indicated that variables should be heap allocated and referenced via references to shared pointers. My initial thought when I learned about std::function was that it creates an object encapsulating the closure, therefore it could provide storage for the closure environment, but my experiments show that this doesn't seem to help.

I think safely returnable closures in C++11 are going to be paramount to using them, does anyone know how this can be accomplished more elegantly?

Answer 1

In f1 you're getting undefined behavior for the reason you say; the lambda contains a reference to a local variable, and after the function returns the reference is no longer valid. To get around this you don't have to allocate on the heap, you simply have to declare that captured values are mutable:

int k = 121;
return std::function<int(void)>([=]() mutable {return k++;});

You will have to be careful about using this lambda though, because different copies of it will be modifying their own copy of the captured variable. Often algorithms expect that using a copy of a functor is equivalent to using the original. I think there's only one algorithm that actually makes allowances for a stateful function object, std::for_each, where it returns another copy of the function object it uses so you can access whatever modifications occurred.

---

In f3 nothing is maintaining a copy of the shared pointer, so the memory is being freed and accessing it gives undefined behavior. You can fix this by explicitly capturing the shared pointer by value and still capture the pointed-to int by reference.

std::shared_ptr<int> o = std::shared_ptr<int>(new int(k));
int &p = *o;
return std::function<int(void)>([&p,o]{return p++;});

f4 is again undefined behavior because you're again capturing a reference to a local variable, o. You should simply capture by value but then still create your int &p inside the lambda in order to get the syntax you want.

std::shared_ptr<int> o = std::shared_ptr<int>(new int(k));
return std::function<int(void)>([o]() -> int {int &p = *o; return p++;});

Note that when you add the second statement C++11 no longer allows you to omit the return type. (clang and I assume gcc have an extension that allows return type deduction even with multiple statement, but you should get a warning at least.)