C++ lambda returning itself

Question

I wanted to write a lambda that returns itself, so I could call it multiple times on the spot. But it looks like inside of a lambda this refers not to the lambda but to the surrounding object's this, if the lambda is defines inside a member function.

Here's an example:

#include <iostream>

int main(int argc, char* argv[]) {
  int a = 5;
  [&](int b) {
    std::cout << (a + b) << std::endl;
    return *this;
  }(4)(6);
}

Is there a way to do something comparable?

@Jarod42 Thanks. I forgot that we can now define classes within functions. — SU3, Sep 04 '18 at 20:48
Related: [A Proposal to Add Y Combinator to the Standard Library](http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html). I quote: "C++11/14 lambdas do not encourage recursion: there is no way to reference the lambda object from the body of the lambda function. A common workaround for this problem is to use std::function" — Ben Voigt, Sep 04 '18 at 20:58
@HolyBlackCat: A bit of difficulty writing out the type parameter to that `std::function`, there is. — Ben Voigt, Sep 04 '18 at 21:06
@SU3: You always could, you just couldn't use local class types as template type parameters before C++11. (Which is not a roadblock to this case) — Ben Voigt, Sep 04 '18 at 21:55

score 8 · Accepted Answer · edited Sep 05 '18 at 19:58

8

With old functor:

int main() {
  int a = 5;
  struct S {
    const S& operator ()(int b) const {
      std::cout << (a + b) << std::endl;
      return *this;
    }
    const int& a;
  };
  S{a}(4)(6);
}

Demo

edited Sep 05 '18 at 19:58

SU3

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answered Sep 04 '18 at 20:58

Jarod42

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score 6 · Answer 2 · answered Sep 04 '18 at 20:53

6

You cant return the lambda itself, but you can return a different one:

#include <iostream>

int main() {
  int a = 5;    
  [&](int b) {
      auto impl = [&](int b){std::cout << (a + b) << std::endl;};
      impl(b);
      return impl;
  }(4)(6);
}

However, this allows only to call it one more time. Not sure if there is some trick to get more out of it...

answered Sep 04 '18 at 20:53

463035818_is_not_a_number

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Now that's clever. – HolyBlackCat Sep 04 '18 at 20:58
2

@HolyBlackCat looks better than it actually is. One can only call it once or twice :/ – 463035818_is_not_a_number Sep 04 '18 at 20:59
1

The "trick" is a fixed-point combinator. See my comment on the question. – Ben Voigt Sep 04 '18 at 21:01

Cássio Renan · Answer 3 · 2018-09-04T22:15:55.240

3

Ben Voigt proposes to use Y combinators (which are a great proposal to the standard library, BTW), but your problem is simpler. You can introduce a small template functor that will be called instead of the lambda:

template<typename T>
struct recallable_impl {
    template<typename... Args>
    recallable_impl& operator()(Args&&... args) {
        f(std::forward<Args>(args)...);
        return *this;
    }

    template<typename F>
    recallable_impl(F&& f)
      :  f{std::forward<F>(f)}
    {}
private:
    T f;
};

template<typename T>
decltype(auto) recallable(T&& f) {
    return recallable_impl<std::decay_t<T>>(std::forward<T>(f));
}

Your lambda will then not even need to explicitly return anything:

int main() {
  int a = 5;  
  recallable([&](int b){std::cout << (a + b) << std::endl;})(4)(5)(6)(7)(8);
}

edited Sep 04 '18 at 22:15

answered Sep 04 '18 at 21:24

Cássio Renan

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How is it deducing `T` in your example? – jaggedSpire Sep 04 '18 at 21:30
1

Ah, I assumed C++17. I will update with a C++14-enabled example. – Cássio Renan Sep 04 '18 at 21:30
2

Nice. And conveniently every call is made on the same object instance, so if the lambda has mutable state, it will propagate correctly. – Ben Voigt Sep 04 '18 at 21:53
@BenVoigt Thanks. I'm on the impression that keeping that reference to a temporary in the functor is UB, though. I'm gonna edit to correct that. – Cássio Renan Sep 04 '18 at 22:13
1

Because of lifetime mismatch? IIRC, temporary lifetimes end at the end of the full-expression they're in, so it should be good on that front so long as the `recallable_impl` isn't stored locally. – jaggedSpire Sep 04 '18 at 22:24
3

@jaggedSpire thanks for the info! Still, from the perspective of a user of this class, accidentally causing UB by simply storing an object sounds like a big no-no. – Cássio Renan Sep 04 '18 at 22:27
2

Absolutely! [it could be quite problematic.](http://coliru.stacked-crooked.com/a/f9e6754783034824) – jaggedSpire Sep 04 '18 at 22:36

score 0 · Answer 4 · answered Sep 04 '18 at 21:01

0

You can call a lambda if you assign it to a predifned type (not auto), and it gets captured:

std::function<void(int)> f =[&f](int n)
{
    if (n>0) f(n-1);
    return;
};

answered Sep 04 '18 at 21:01

Robert Andrzejuk

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While recursion and returning self both require naming the lambda from inside the lambda, they are still different operations. Recursion is in a sense "easier" because the return type (which has to be named inside of the `std::function` type argument) is simpler. Much simpler. – Ben Voigt Sep 04 '18 at 21:03
1

Yes.So I would have to declare a type which returns a type, which returns a type, ... OMG it's turtle types all the way down ;-) – Robert Andrzejuk Sep 04 '18 at 21:11

C++ lambda returning itself

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