Recursive Lambda Functions in C++14

Recursive lambda functions in C++14

The crux of the issue is that in a C++ lambda expression the implicit this parameter will always refer to the object of the enclosing context of the expression, if present at all, and not the functor object resulting from the lambda expression.

Borrowing a leaf from anonymous recursion (sometimes also known as 'open recursion'), we can use the generic lambda expressions of C++14 to re-introduce an explicit parameter to refer to our would-be recursive functor:

auto f = [](auto&& self, int n) -> int
{ return n < 2 ? 1 : n * self(/* hold on */); };

The caller now has a new burden of making calls of the form e.g. f(f, 5). Since our lambda expression is self-referential, it is in fact a caller of itself and thus we should have return n < 2 ? 1 : n * self(self, n - 1);.

Since that pattern of explicitly passing the functor object itself in the first position is predictable, we can refactor this ugly wart away:

template<typename Functor>
struct fix_type {
    Functor functor;

    template<typename... Args>
    decltype(auto) operator()(Args&&... args) const&
    { return functor(functor, std::forward<Args>(args)...); }

    /* other cv- and ref-qualified overloads of operator() omitted for brevity */
};

template<typename Functor>
fix_type<typename std::decay<Functor>::type> fix(Functor&& functor)
{ return { std::forward<Functor>(functor) }; }

This allows one to write:

auto factorial = fix([](auto&& self, int n) -> int
{ return n < 2 ? 1 : n * self(self, n - 1); });

assert( factorial(5) == 120 );

Did we succeed? Since the fix_type<F> object contains its own functor which it passes to it for each call, there is never a risk of a dangling reference. So our factorial object can truly be endless copied, moved from, in and out of functions without hassle.

Except... while the 'external' callers can readily make calls of the form factorial(5), as it turns out inside our lambda expression the recursive call still looks like self(self, /* actual interesting args */). We can improve on this by changing fix_type to not pass functor to itself, but by passing *this instead. That is, we pass in the fix_type object which is in charge of passing the correct 'implicit-as-explicit' argument in the first position: return functor(*this, std::forward<Args>(args)...);. Then the recursion becomes n * self(n - 1), as it should be.

Finally, this is the generated code for a main that uses return factorial(5); instead of the assertion (for either flavour of fix_type):

00000000004005e0 <main>:
  4005e0:       b8 78 00 00 00          mov    eax,0x78
  4005e5:       c3                      ret    
  4005e6:       66 90                   xchg   ax,ax

The compiler was able to optimize everything away, as it would have done with a run-off-the-mill recursive function.

What are the costs?

The astute reader may have noticed one curious detail. In the move from a non-generic to a generic lambda, I added an explicit return type (i.e. -> int). How come?

This has to do with the fact that the return type to be deduced is the type of the conditional expression, which type depends on the call to self, which type is being deduced. A quick reading of Return type deduction for normal functions would suggest that rewriting the lambda expression as follows should work:

[](auto&& self, int n)
{
    if(n < 2) return 1;               // return type is deduced here
    else return n * self(/* args */); // this has no impact
}

GCC will in fact accept this code with the first form of fix_type only (the one that passes functor). I'm not able to determine if it is right to complain about the other form (where *this is passed). I leave it to the reader to choose what trade-off to make: less type deduction, or less ugly recursive calls (it's also of course completely possible to have access to either flavour anyway).

GCC 4.9 examples

Complete code, first flavour
Complete code, second flavour
Complete code, first flavour, C++11
An example of a variadic fix for a group of mutually recursive lambda expressions

Recursion with generic lambda functions in C++14

auto& is lvalue only.

This matters little until you refactor and replace the lvalue recursive object with a temporary proxy memoizer, for example.

auto&& is harmless, and means "I do not mind if this is a temprary or whatever, just don't make a copy", which expresses meaning well here. auto& states "No temporaries allowed!" Sometimes you want to exclude temporaries when making a reference, but it is rare.

auto const&, auto and auto&& should be your bread and butter.

Only use auto& if your operation is explicitly about writing and you are ok with excluding proxy references.

Recursive lambda functions in C++11

Think about the difference between the auto version and the fully specified type version. The auto keyword infers its type from whatever it's initialized with, but what you're initializing it with needs to know what its type is (in this case, the lambda closure needs to know the types it's capturing). Something of a chicken-and-egg problem.

On the other hand, a fully specified function object's type doesn't need to "know" anything about what is being assigned to it, and so the lambda's closure can likewise be fully informed about the types its capturing.

Consider this slight modification of your code and it may make more sense:

std::function<int(int, int)> sum;

sum = [term, next, &sum](int a, int b) -> int {
    if (a > b)
        return 0;
    else
        return term(a) + sum(next(a), b);
};

Obviously, this wouldn't work with auto. Recursive lambda functions work perfectly well (at least they do in MSVC, where I have experience with them), it's just that they aren't really compatible with type inference.

Return recursive lambda from function in C++

If fibonacci (line 2) is a local of the makeFibonacci() function, and therefore goes out of scope when the function exits, how can it be captured by reference and used recursively?

It's just chance that the function is working as expected. What you have is undefined behavior. You are referencing an object that goes out of scope in the function.

Also, why does the program segfault when I capture the lambda by copy?

This happens because of how the std::function is initialized. The lambda is initialized first, the std::function is initialized with the lambda afterwards. Which means that you are copying an instance of std::function that is not initialized, and therefore it is probably not in a state which can allow good copies. Invariants are broken inside, which are likely causing the segmentation fault.

You can make a recursive lambda function more efficiently without std::function by using a polymorphic lambda as follows

auto makeFibonacci() {
    auto fib = [](int n, auto& self) {
        if (n == 1) {
            return 1;
        }
        if (n == 2) {
            return 1;
        }
        return self(n - 1, self) + self(n - 2, self);
    };
    return [fib](int n) {
        return fib(n, fib);
    };
};

Here the lambda owns all the state it needs. You can then use it like this

auto fibonacci = makeFibonacci();
cout << fibonacci(6) << endl;

Also note that this is probably the worst way to calculate fibonacci numbers.

Can lambda functions be recursive?

Yes, they can. Starting with C++23 you can use the explicit this parameter:

auto factorial = [](this auto self, int i) 
{ 
    return (i == 1) ? 1 : i * self(i - 1); 
};

With previous C++ standards, you can store the lambda in a variable and reference that variable (although you cannot declare the type of that variable as auto, you would have to use an std::function object instead). For instance:

std::function<int (int)> factorial = [&] (int i) 
{ 
    return (i == 1) ? 1 : i * factorial(i - 1); 
};

Call C++ recursive lambda in the same line where it is declared

Let me offer a glimpse into the functional programming world, where people usually use combinators to deal with recursive lambdas. There was a proposal (P0200r0) last year to add a simple Y-combinator to the standard library.

Leaving aside the question whether it is a good idea to do this, this would allow you to write and invoke a recursive lambda like this:

y_combinator([](auto self, int i){
    if (i>1) {
        std::cout << "func(a, ";
        self(i-1);
        std::cout << ")";
    } else {
        std::cout << "a";
    }
})(6);

The basic idea here is that the y-combinator is a higher order function that wraps a lambda which is passed 'itself' as a first argument. The combinator takes care of wrapping the self argument away for all invocations of the lambda.

You can try it in coliru.

constexpr recursive function defined via lambda

There's a trick due to a blog post by Pedro Melendez, which circumvents direct recursion, and can be used in a constexpr context. Thanks @HolbyBlackCat for the reference and the idea.

constexpr auto factorial = [](int n) {
    auto factorial_impl = [](int n, auto& factorial_ref) {
        if(n <= 1) { return 1; }
        return n * factorial_ref(n-1, factorial_ref);
    };
    return factorial_impl(n,factorial_impl);
};

See it on GodBolt.

The (external) lambda is a "closure type", which became "literal" and usable with constexpr only in C++17 (so this won't work with C++14).

PS - I simplified your factorial function a bit and am using integers because IMHO the use of doubles just distracts from what the question is about.

A Problem with Recursive Void Lambda Expressions in C++

See this question for recursive lamda-functions explanation. You can declare your Lambda as std::function<void(std::vector<int>&)> and capture it inside lambda:

std::function<void(std::vector<int>&)> Lambda;
Lambda = [&Lambda](std::vector<int>& A) {
    while (A.size() < 5) {
        A.push_back(1);
        Lambda(A);
    }
};
Lambda(A);

Recursive Lambda Functions in C++14