How do I capture a closure in another closure? - rust

type Time = f32;
type Behaviour<'a,T> = |Time|: 'a -> T;
fn map_one<'a,T,R>(f: |T| -> R
,b: Behaviour<'a,T>)
-> Behaviour<'a,R>{
|time| -> R {
f(b(time))
}
}
Err:
<anon>:8:9: 8:10 error: captured variable `b` does not outlive the enclosing closure
<anon>:8 f(b(time))
^
<anon>:6:38: 10:2 note: captured variable is valid for the block at 6:37
<anon>:6 -> Behaviour<'a,R>{
<anon>:7 |time| -> R {
<anon>:8 f(b(time))
<anon>:9 }
<anon>:10 }
I think the error means that the life time of 'a expires when I try to move b into another closure.
How would I express something like this?

Your code is using "boxed" closures. Those closures capture values by reference, so the captured value must live longer than the closure. You're trying to capture b, which is a parameter, so its lifetime is the duration of the function call. You cannot do what you want to do with "boxed closures.
Unboxed closures were added to the language to solve issues like this. Here's how map_one would be defined:
#![feature(overloaded_calls, unboxed_closures, unboxed_closure_sugar)]
type Time = f32;
fn map_one<'a, T, R, F: Fn<(T,), R>, B: Fn<(Time,), T>>(
f: F, b: B) -> Box<Fn<(Time,), R>+'a> {
box |&: time| -> R {
f(b(time))
}
}
Fn is a trait with a call method that takes self by immutable reference (FnMut takes self by mutable reference and FnOnce takes self by value). The captured values are moved into the closure, rather than referenced.
The input closure parameters are defined with a type parameter that implements the Fn trait, which is the only way to pass closures by value. The return value, however, must be boxed, because the concrete result type (which implements Fn) is created by the compiler, and we can't name it.
I tried writing a main that uses map_one, but I'm getting an internal compiler error... At that point, I can't tell if I made a mistake or if the code is supposed to be valid. (These Rust issues may be related to this error: #16672, #16791, #17060.)
fn main() {
let t = 30f32;
let fa = |&: a: String| -> uint { a.len() };
let fb = |&: b: f32| -> String { b.to_string() };
let m = map_one(fa, fb);
let r = m.call((t,));
println!("{}", r);
}
Note: I couldn't use your Behaviour generic type alias to bound B.
#![feature(overloaded_calls, unboxed_closures, unboxed_closure_sugar)]
type Time = f32;
type Behaviour<T> = Fn<(Time,), T>;
fn map_one<'a, T, R, F: Fn<(T,), R>, B: Behaviour<T>>(
f: F, b: B) -> Box<Behaviour<R>+'a> {
box |&: time| -> R {
f(b(time))
}
}
The compiler complains:
<anon>:6:41: 6:53 error: `Behaviour` is not a trait
<anon>:6 fn map_one<'a, T, R, F: Fn<(T,), R>, B: Behaviour<T>>(
^~~~~~~~~~~~
<anon>:6:41: 6:53 note: `type` aliases cannot be used for traits
<anon>:6 fn map_one<'a, T, R, F: Fn<(T,), R>, B: Behaviour<T>>(
^~~~~~~~~~~~

Related

Why does Rust closure not follow the lifetime elision rule? [duplicate]

fn main() {
let _ref_in_ref_out = |var: &i64| var;
}
This does not compile:
error: lifetime may not live long enough
--> src/main.rs:2:39
|
2 | let _ref_in_ref_out = |var: &i64| var;
| - - ^^^ returning this value requires that `'1` must outlive `'2`
| | |
| | return type of closure is &'2 i64
| let's call the lifetime of this reference `'1`
Apparently the compiler infers two different lifetimes (for the argument and the return type), instead of it being the same.
Is it possible to write a closure so that the input lifetime is the same as the output lifetime?
Something like
fn ref_in_ref_out<'a> (var: &'a i64) -> &'a i64 { var }
but as a closure
Lifetime elisions rules do not apply to closures, neither can you explicitly specify lifetimes for closures. There are several ways to make this code work, though.
The easiest solution is to simply omit the type annotations and let the compiler infer everything:
let ref_in_ref_out = |var| var;
let i: i64 = 42;
ref_in_ref_out(&i);
Alternatively, it's actually fine to specify the return type. This compiles:
let _ref_in_ref_out = |var| -> &i64 { var };
Yet another option for the case that your closure does not close over any local variables is to convert it to a function pointer, since lifetime elision rules apply to function pointers:
let ref_in_ref_out: fn(&i64) -> &i64 = |var| var;
And finally the most general solution is to use a helper function to apply a function trait bound to your closure:
fn constrain_closure<F: Fn(&i64) -> &i64>(f: F) -> F {
f
}
let _ref_in_ref_out = constrain_closure(|var| var);
I've just found out a way to solve this issue, but It'd be great if there is an easier solution:
fn infer_lifetime<'a, T: 'a, F: Fn(&'a T) -> &'a T>(f: F) -> F {
f
}
fn main() {
let _ref_in_ref_out = infer_lifetime(|var: &i64| var);
}

Why can I not return a reference from a closure?

fn main() {
let _ref_in_ref_out = |var: &i64| var;
}
This does not compile:
error: lifetime may not live long enough
--> src/main.rs:2:39
|
2 | let _ref_in_ref_out = |var: &i64| var;
| - - ^^^ returning this value requires that `'1` must outlive `'2`
| | |
| | return type of closure is &'2 i64
| let's call the lifetime of this reference `'1`
Apparently the compiler infers two different lifetimes (for the argument and the return type), instead of it being the same.
Is it possible to write a closure so that the input lifetime is the same as the output lifetime?
Something like
fn ref_in_ref_out<'a> (var: &'a i64) -> &'a i64 { var }
but as a closure
Lifetime elisions rules do not apply to closures, neither can you explicitly specify lifetimes for closures. There are several ways to make this code work, though.
The easiest solution is to simply omit the type annotations and let the compiler infer everything:
let ref_in_ref_out = |var| var;
let i: i64 = 42;
ref_in_ref_out(&i);
Alternatively, it's actually fine to specify the return type. This compiles:
let _ref_in_ref_out = |var| -> &i64 { var };
Yet another option for the case that your closure does not close over any local variables is to convert it to a function pointer, since lifetime elision rules apply to function pointers:
let ref_in_ref_out: fn(&i64) -> &i64 = |var| var;
And finally the most general solution is to use a helper function to apply a function trait bound to your closure:
fn constrain_closure<F: Fn(&i64) -> &i64>(f: F) -> F {
f
}
let _ref_in_ref_out = constrain_closure(|var| var);
I've just found out a way to solve this issue, but It'd be great if there is an easier solution:
fn infer_lifetime<'a, T: 'a, F: Fn(&'a T) -> &'a T>(f: F) -> F {
f
}
fn main() {
let _ref_in_ref_out = infer_lifetime(|var: &i64| var);
}

Can't closure return a reference?

fn main() {
let a = String::from("foo");
let f = || &a;
fn_immut(f);
println!("{}", a);
}
fn fn_immut<F>(f: F)
where F: Fn() -> &String
{
println!("calling Fn closure from fn, {}", f());
}
This code can't compile, rustc tells me that I should add a 'static like this:
fn fn_immut<F>(f: F)
where F: Fn() -> &'static String
I tried to do it, but it still doesn't work. And rustc also tells me " this function's return type contains a borrowed value, but there is no value for it to be borrowed from".
My question is: in this code,the closure already captures the reference of variable a in its scope, why does rustc still tell me that "there is no value for it to be borrowed from"?
The key message from the compiler is indeed the lack of a lifetime specifier for the closure returning a string. Since the signature defined by the trait Fn() -> &String does not have any function parameters, there are no values from which the compiler can infer the lifetime of the returned reference.
error[E0106]: missing lifetime specifier
--> src/main.rs:10:16
|
10 | F: Fn() -> &String,
| ^ help: consider giving it a 'static lifetime: `&'static`
|
= help: this function's return type contains a borrowed value, but there is no value for it to be borrowed from
Adding 'static here does not contribute to solving the problem, because in practice the returned string won't have that lifetime. A new lifetime parameter needs to be introduced in fn_immut, from which can be transferred to F's constraint.
fn fn_immut<'a, F>(f: F)
where
F: Fn() -> &'a String,
You can also return a string slice (&str) instead of &String. The full code:
fn main() {
let a = String::from("foo");
let f = || &*a;
fn_immut(f);
println!("{}", a);
}
fn fn_immut<'a, F>(f: F)
where
F: Fn() -> &'a str,
{
println!("calling Fn closure from fn, {}", f());
}
Playground

Rust function which takes function with arg a function

I want to write a generic function count_calls which calls a function f which takes a function pointer (lambda) where count_calls counts how often function f called the given lambda function.
I struggle with the approach (Playground).
fn count_calls<S, F>(s: S, f: F) -> u32
where
S: Clone,
F: Sized + FnMut(Fn() -> S) -> (),
{
let mut counter: u32 = 0;
f(|| {
counter += 1;
s.clone()
});
counter
}
#[cfg(test)]
mod stackoverflow {
use super::*;
fn f(p: fn() -> i32) {
p();
p();
}
#[test]
fn test() {
let counts = count_calls(3, f);
assert_eq!(counts, 2);
}
}
Here I get the error:
error[E0277]: the size for values of type `(dyn std::ops::Fn() -> S + 'static)` cannot be known at compilation time
--> src/lib.rs:1:1
|
1 | / fn count_calls<S, F>(s: S, f: F) -> u32
2 | | where
3 | | S: Clone,
4 | | F: Sized + FnMut(Fn() -> S) -> (),
... |
12 | | counter
13 | | }
| |_^ doesn't have a size known at compile-time
|
= help: within `((dyn std::ops::Fn() -> S + 'static),)`, the trait `std::marker::Sized` is not implemented for `(dyn std::ops::Fn() -> S + 'static)`
= note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
= note: required because it appears within the type `((dyn std::ops::Fn() -> S + 'static),)`
= note: required by `std::ops::FnMut`
Does someone know how to fix this?
[Edit]
I think using Box<Fn()->S> might be a solution. But I would prefer a stack only solution, if possible.
The error "the size for values of type (dyn std::ops::Fn() -> S + 'static) cannot be known at compilation time" is caused by your trait bound for F:
F: Sized + FnMut(Fn() -> S) -> ()
This is equivalent to F: Sized + FnMut(dyn Fn() -> S). This means that the closure F would take a trait object (dyn Fn() -> S) by value. But trait objects are unsized and cannot be passed by value (yet).
One solution would be to pass the trait object by reference or in a Box. The answer by rodrigo explains and discusses these solutions.
Can we avoid trait objects and dynamic dispatch?
Not properly, I think.
Non solutions
One idea would be to add another type parameter to count_calls:
fn count_calls<S, F, G>(s: S, f: F) -> u32
where
S: Clone,
F: Sized + FnMut(G),
G: Fn() -> S,
However, this doesn't work:
error[E0308]: mismatched types
--> src/lib.rs:9:7
|
9 | f(|| {
| _______^
10 | | counter += 1;
11 | | s.clone()
12 | | });
| |_____^ expected type parameter, found closure
|
= note: expected type `G`
found type `[closure#src/lib.rs:9:7: 12:6 counter:_, s:_]`
The problem here is that type arguments of count_calls are chosen by the caller of count_calls. But we actually want G to always be the type of our own closure. So that doesn't work.
What we want is a generic closure (where we can choose it's type parameters). Something similar is possible, but restricted to lifetime parameters. It's called HRTBs and looks like F: for<'a> Fn(&'a u32). But it doesn't help here because we need a type parameter and for<T> doesn't exist (yet?).
Sub-optimal, nightly solution
One solution would be to not use a closure, but a type with a known name which implements FnMut. Sadly, you can't implement the Fn* traits for your own type on stable yet. On nightly, this works.
struct CallCounter<S> {
counter: u32,
s: S,
}
impl<S: Clone> FnOnce<()> for CallCounter<S> {
type Output = S;
extern "rust-call" fn call_once(self, _: ()) -> Self::Output {
// No point in incrementing the counter here
self.s
}
}
impl<S: Clone> FnMut<()> for CallCounter<S> {
extern "rust-call" fn call_mut(&mut self, _: ()) -> Self::Output {
self.counter += 1;
self.s.clone()
}
}
fn count_calls<S, F>(s: S, mut f: F) -> u32
where
S: Clone,
F: Sized + FnMut(&mut CallCounter<S>), // <----
{
let mut counter = CallCounter {
counter: 0,
s,
};
f(&mut counter); // <-------
counter.counter
}
Unfortunately, now you have this strange type in your public interface (which should be implementation detail).
Apart from that, I can't think of any real solution (only other super verbose solutions with plenty of disadvantages). The developments in the type system corner (in particular in regards to GATs and HKTs) could solve this properly in the future. However, I think there are still a few different features lacking; in particular, I don't think that GATs as proposed would already solve this.
So if this is a real life problem which needs to be solved right now, I would:
step back and rethink the problem in a bigger scope to maybe avoid this Rust limitation, or
just use dynamic dispatch.
This is the simplest code that I managed to get working (playground):
fn count_calls<S, F>(s: S, mut f: F) -> u32
where
S: Clone,
F: FnMut(&mut dyn FnMut() -> S) -> (),
{
let mut counter: u32 = 0;
f(&mut || {
counter += 1;
s.clone()
});
counter
}
#[cfg(test)]
mod stackoverflow {
use super::*;
fn f(p: &mut dyn FnMut() -> i32) {
p();
p();
}
#[test]
fn test() {
let counts = count_calls(3, f);
assert_eq!(counts, 2);
}
}
The key change is that the function argument for F is changed from Fn() -> S into &mut dyn FnMut() -> S. You need a reference because you are using dynamic dispatching. Also you need FnMut because you are capturing counter and changing it inside, and a Fn will not allow it.
Note that you cannot use Box<FnMut() -> S. It will not allow capturing a reference to counter because boxed functions must be 'static.
If you find that changing your Fn to FnMut is undesirable (because you are changing your public API) you could go back to F: FnMut(&mut dyn Fn() -> S) -> () by defining counter as a Cell<u32>:
fn count_calls<S, F>(s: S, mut f: F) -> u32
where
S: Clone,
F: FnMut(&dyn Fn() -> S) -> (),
{
let counter: Cell<u32> = Cell::new(0);
f(&|| {
counter.set(counter.get() + 1);
s.clone()
});
counter.into_inner()
}

How to implement the Y combinator in Rust? [duplicate]

I've just started Rust tutorial and ended with such code using recursion
extern crate rand;
use std::io;
use rand::Rng;
use std::cmp::Ordering;
use std::str::FromStr;
use std::fmt::{Display, Debug};
fn try_guess<T: Ord>(guess: T, actual: T) -> bool {
match guess.cmp(&actual) {
Ordering::Less => {
println!("Too small");
false
}
Ordering::Greater => {
println!("Too big");
false
}
Ordering::Equal => {
println!("You win!");
true
}
}
}
fn guess_loop<T: Ord + FromStr + Display + Copy>(actual: T)
where <T as FromStr>::Err: Debug
{
println!("PLease input your guess.");
let mut guess = String::new();
io::stdin()
.read_line(&mut guess)
.expect("Failed to read line");
let guess_int: T = guess.trim()
.parse()
.expect("Should enter integer number");
println!("You guessed {} !", guess_int);
if !try_guess(guess_int, actual) {
guess_loop(actual)
}
}
fn main() {
println!("Guess the number!!!");
let secret_number = rand::thread_rng().gen_range(1, 51);
guess_loop(secret_number);
}
I was hoping to factor-out the recursion from the guess_loop function and introduced a fix point operator:
fn guess_loop<T: Ord + FromStr + Display + Copy>(actual: T, recur: fn(T) -> ()) -> ()
where <T as FromStr>::Err: Debug
{
println!("PLease input your guess.");
let mut guess = String::new();
io::stdin()
.read_line(&mut guess)
.expect("Failed to read line");
let guess_int: T = guess.trim()
.parse()
.expect("Should enter integer number");
println!("You guessed {} !", guess_int);
if !try_guess(guess_int, actual) {
recur(actual)
}
}
fn fix<T, R>(func: fn(T, fn(T) -> R) -> R) -> fn(T) -> R {
fn fixed(val: T) -> R {
func(val, fixed)
}
fixed
}
fn main() {
println!("Guess the number!!!");
let secret_number = rand::thread_rng().gen_range(1, 51);
fix(guess_loop)(secret_number);
}
but this led to numerous errors, such as
error[E0401]: can't use type parameters from outer function; try using a local type parameter instead
--> src/main.rs:49:19
|
49 | fn fixed(val: T) -> R {
| ^ use of type variable from outer function
error[E0401]: can't use type parameters from outer function; try using a local type parameter instead
--> src/main.rs:49:25
|
49 | fn fixed(val: T) -> R {
| ^ use of type variable from outer function
error[E0434]: can't capture dynamic environment in a fn item; use the || { ... } closure form instead
--> src/main.rs:50:9
|
50 | func(val, fixed)
| ^^^^
My next attempt was changing guess_loop's definition to
fn guess_loop<T: Ord + FromStr + Display + Copy, F>(actual: T, recur: F) -> ()
where <T as FromStr>::Err: Debug,
F: Fn(T) -> ()
{ ... }
and redefine fix as
fn fix<T, R, F>(func: fn(T, F) -> R) -> F
where F: Fn(T) -> R
{
let fixed = |val: T| func(val, fix(func));
fixed
}
this led to
error[E0308]: mismatched types
--> src/main.rs:53:5
|
53 | fixed
| ^^^^^ expected type parameter, found closure
|
= note: expected type `F`
= note: found type `[closure#src/main.rs:52:17: 52:46 func:_]`
error: the type of this value must be known in this context
--> src/main.rs:61:5
|
61 | fix(guess_loop)(secret_number);
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
How can I write a similar fix function?
Firstly, variable names don't exist until after they're initialised. You can't have fixed refer to itself like that.
Secondly, you can't return closures by-value from a function, period. Generic parameters are chosen by the caller, and the caller has no idea what the type of a closure inside the function is going to be.
I'm not claiming that what follows is the best way of doing this, but it was the simplest I was able to come up with that type-checks.
fn guess_loop<T>(actual: T, recur: &Fn(T)) -> ()
where T: Ord + FromStr + Display + Copy,
<T as FromStr>::Err: Debug
{
// ...
}
fn fix<T, R, F>(func: F) -> Box<Fn(T) -> R>
where T: 'static,
R: 'static,
F: Fn(T, &Fn(T) -> R) -> R + 'static
{
use std::cell::RefCell;
use std::rc::Rc;
let fixed = Rc::new(RefCell::new(None));
let fixed_fn = {
let fixed = fixed.clone();
move |val: T| -> R {
let fixed_ref = fixed.borrow();
let fixed_ref: &Box<_> = fixed_ref.as_ref().unwrap();
func(val, &**fixed_ref)
}
};
*fixed.borrow_mut() = Some(Box::new(fixed_fn));
Box::new(move |val: T| -> R {
let fixed_ref = fixed.borrow();
let fixed_ref: &Box<_> = fixed_ref.as_ref().unwrap();
fixed_ref(val)
})
}
In order for fixed_fn to refer to itself, we have to create something for it to read from before it exists. Unfortunately, this means having a cycle, and Rust hates cycles. So, we do this by constructing a reference-counted RefCell<Option<_>> that starts with None, and which will be mutated later to contain the fixed-point closure.
Secondly, we can't use this handle as a callable, so we have to explicitly pull a pointer to the closure out so that we can pass it to func.
Third, the compiler doesn't seem to be able to infer the type of fixed correctly. I was hoping it would be able to work out that it is Rc<RefCell<Option<{closure}>>>, but it refused to do so. As a result, we have to resort to storing a Box<Fn(T) -> R>, since we can't name the type of the closure explicitly.
Finally, we have to construct a new closure that takes a second handle to fixed, unpacks it, and calls it. Again, we can't use fixed as a callable directly. We also can't re-use the closure inside fixed, because to do that we'd have to put that inside its own Rc and at that point, things are starting to get crazy.
... more crazy.
Finally, we have to return this second closure in a Box because, as I said before, we can't return closures by value because we can't name their types in the signature.
*deep breath*
If someone has a simpler solution, I'd love to see it. :P
This is an answer to my own question about implementing the Y combinator which is a subset of this question. In pure lambda expression, a version of the Y combinator looks like
λf.(λw.w w)(λw.f (w w))
The solution in Rosetta Code is too complicated and used Box to allocate memory in the heap. I want to simplify this.
First, let's implement the type Mu<T> as a trait instead.
trait Mu<T> {
fn unroll(&self, &Mu<T>) -> T;
}
Note that we need this trait to be object safe, which means we cannot ask for Self in any of its definition so the second parameter is typed &Mu<T> and it is a trait object.
Now we can write a generic trait implementation:
impl<T, F: Fn(&Mu<T>) -> T> Mu<T> for F {
fn unroll(&self, o: &Mu<T>) -> T {
self(o)
}
}
With this, we can now write the y combinator as the following:
fn y<T, F: Fn(T) -> T>(f: &F) -> T {
(&|w: &Mu<T>| w.unroll(w))(&|w: &Mu<T>| f(w.unroll(w)))
}
The above compiles in the Rust playground without enabling any features and using only the stable channel so this is a pretty good answer to my question.
However, the above would not work in practice because Rust is call-by-value but the code above is the call-by-name Y combinator.
The call-by-value solution
To work with the stable channel without requiring any features, we cannot return closures (which requires impl Trait). Instead, I came up with making another Mu2 type that takes two type parameters:
trait Mu2<T, R> {
fn unroll(&self, &Mu2<T, R>, t: T) -> R;
}
As above, let's implement this new trait.
impl<T, R, F> Mu2<T, R> for F
where
F: Fn(&Mu2<T, R>, T) -> R,
{
fn unroll(&self, o: &Mu2<T, R>, t: T) -> R {
self(o, t)
}
}
The new Y combinator:
fn y<T, R, F>(f: &F, t: T) -> R
where
F: Fn(&Fn(T) -> R, T) -> R,
{
(&|w: &Mu2<T, R>, t| w.unroll(w, t))((&|w: &Mu2<T, R>, t| f(&|t| w.unroll(w, t), t)), t)
}
Now it is time to test our new facility.
fn main() {
let fac = &|f: &Fn(i32) -> i32, i| if i > 0 { i * f(i - 1) } else { 1 };
println!("{}", y(fac, 10))
}
Results in:
3628800
All done!
You can see that the y function has a slightly different signature than the questioner's fix, but it shouldn't matter.
The direct recurring version
The same technology to avoid returning a closure can be used for the normal direct recurring version as well:
fn fix<T, R, F>(f: &F, t: T) -> R
where
F: Fn(&Fn(T) -> R, T) -> R,
{
f(&|t| fix(f, t), t)
}
fn fib(i: i32) -> i32 {
let fn_ = &|f:&Fn(i32) -> i32, x| if x < 2 { x } else { f(x-1) + f(x-2) };
fix(fn_, i)
}
Basically, whenever you need to return a closure from a function, you can add the closure's parameter to the function, and change the return type to the closure's return type. Later on when you need a real closure, just create the closure by partial evaluating that function.
Further discussions
Compare to other languages, in Rust there is a big difference: the function given to find fix point must not have any internal states. In Rust this is a requirement that the F type parameter of y must be Fn, not FnMut or FnOnce.
For example, we cannot implement a fix_mut that would be used like
fn fib1(i: u32) -> u32 {
let mut i0 = 1;
let mut i1 = 1;
let fn_ = &mut |f:&Fn(u32) -> u32, x|
match x {
0 => i0,
1 => i1,
_ => {
let i2 = i0;
i0 = i1;
i1 = i1 + i2;
f(x)
}
};
fix_mut(fn_, i)
}
without unsafe code whilst this version, if it works, performs much better (O(N)) than the version given above (O(2^N)).
This is because you can only have one &mut of one object at a single time. But the idea of Y combinator, or even the fix point function, requires capturing/passing the function at the same time when calling it, that's two references and you can't just mark any of them immutable without marking another so.
On the other hand, I was wonder if we could do something that other languages usually not able to but Rust seems to be able. I was thinking restricting the first argument type of F from Fn to FnOnce (as y function will provide the implementation, change to FnMut does not make sense, we know it will not have states, but change to FnOnce means we want it to be used only once), Rust would not allow at the moment as we cannot pass unsized object by value.
So basically, this implementation is the most flexible solution we could think of.
By the way, the work around of the immutable restriction is to use pseudo-mutation:
fn fib(i: u32) -> u32 {
let fn_ = &|f:&Fn((u32,u32,u32)) -> u32, (x,i,j)|
match x {
0 => i,
1 => j,
_ => {
f((x-1,j,i+j))
}
};
fix(&fn_, (i,1,1))
}
Starting at where you left off:
fn fix<T, R, F>(func: fn(T, F) -> R) -> F
where F: Fn(T) -> R
{
|val: T| func(val, fix(func))
}
The returned object has an unnameable closure type. Using a generic type won’t help here, since the type of the closure is decided by the callee, not the caller. Here’s where impl traits come in handy:
fn fix<T, R, F>(func: fn(T, F) -> R) -> impl Fn(T) -> R
where F: Fn(T) -> R
{
|val: T| func(val, fix(func))
}
We can’t pass fix(func) to func because it expects a nameable type for F. We’ll have to settle for a trait object instead:
fn fix<T, R>(func: fn(T, &Fn(T) -> R) -> R) -> impl Fn(T) -> R {
|val: T| func(val, &fix(func))
}
Now it’s time to fight the lifetime checker. The compiler complains:
only named lifetimes are allowed in `impl Trait`, but `` was found in the type `…`
This is a somewhat cryptic message. Since impl traits are always 'static by default, this is a roundabout way of saying: “the closure does not live long enough for 'static”. To get the real error message, we append + 'static to the impl Fn(T) -> R and recompile:
closure may outlive the current function, but it borrows `func`, which is owned by the current function
So that was the real problem. It is borrowing func. We don’t need to borrow func because fn is Copy, so we can duplicate it as much as we want. Let’s prepend the closure with move and get rid of the + 'static from earlier:
fn fix<T, R>(func: fn(T, &Fn(T) -> R) -> R) -> impl Fn(T) -> R {
move |val: T| func(val, &fix(func))
}
And voila, it works! Well, almost … you’ll have to edit guess_loop and change fn(T) -> () to &Fn(T) -> (). I’m actually quite amazed that this solution doesn’t require any allocations.
If you can’t use impl traits, you can instead write:
fn fix<T, R>(func: fn(T, &Fn(T) -> R) -> R) -> Box<Fn(T) -> R>
where T: 'static,
R: 'static
{
Box::new(move |val: T| func(val, fix(func).as_ref()))
}
which is unfortunately not allocation-free.
Also, we can generalize the result a bit to allow arbitrary closures and lifetimes:
fn fix<'a, T, R, F>(func: F) -> impl 'a + Fn(T) -> R
where F: 'a + Fn(T, &Fn(T) -> R) -> R + Copy
{
move |val: T| func(val, &fix(func))
}
In the process of figuring out a solution for your problem, I ended up writing a simpler version of fix, which actually ended up guide me towards a solution to your fix function:
type Lazy<'a, T> = Box<FnBox() -> T + 'a>;
// fix: (Lazy<T> -> T) -> T
fn fix<'a, T, F>(f: F) -> T
where F: Fn(Lazy<'a, T>) -> T + Copy + 'a
{
f(Box::new(move || fix(f)))
}
Here’s a demonstration of how this fix function could be used to calculate the factorial:
fn factorial(n: u64) -> u64 {
// f: Lazy<u64 -> u64> -> u64 -> u64
fn f(fac: Lazy<'static, Box<FnBox(u64) -> u64>>) -> Box<FnBox(u64) -> u64> {
Box::new(move |n| {
if n == 0 {
1
} else {
n * fac()(n - 1)
}
})
}
fix(f)(n)
}
This can be done at zero runtime cost if you're willing to use unstable features (i.e. a nightly compiler) and willing to... obfuscate your code slightly.
First, we need to turn the result of fix into a named struct. This struct needs to implement Fn, so we'll implement it manually (this is an unstable feature).
#![feature(fn_traits)]
#![feature(unboxed_closures)]
extern crate rand;
use rand::Rng;
use std::cmp::Ordering;
fn try_guess<T: Ord>(guess: T, actual: T) -> bool {
match guess.cmp(&actual) {
Ordering::Less => {
println!("Too small");
false
}
Ordering::Greater => {
println!("Too big");
false
}
Ordering::Equal => {
println!("You win!");
true
}
}
}
struct Fix<F>
where F: Fn(i32, &Fix<F>)
{
func: F,
}
impl<F> FnOnce<(i32,)> for Fix<F>
where F: Fn(i32, &Fix<F>)
{
type Output = ();
extern "rust-call" fn call_once(self, args: (i32,)) -> Self::Output {
self.call(args)
}
}
impl<F> FnMut<(i32,)> for Fix<F>
where F: Fn(i32, &Fix<F>)
{
extern "rust-call" fn call_mut(&mut self, args: (i32,)) -> Self::Output {
self.call(args)
}
}
impl<F> Fn<(i32,)> for Fix<F>
where F: Fn(i32, &Fix<F>)
{
extern "rust-call" fn call(&self, (val,): (i32,)) -> Self::Output {
(self.func)(val, self);
}
}
fn fix<F>(func: F) -> Fix<F>
where F: Fn(i32, &Fix<F>)
{
Fix { func: func }
}
fn guess_loop<F>(actual: i32, recur: &F)
where F: Fn(i32)
{
let guess_int = rand::thread_rng().gen_range(1, 51);
if guess_int != actual {
recur(actual)
}
}
fn main() {
let secret_number = rand::thread_rng().gen_range(1, 51);
fix(guess_loop)(secret_number);
}
However, we're not done yet. This fails to compile with the following error:
error[E0281]: type mismatch: the type `fn(i32, &_) {guess_loop::<_>}` implements the trait `for<'r> std::ops::Fn<(i32, &'r _)>`, but the trait `for<'r> std::ops::Fn<(i32, &'r Fix<fn(i32, &_) {guess_loop::<_>}>)>` is required (cyclic type of infinite size)
--> src/main.rs:77:5
|
77 | fix(guess_loop)(secret_number);
| ^^^
|
= note: required by `fix`
Note: In case you're not aware, in Rust, each function has its own, zero-sized type. If a function is generic, then each instantiation of that function will have its own type as well. For example, the type of guess_loop::<X> will be reported by the compiler as fn(i32, &X) {guess_loop::<X>} (as you can see in the error message above, except with underscores where the concrete type hasn't been resolved yet). That type can be coerced to a function pointer type implicitly in some contexts or explicitly with a cast (as).
The problem is that, in the expression fix(guess_loop), the compiler needs to instantiate guess_loop, which is a generic function, and it looks like the compiler isn't able to figure out the proper type to instantiate it with. In fact, the type we would like to set for type parameter F references the type of guess_loop. If we were to write it out in the style reported by the compiler, the type would look like fn(i32, &Fix<X>) {guess_loop::<Fix<&X>>}, where X is replaced by the type itself (you can see now where the "cyclic type of infinite size" comes from).
We can solve this by replacing the guess_loop function by a non-generic struct (we'll call it GuessLoop) that implements Fn by referring to itself. (You can't do this with a normal function because you can't name a function's type.)
struct GuessLoop;
impl<'a> FnOnce<(i32, &'a Fix<GuessLoop>)> for GuessLoop {
type Output = ();
extern "rust-call" fn call_once(self, args: (i32, &Fix<GuessLoop>)) -> Self::Output {
self.call(args)
}
}
impl<'a> FnMut<(i32, &'a Fix<GuessLoop>)> for GuessLoop {
extern "rust-call" fn call_mut(&mut self, args: (i32, &Fix<GuessLoop>)) -> Self::Output {
self.call(args)
}
}
impl<'a> Fn<(i32, &'a Fix<GuessLoop>)> for GuessLoop {
extern "rust-call" fn call(&self, (actual, recur): (i32, &Fix<GuessLoop>)) -> Self::Output {
let guess_int = rand::thread_rng().gen_range(1, 51);
if !try_guess(guess_int, actual) {
recur(actual)
}
}
}
fn main() {
let secret_number = rand::thread_rng().gen_range(1, 51);
fix(GuessLoop)(secret_number);
}
Notice that GuessLoop's implementation of Fn is no longer generic on the type of the recur parameter. What if we tried to make the implementation of Fn generic (while still leaving the struct itself non-generic, to avoid cyclic types)?
struct GuessLoop;
impl<'a, F> FnOnce<(i32, &'a F)> for GuessLoop
where F: Fn(i32),
{
type Output = ();
extern "rust-call" fn call_once(self, args: (i32, &'a F)) -> Self::Output {
self.call(args)
}
}
impl<'a, F> FnMut<(i32, &'a F)> for GuessLoop
where F: Fn(i32),
{
extern "rust-call" fn call_mut(&mut self, args: (i32, &'a F)) -> Self::Output {
self.call(args)
}
}
impl<'a, F> Fn<(i32, &'a F)> for GuessLoop
where F: Fn(i32),
{
extern "rust-call" fn call(&self, (actual, recur): (i32, &'a F)) -> Self::Output {
let guess_int = rand::thread_rng().gen_range(1, 51);
if !try_guess(guess_int, actual) {
recur(actual)
}
}
}
Unfortunately, this fails to compile with the following error:
error[E0275]: overflow evaluating the requirement `<Fix<GuessLoop> as std::ops::FnOnce<(i32,)>>::Output == ()`
--> src/main.rs:99:5
|
99 | fix(GuessLoop)(secret_number);
| ^^^
|
= note: required because of the requirements on the impl of `for<'r> std::ops::Fn<(i32, &'r Fix<GuessLoop>)>` for `GuessLoop`
= note: required by `fix`
Essentially, the compiler is unable to verify that Fix<GuessLoop> implements Fn(i32), because in order to do that, it needs to verify that GuessLoop implements Fn(i32, &Fix<GuessLoop>), but that is only true if Fix<GuessLoop> implements Fn(i32) (because that impl is conditional), which is only true if GuessLoop implements Fn(i32, &Fix<GuessLoop>) (because that impl is conditional too), which... you get the idea. In order words, the two implementations of Fn here are dependent on each other, and the compiler is unable to resolve that.

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