I have a function that expects a short lived object. I would expect that I would be able to always pass it a long lived object. But I am getting a strange error when I try to encode that:
type F<'arg> = Box<dyn FnOnce(&'arg ())>;
fn contravar<'small, 'large: 'small>(f: F<'small>) -> F<'large> {
f
}
playground
Particularly:
error: lifetime may not live long enough
--> src/lib.rs:3:5
|
2 | fn contravar<'small, 'large: 'small>(f: F<'small>) -> F<'large> {
| ------ ------ lifetime `'large` defined here
| |
| lifetime `'small` defined here
3 | f
| ^ function was supposed to return data with lifetime `'large` but it is returning data with lifetime `'small`
|
= help: consider adding the following bound: `'small: 'large`
It seems like F is invariant for its argument but I would have guessed that it's contravariant. Am I missing something? Is there a way to make F<'arg> really contravariant for 'arg?
Edit: it looks like the "problem" is that rust wants to treat all generic traits the same (including Fn/FnMut/FnOnce). My opinion is that those 3 are and should be treated special especially given that they are the only way to refer to closures. For that reason I opened an issue
The Rust Reference's page on Subtyping and Variance documents that, as of Rust 1.63.0, fn(T) -> () is contravariant over T and that dyn Trait<T> + 'a is invariant over T.
FnOnce, FnMut and Fn are traits, so that means dyn FnOnce(&'a ()) is unfortunately invariant over &'a ().
// Compiles
pub fn contravariant<'a, 'b: 'a>(x: fn(&'a ())) -> fn(&'b ()) { x }
// Doesn't compile
pub fn contravariant2<'a, 'b: 'a>(x: Box<dyn FnOnce(&'a ())>) -> Box<dyn FnOnce(&'b ())> { x }
Is there a way to wrap FnOnce somehow to convince the compiler of the correct variance?
Here's what I could come up with using unsafe code. Note that I'm not making any guarantees as to whether this is sound or not. I don't know of any way to do this without unsafe code.
use std::marker::PhantomData;
trait Erased {}
impl<T> Erased for T {}
pub struct VariantBoxedFnOnce<Arg, Output> {
boxed_real_fn: Box<dyn Erased + 'static>,
_phantom_fn: PhantomData<fn(Arg) -> Output>,
}
impl<Arg, Output> VariantBoxedFnOnce<Arg, Output> {
pub fn new(real_fn: Box<dyn FnOnce(Arg) -> Output>) -> Self {
let boxed_real_fn: Box<dyn Erased + '_> = Box::new(real_fn);
let boxed_real_fn: Box<dyn Erased + 'static> = unsafe {
// Step through *const T because *mut T is invariant over T
Box::from_raw(Box::into_raw(boxed_real_fn) as *const (dyn Erased + '_) as *mut (dyn Erased + 'static))
};
Self {
boxed_real_fn,
_phantom_fn: PhantomData,
}
}
pub fn call_once(self, arg: Arg) -> Output {
let boxed_real_fn: Box<Box<dyn FnOnce(Arg) -> Output>> = unsafe {
// Based on Box<dyn Any>::downcast()
Box::from_raw(Box::into_raw(self.boxed_real_fn) as *mut Box<dyn FnOnce(Arg) -> Output>)
};
boxed_real_fn(arg)
}
}
pub fn contravariant<'a, 'b: 'a>(x: VariantBoxedFnOnce<&'a (), ()>) -> VariantBoxedFnOnce<&'b (), ()> { x }
#[cfg(test)]
mod tests {
use super::*;
fn foo(_x: &()) {}
#[test]
pub fn check_fn_does_not_require_static() {
let f = VariantBoxedFnOnce::new(Box::new(foo));
let x = ();
f.call_once(&x);
}
#[test]
pub fn check_fn_arg_is_contravariant() {
let f = VariantBoxedFnOnce::new(Box::new(foo));
let g = contravariant(f);
let x = ();
g.call_once(&x);
}
}
Here, VariantBoxedFnOnce is limited to functions taking one argument.
The trick is to store a type-erased version of the Box<dyn FnOnce(Arg) -> Output> such that Arg disappears, because we don't want the variance of VariantBoxedFnOnce<Arg, Output> to depend on Box<dyn FnOnce(Arg) -> Output> (which is invariant over Arg). However, there's also a PhantomData<fn(Arg) -> Output> field to provide the proper contravariance over Arg (and covariance over Output).
We can't use Any as our erased type, because only 'static types implement Any, and we have a step in VariantBoxedFnOnce::new() where we have a Box<dyn Erased + '_> where '_ is not guaranteed to be 'static. We then immediately "transmute" it into 'static, to avoid having a redundant lifetime parameter on VariantBoxedFnOnce, but that 'static is a lie (hence the unsafe code). call_once "downcasts" the erased type to the "original" Box<dyn FnOnce(Arg) -> Output>, except that Arg and Output may be different from the original due to variance.
I'm trying to wrap an async function in a struct. For example:
use std::future::Future;
struct X;
struct Y;
async fn f(x: &X) -> Y {
Y
}
struct MyStruct<F, Fut>(F)
where
F: Fn(&X) -> Fut,
Fut: Future<Output = Y>;
fn main() {
MyStruct(f);
}
The compiler complains about this with the following (unhelpful) error:
error[E0308]: mismatched types
--> src/main.rs:16:5
|
16 | MyStruct(f);
| ^^^^^^^^ one type is more general than the other
|
= note: expected associated type `<for<'_> fn(&X) -> impl Future {f} as FnOnce<(&X,)>>::Output`
found associated type `<for<'_> fn(&X) -> impl Future {f} as FnOnce<(&X,)>>::Output`
Is something like this actually possible? As I understand it, f desugars to something like:
fn f<'a>(x: &'a X) -> impl Future<Output = Y> + 'a {
Y
}
so I'd need to somehow express in MyStruct that Fut has the same lifetime as x.
I actually don't know much about async.
However, when it comes to trait-bounds, typically, fewer trait-bound are better. In other words, only declare those trait-bounds that you really need.
In the case of a struct, as long as you don't need an associated type within your struct, you are mostly good without any bounds. This is pretty much to what #George Glavan wrote in his answer.
When you add methods to your struct, you are more likely to use traits and thus require trait-bounds. Sometimes it is useful to combine the trait-bound of multiple functions by declaring it on the impl-block itself. Tho, this has some restrictions. You should also consider whether each function really needs all these constraints.
For instance, consider the following code:
struct X;
struct Y;
struct MyStruct<F>(F);
impl<F> MyStruct<F> {
pub fn new(f: F) -> Self {
MyStruct(f)
}
pub fn invoke<'a, Fut>(&self) -> Fut
where
F: Fn(&'a X) -> Fut,
{
(self.0)(&X)
}
}
I added a new and a invoke function. The former doesn't require any traits, thus it doesn't have trait-bounds. The latter only calls the function, so it bounds F by Fn. And this is good enough, because in the end, the caller must already know what the return type is, i.e. whether it is some Future or not.
However, there are a few cases where one really needs additional trait-bounds, which involves additional generics such as for a function return type. In such a case, you can declare additional (phantom) generics on a struct, e.g.:
use std::future::Future;
use std::marker::PhantomData;
struct X;
struct Y;
struct MyStruct<F, Fut> {
func: F,
_fut: PhantomData<Fut>,
}
impl<'a, F, Fut> MyStruct<F, Fut>
where
F: Fn(&'a X) -> Fut,
Fut: Future<Output = Y> + Send + Sync + 'a,
{
pub fn new(f: F) -> Self {
MyStruct {
func: f,
_fut: PhantomData,
}
}
pub fn invoke(&self) {
(self.func)(&X);
}
}
Notice, that in this example the trait-bounds apply to both function new and invoke, and both are over-restricted. Still, you don't need to over-restrict the struct itself.
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.