Re:Rust vs. Go
by iopq on Wednesday January 18, 2017 @05:07PM
(#13430800)
Joke's on you: it took me a month of asking around why my program won't compile to figure out I really CAN'T do it without dereferencing inside the function. This is because of higher ranked lifetime bounds... there's no way to express the correct lifetime bound in this case (in Rust, maybe in Haskell it's possible), so you just have to deref first.
[ Parent
| on HN ]
Re:Rust vs. Go
by openasocket on Wednesday January 18, 2017 @07:29PM
(#13431865)
> This is because of higher ranked lifetime bounds...
No, it's because the function you're returning is of type FnMut(&B) -> C, but your F function is of type FnMut(B) -> G. It expects an owned value of B, but you only have a borrowed reference to a B. Just make F of type FnMut(&B) -> G and you no longer have to dereference and no longer need the restriction that B: Copy. Higher-ranked lifetime values don't have anything to do with it.
Here's the new version:
fn apply<A, B, C, F, G>(mut f: F, a: A)
-> impl FnMut(&B) -> C
where F: FnMut(&B) -> G,
G: FnMut(A) -> C,
A: Clone {
move |b| f(b)(a.clone())
}You think I didn't try that one before? It doesn't work in MY case:
error[E0281]: type mismatch: the type `fn(_) -> _ {tool::second::<_>}` implements the trait `std::ops::FnMut<(_,)>`, but the trait `for<'r> std::ops::FnMut<(&'r _,)>` is required (expected concrete lifetime, found bound lifetime parameter
--> src\lib.rs:50:17
|
50 | .filter(apply(second, i))
| ^^^^^
|
= note: required by `apply`
error[E0271]: type mismatch resolving `for<'r> <fn(_) -> _ {tool::second::<_>} as std::ops::FnOnce<(&'r _,)>>::Output == _`
--> src\lib.rs:50:17
|
50 | .filter(apply(second, i))
| ^^^^^ expected bound lifetime parameter , found concrete lifetime
|
= note: concrete lifetime that was found is lifetime '_#11r
= note: required by `apply`
error: aborting due to 2 previous errors
error: Could not compile `fizzbuzz`.You probably have a simpler, clearer way of solving that problem than by torturing the language.
I do, but the problem is FizzBuzz. What's the point of solving FizzBuzz in a SIMPLE way? The whole point of FizzBuzz is to torture the language.
fn main() {
for i in 1..101 {
if i % 15 == 0 {
println!("FizzBuzz");
} else if i % 3 == 0 {
println!("Fizz");
} else if i % 5 == 0 {
println!("Buzz");
} else {
println!("{}", i);
}
}
}I don't know why people over complicate things. I was searching for other solutions and found this abomination:
https://bitbucket.org/iopq/fizzbuzz-in-rust/src/bf4968973d73...
I hope their production code isn't like this.
Wut? It's about as representative of Rust as writing FizzBuzz using Monads. In Java.
If you want production code look at Rust and Servo.
I wrote this as an exercise. This is specifically written to be point-free code without any if or match expressions.
.unwrap_or_else(|| i.to_string().into())
If the problem is FizzBuzz then it's you that's made it too complicated. Here is a simple version:
https://github.com/the-guitarman/rust_fizzbuzz/blob/master/s...