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Rust is hard to write but I'll grant that once written it is relatively easy to understand. That's kinda the opposite of C++ where it's easy to write and hard to understand (although it's easy to think you understand).

I appreciate the safety+performance value proposition in Rust. More cathedral, less bazaar.



Not THAT easy

    fn apply<A, B, C, F, G>(mut f: F, a: A) 
    -> impl FnMut(&B) -> C // must still be `for<'r> impl FnMut(&'r B) -> C`, because that’s what filter requires
             where F: FnMut(B) -> G, // must not be `for<'r> FnMut(&'r B) -> G`, because regular functions do not implement it
                   G: FnMut(A) -> C,
                   B: Copy, // for dereferencing
                   A: Clone {
    
        move |b| f(*b)(a.clone()) // this must do any bridging necessary to satisfy the requirements
    }
this function signature must be EXACTLY like what I wrote or it won't work


No, not that easy especially when you get to the functional programming stuff. But I find FP hard to understand since I don't use it much.

> this function signature must be EXACTLY like what I wrote or it won't work

This can be seen as a feature in the sense that you can't program by luck: well, this incantation seems about right and the compiler nods its head that it understands and that means something, so it must be right.


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.


> 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())
this is the only one that was too ugly to make point-free


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...


It doesn't do the same exact thing as mine does. For one thing, mine does FizzBuzzBazz (factors of 3, 5, 7) or even FizzBuzzBazzQuux if you want (actually arbitrary strings and arbitrary conditions)


Equivalent [1] C++14 for comparison:

   template<class F, class A>
   auto apply(F f, A a) {
      return [f=std::move(f), a=std::move(a)](auto&& b) {
        return f(std::forward<dectlype(b)>(b))(a);
      }
   }
The forward<decltype> is an abomination and is really in need of a language based solution, but otherwise is fairly straightforward.

[1] No type checking of the definition of course, only of instantiations. On the plus side, the returned function is polymorphic and can be called for all Bs that f can be called.


Not sure what you're talking about:

https://is.gd/Xnu0YJ


it doesn't work in my program:

https://bitbucket.org/iopq/fizzbuzz-in-rust/src/bf4968973d73...

it gives an error when I use a generic `second` method from tool.rs


Why ask that B: Copy instead of B: Clone?


To deref it


b.clone(), of course, derefs b as well.


I don't know Rust, but code looks readable. Function apply receives two arguments and returns function of one argument (address of b) which returns value of type C. I assume function apply() creates closure, so a.clone() is necessary to create fresh copy of a every time when closure is called. b is passed by reference, so code is generic. Not bad.


Is Rust harder to write than safe, readable C++ code?

EDIT: Reworded it to be more specific


I don't know that good C++ has been standardized or is enforceable by the compiler. But Rust is hard to write and that's doubly especially the case coming from an anything goes language.


Heavily disagree. Usually, compiler tells you how to fix your errors. You still need solid grasp of language.


The borrow checker can be hard to follow occasionally.

Your solid grasp of the language may be different or the same as my hard to write. But neither would be confused with utterly ambiguous C. The counterpart to borrowing in C would be it's antithesis in Rust, aliasing.

That said I'm on my learning curve and my opinion could change.


When was the last time you used Rust?

I ask this, because newer versions of Rust have suggestions how to fix your code, which may get you out of borrow checker fight.




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