Conditionally iterate over one of several possible iterators

Question

I'm trying to switch behavior based on an Option input to a function. The idea is to iterate based on whether or not a given Option is present. Here's a minimal, if silly, example:

use std::iter;

fn main() {
    let x: Option<i64> = None;

    // Repeat x 5 times if present, otherwise count from 1 to 5
    for i in match x {
        None => 1..5,
        Some(x) => iter::repeat(x).take(5),
    } {
        println!("{}", i);
    }
}

I get an error:

error[E0308]: match arms have incompatible types
  --> src/main.rs:7:14
   |
7  |       for i in match x {
   |  ______________^
8  | |         None => 1..5,
9  | |         Some(x) => iter::repeat(x).take(5),
   | |                    ----------------------- match arm with an incompatible type
10 | |     } {
   | |_____^ expected struct `std::ops::Range`, found struct `std::iter::Take`
   |
   = note: expected type `std::ops::Range<{integer}>`
              found type `std::iter::Take<std::iter::Repeat<i64>>`

This makes perfect sense, of course, but I'd really like to choose my iterator based on a condition, since the code in the for-loop is non-trivial and copy-pasting all of that just to change iterator selection would be pretty ugly and unmaintainable.

I tried using as Iterator<Item = i64> on both arms, but that gives me an error about unsized types because it's a trait object. Is there an easy way to go about this?

I could, of course, use .collect() since they return the same type and iterate over that vector. Which is a good quick fix, but for large lists seems a bit excessive.

Answer 1

The most straightforward solution is to use a trait object:

use std::iter;

fn main() {
    let mut a;
    let mut b;

    let x: Option<i64> = None;

    // Repeat x 5 times if present, otherwise count from 1 to 5
    let iter: &mut dyn Iterator<Item = i64> = match x {
        None => {
            a = 1..5;
            &mut a
        }
        Some(x) => {
            b = iter::repeat(x).take(5);
            &mut b
        }
    };

    for i in iter {
        println!("{}", i);
    }
}

The main downside for this solution is that you have to allocate stack space for each concrete type you have. This also means variables for each type. A good thing is that only the used type needs to be initialized.

The same idea but requiring heap allocation is to use boxed trait objects:

use std::iter;

fn main() {
    let x: Option<i64> = None;

    // Repeat x 5 times if present, otherwise count from 1 to 5
    let iter: Box<dyn Iterator<Item = i64>> = match x {
        None => Box::new(1..5),
        Some(x) => Box::new(iter::repeat(x).take(5)),
    };

    for i in iter {
        println!("{}", i);
    }
}

This is mostly useful when you want to return the iterator from a function. The stack space taken is a single pointer, and only the needed heap space will be allocated.

You can also use an enum for each possible concrete iterator.

Answer 2

The either crate provides the Either type. If both halves of Either are iterators, then so is the Either:

extern crate either;

use either::Either;
use std::iter;

fn main() {
    let x: Option<i64> = None;

    // Repeat x 5 times if present, otherwise count from 1 to 5
    let iter = match x {
        None => Either::Left(1..5),
        Some(x) => Either::Right(iter::repeat(x).take(5)),
    };

    for i in iter {
        println!("{}", i);
    }
}

Like a previous answer, this still takes stack space for each concrete type you have. However, you don't need individual variables for each concrete value.

This type can also be returned from a function, unlike the trait object references. Compared to boxed trait objects, it will always use a fixed size on the stack, regardless of which concrete type was chosen.

You'll find this type (or semantic equivalent) in other places as well, such as futures::Either

Answer 3

Personally, rather than use Either, I often prefer to create a series of Option<Iterator> values that get chained together. Something like this:

playground

use std::iter;

fn main() {
    let x: Option<i64> = None;

    // Repeat x 5 times if present, otherwise count from 1 to 5
    for i in pick(x) {
        println!("{}", i);
    }
}

fn pick(opt_x: Option<i64>) -> impl Iterator<Item = i64> {
    let iter_a = if let None = opt_x {
        Some(1..5)  
    } else {
        None
    };

    let iter_b = if let Some(x) = opt_x {
        Some(iter::repeat(x).take(5))
    } else {
        None
    };

    iter_a.into_iter().flatten().chain(iter_b.into_iter().flatten())
}

It's a bit less obvious than using Either, but it avoids another crate, and sometimes it works out quite elegantly.

Answer 4

Here is a variation of @Niko's brilliant solution using a single match expression instead of several if let expressions, which may be more convenient when dealing with more conditional cases:

use std::iter;

fn main() {
    let x: Option<i64> = None;

    // Repeat x 5 times if present, otherwise count from 1 to 5
    for i in pick(x) {
        println!("{}", i);
    }
}

fn pick(opt_x: Option<i64>) -> impl Iterator<Item = i64> {
    let mut iter_a = None;
    let mut iter_b = None;

    match opt_x {
        None => iter_a = Some(1..5),
        Some(x) => iter_b = Some(iter::repeat(x).take(5)),
    }

    iter_a.into_iter().flatten().chain(iter_b.into_iter().flatten())
}