this post was submitted on 10 Apr 2024
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I just had a random thought: a common pattern in Rust is to things such as:

let vec_a: Vec<String> = /* ... */;
let vec_b: Vec<String> = vec_a.into_iter().filter(some_filter).collect();

Usually, we need to be aware of the fact that Iterator::collect() allocates for the container we are collecting into. But in the snippet above, we've consumed a container of the same type. And since Rust has full ownership of the vector, in theory the memory allocated by vec_a could be reused to store the collected results of vec_b, meaning everything could be done in-place and no additional allocation is necessary.

It's a highly specific optimization though, so I wonder if such a thing has been implemented in the Rust compiler. Anybody who has an idea about this?

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[–] arendjr@programming.dev 5 points 4 months ago (8 children)

I mean, the actual operation is just an example, of course. Feel free to make it a .map() operation instead. The strings couldn’t be reused then, but the vector’s allocation still could… in theory.

[–] myotherself@lemmings.world 7 points 4 months ago* (last edited 4 months ago) (7 children)

map() can still be used with Vec::iter_mut(), filter_map() can be replaced with Vec::retain_mut().

[–] arendjr@programming.dev 4 points 4 months ago* (last edited 4 months ago) (6 children)

Yeah, that's helpful if I would be currently optimizing a hot loop now. But I was really just using it as an example. Also, retain_mut() doesn't compose as well.

I'd much rather write:

let vec_a: Vec<String> = /* ... */;
let vec_b: Vec<String> = vec_a
    .into_iter()
    .filter(some_filter)
    .map(some_map_fn)
    .collect();

Over:

let mut vec_a: Vec<String> = /* ... */;
vec_a.retain_mut(|x| if some_filter(x) {
    *x = some_map_fn(*x); // Yikes, cannot move out of reference.
    true
} else {
    false
});

And it would be nice if that would be optimized the same. After all, the point of Rust's iterators is to provide zero-cost abstractions. In my opinion, functions like retain_mut() represent a leakiness to that abstraction, because the alternative turns out to not be zero cost.

[–] myotherself@lemmings.world 1 points 4 months ago

To be fair, these alternatives are also limited to the case where the item type stays the same.

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