this post was submitted on 14 Feb 2024
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[–] PoolloverNathan@programming.dev 8 points 7 months ago (6 children)

Neither does Haskell, and Haskell won't waste time doing something that doesn't matter.

[–] anton@lemmy.blahaj.zone 1 points 7 months ago (5 children)

Imagine using a linked list as your default sequential container.
Rust iterators are lazy btw.

[–] PoolloverNathan@programming.dev 1 points 7 months ago (4 children)

You can't random-access an iterator and use it again later. Can Rust compute the value of calling a function an infinite number of times?

— former rustacean

[–] Lauchmelder@feddit.de 2 points 7 months ago* (last edited 7 months ago) (1 children)

it can compute how often I needed to compute the value of calling a function an infinite number of times.

println!("0");
[–] PoolloverNathan@programming.dev 1 points 7 months ago

If you've used a parser library's recursive parser, you have infinite calls right there. If it supplies a recursive-parser function, that function is a type-limited equivalent to fix, which performs the infinite call operation. Your Rust library most likely implements recursion using hidden mutability, but in Haskell, your parsers can remain infinitely-recursive while still referencing themselves and immutable.

Also, we get to ask people if they know what a monad is.

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