this post was submitted on 14 Feb 2024
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submitted 9 months ago* (last edited 6 months ago) by [email protected] to c/[email protected]
 
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[–] [email protected] 17 points 9 months ago (1 children)

Imagine getting segmentation faults at runtime

This post was brought to you by the Rust crew

[–] [email protected] 7 points 9 months ago (1 children)

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

[–] [email protected] 2 points 9 months ago (1 children)

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

[–] [email protected] 1 points 9 months ago (2 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

[–] [email protected] 2 points 9 months ago* (last edited 9 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");
[–] [email protected] 0 points 9 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.

[–] [email protected] 1 points 9 months ago (1 children)

You can't random-access an iterator and use it again later.

If your specific use case really needs random access to a list while lazy computing the elements just wrap them in Lazy and put them in a vector.

Can Rust compute the value of calling a function an infinite number of times?

The return type of an infinitely recursive function / infinite loops is ⊥, a type that by definition has no values. (Known in rust as !)

[–] [email protected] 1 points 9 months ago

Haskell lets you infinitely recurse while still completing in finite time, and there's even a function (fix) for that. Doing e.g. fix (+ 2) would be an infinite loop if evaluated, yes, but fix (2 :) would give you a useful value that's an infinite stream of 2s. (it's also useful for other things too)