You're viewing a single thread.
Imagine getting segmentation faults at runtime
This post was brought to you by the Rust crew
Neither does Haskell, and Haskell won't waste time doing something that doesn't matter.
Imagine using a linked list as your default sequential container.
Rust iterators are lazy btw.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
it can compute how often I needed to compute the value of calling a function an infinite number of times.
println!("0");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.
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
Lazyand 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
!)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, butfix (2 :)would give you a useful value that's an infinite stream of 2s. (it's also useful for other things too)