Comment on Give me Options or give me death

• anton@lemmy.blahaj.zone ⁨9⁩ ⁨months⁩ ago

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

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  • PoolloverNathan@programming.dev ⁨9⁩ ⁨months⁩ ago

    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

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    • Lauchmelder@feddit.de ⁨9⁩ ⁨months⁩ ago

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

      println!("0");
      

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      • PoolloverNathan@programming.dev ⁨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.

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    • anton@lemmy.blahaj.zone ⁨9⁩ ⁨months⁩ ago

      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 !)

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      • PoolloverNathan@programming.dev ⁨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)

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