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Comment by oneshtein | Hacker News Reader

  fn f() {
    let mut v = vec![1, 2, 3, 4, 5];
    let (header, tail) = v.split_at_mut(1);
    b(&header[0], &mut tail[0]);
  }

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split_at_mut is just unsafe code (and sibling comment mentioned it hours before you did). The borrow checker doesn't natively understand that.

It is safe btw. The difference is that it returns two mutable references vs. one shared ref and one mutable ref. But as they noted, a mutable ref can always be “downgraded” into a shared ref.

The implementation is unsafe, as I said:

> split_at_mut is just unsafe code (and sibling comment mentioned it hours before you did). The borrow checker doesn't natively understand that.

https://doc.rust-lang.org/src/core/slice/mod.rs.html#2086

No, that’s the unchecked version. Two people are telling you that this method exists and is safe, so I am not sure why you’re still doubting this lol.

The checked variant just calls the unchecked, and the panicking variant calls the checked variant. They all need to call unsafe code. See here for details: https://doc.rust-lang.org/nomicon/borrow-splitting.html

Then you misunderstand what unsafe means in Rust. Every single Rust binary needs to eventually call unsafe code at some layer of the callstack.

Is creating a TCP socket using stdlib functions unsafe? How about writing to a file? Or acquiring a mutex?

I would suggest doing some more reading before chiming in here :)

You have totally misunderstood what the person you are talking with means by unsafe. Perhaps you should resolve that prior to such condescensions.

Indeed I have haha - my bad :)

Easier to lose context with longer comment chains...

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