What are pipes and underscores doing in Rust?

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16

No, it doesn't - that's not how _ works. let _ = <expr>; is exactly equivalent to <expr>;. There are some value category things going on here. I'd recommend trying out the following examples, to get more intuition:

struct D(i32);
impl Drop for D {
  fn drop(&mut self) { println!("dropped {}", self.0); }
}

fn main() {
  {
    D(0);
    let x = D(1);
    let _ = D(2);
    D(3);
    let y = D(4);
    let _ = D(5);
  }

  {
    let var = D(0);
    let _ = var;
    var;
    D(1);
  }
}

I don't really feel like explaining what's going on here with the value categories, because that would require synthesis of difficult wording on my part. However, if you know C++, the rules are similar.