Hacker News Reader

Comment by tormeh | Hacker News Reader

And is the behavior exactly the same as it would be if I omitted "let" on the second code line? If the answer is yes, isn't it just mutability with different syntax then?

I would guess shadowing wouldn't affect values shared with a different thread, because it's technically a new value, but in single-thread use it would be equivalent to assignment.

View on HN

The behavior is not the same. It's a new variable, entirely. So you'll have two integers on the stack, one of them is not accessible any more. You can see this difference if you add a scope around the second x, and then print the value of the original x after it goes out of scope.

Consider this program:

    fn main() {
        let x = 5;
        let r1 = &x;
        let x = x + 1;
        let r2 = &x;
        
        println!("r1: {:p}", r1);
        println!("r2: {:p}", r2);
    }

This will print something like

  r1: 0x7fff955e3dfc
  r2: 0x7fff955e3dec

Note that while you can't access the original x, it's still there, and references to it still work. Because the value of x was copied, not mutated.

Also, given compiler details like SSA, I'd argue that in a certain sense, mutation is similar to this, rather than this being similar to mutation...

It doesn't have the same semantics, because even after introducing the new variable, it is still immutable. So you cannot accidentally modify the new variable. e.g.,

  let x = 1;
  let x = x + 1;
  x = 99; // Error: you can't accidentally modify the immutable variable `x`

  let mut x = 1;
  x = x + 1;
  x = 99; // Ok

You can even turn back a mutable varaible into an immutable variable, like this:

  let mut x = 1;
  x = x + 1;
  let x = x;
  x = 99; // Error: `x` became immutable

This can be useful to "pin" the variable after a large number of calculations done to it.

  > isn't it just mutability with different syntax then?

For one thing, in order to take a mutable reference to a binding, that binding must be declared as `mut`.

    let mut a = 2;
    let b = 3;
    let x = &mut a;
    // let y = &mut b;  // error: cannot borrow immutable local variable `b` as mutable

> I would guess shadowing wouldn't affect values shared with a different thread...

That is actually a big deal, though.

Reply on news.ycombinator.com