Why does this Rust program ignore immutability

问题

I have the following Rust program and I expect it to result in an compilation error since x is reassigned later. But it complies and gives output. Why?

fn main() {
   let (x, y) = (1, 3);
   println!("X is {} and Y is {}", x, y);

   let x: i32 = 565;
   println!("Now X is {}", x);
}

回答1:

Rust actually lets you shadow other variables in a block, so let x: i32 = 565; is defining a new variable x that shadows the x defined earlier with let (x,y) = (1,3);. Note that you could even have redefined x to have a different type since the second x is a whole new variable!

fn main(){
   let x = 1;
   println!("Now X is {}",x);

   let x = "hi";
   println!("Now X is {}",x);
}

This reddit thread goes into more detail about why this is useful. The two things that are mentioned that seem interesting are:

• For operations which take ownership of the variable, but return another variable of the same type, it sometimes "looks nice" to redefine the returned variable to have the same name. From here:

  let iter = vec.into_iter();
  let iter = modify(iter);
  let iter = double(iter);
  

• Or to make a variable immutable:

  let mut x;
  // Code where `x` is mutable
  let x = x;
  // Code where `x` is immutable
  

来源：https://stackoverflow.com/questions/42991325/why-does-this-rust-program-ignore-immutability