The rule only applies at the ends of blocks/functions. The semicolon can only be omitted on trailing expressions. And, when it's omitted, the block/function takes on the value of that trailing expression. E.g. this is not valid Rust:

  fn foo(x: bool) -> i32 {
      if x { 
          1 // no semicolon on the 1
      }

      let a = 2 + 3;
      return a * 4;
  }

It's not implicitly returning the 1 from `foo`. An early return would need to be written `if x { return 1 }`. On the other hand, the `return a * 4;` could just be written `a * 4` since it's at the end of the function.

This rule means the scanning required is just looking at the end of the function.

In any case, having a static type system means I have never personally encountered a bug caused by this sort of implicit return in Rust.