Why does Rust prohibit the implementation of existing types?

Question

Like this...

impl String {
    fn foo(&self) {
        //...
    }
}

... any other?

fn foo(s: &String) {
    //...
}

Again, you can extend the type implementation if you define a trait in your box. Why?

Answer 1

There are several different arguments pointed out in the following source regarding why it is not possible to implement existing types that are outside the same box,

• impl . , " Add on Vec<T> concat,..., ... ... mathy [is]. impl upgrade, ... ²."

• , .. " " ³."

• . , , , " [] [] , , , " ", " " impl impl, , pull . impl create ⁴."

• , impl , impls . " [ ] ⁵." "HashTable" ⁵.

  mod foo {
      impl Hash for i32 { ... }
  
      fn f(mut table: HashMap<i32, &'static str>) {
          table.insert(0, "hello");
          ::bar::f(&table);
      }
  }
  
  mod bar {
      impl Hash for i32 { ... }
  
      fn f(table: &HashMap<i32, &'static str>) {
          assert_eq!(table.get(0), Some("hello"));
      }
  }