为什么Rust的借用检查器在使用从方法返回的迭代器时会抱怨,而在直接使用Vec的迭代程序时却不会抱怨?

https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=ba9255e504174afa0f31f20467caf0a9

考虑以下Rust程序

struct Foo {
    bar: Vec<i32>,
    baz: i32
}

impl Foo {

    fn get_an_iter(&self) -> impl Iterator<Item=&i32> {
        self.bar.iter()    
    }
    
    fn do_mut_stuff_works(&mut self) {
        for num in self.bar.iter() {
            self.baz += num;
        }
    }
    
    fn do_mut_stuff_does_not_work(&mut self) {
        for num in self.get_an_iter() {
            self.baz += num;
        }
    }
}

在里面 do_mut_stuff_works 我在向量上迭代 bar 并更改字段 baz 。要做到这一点,该方法需要采取 &mut self .

现在,如果我将迭代器的创建从内联转移到一个方法 get_an_iter ,并尝试在中对此进行迭代 do_mut_stuff_does_not_work ,借款检查人员抱怨。

使用时是怎么回事 get_an_ter ,不可变的借位在整个循环中保持有效,因此我的可变借位在 self.baz += num 失败,但当我只是在方法内部创建迭代器时 for num in self.bar.iter() ,借款检查员高兴吗?

完整的编译器输出为:

Compiling playground v0.0.1 (/playground)
error[E0502]: cannot borrow `self.baz` as mutable because it is also borrowed as immutable
  --> src/lib.rs:20:13
   |
19 |         for num in self.get_an_iter() {
   |                    ------------------
   |                    |
   |                    immutable borrow occurs here
   |                    immutable borrow later used here
20 |             self.baz += num;
   |             ^^^^^^^^^^^^^^^ mutable borrow occurs here

For more information about this error, try `rustc --explain E0502`.
error: could not compile `playground` (lib) due to 1 previous error