请看下面的代码：

struct Foo<'a> {
    borrowed: &'a u8,
    owned_one: Vec<u8>,
    owned_two: Vec<u8>,
    output: usize
}

impl<'a> Foo<'a> {
    fn do_stuff(&mut self) {
        self.output = self.owned_one.len();
        let zipped = self.owned_one.iter().zip(self.owned_two.iter());
        Self::subroutine(&zipped);
    }
    
    fn subroutine<Arg: Iterator<Item=(&'a u8, &'a u8)>>(_zipped: &Arg) {}
}

fn main() {
    let num = 0u8;
    let mut foo = Foo {
        borrowed: &num,
        owned_one: vec![0],
        owned_two: vec![1],
        output: 0
    };
    foo.do_stuff();
    let _out = &foo.output;
}

（playground link）
它不会编译，并产生以下错误：

error: lifetime may not live long enough
  --> src/lib.rs:12:9
   |
8  | impl<'a> Foo<'a> {
   |      -- lifetime `'a` defined here
9  |     fn do_stuff(&mut self) {
   |                 - let's call the lifetime of this reference `'1`
...
12 |         Self::subroutine(&zipped);
   |         ^^^^^^^^^^^^^^^^^^^^^^^^^ argument requires that `'1` must outlive `'a`

我并不完全理解这个抱怨--self肯定总是有一个生命周期被赋给我们正在使用impl的类吗？--但是我可以理解zip()的两个参数需要持续相同的时间，所以我将do_stuff改为&'a mut self。

struct Foo<'a> {
    borrowed: &'a u8,
    owned_one: Vec<u8>,
    owned_two: Vec<u8>,
    output: usize
}

impl<'a> Foo<'a> {
    fn do_stuff(&'a mut self) {
        self.output = self.owned_one.len();
        let zipped = self.owned_one.iter().zip(self.owned_two.iter());
        Self::subroutine(&zipped);
    }
    
    fn subroutine<Arg: Iterator<Item=(&'a u8, &'a u8)>>(_zipped: &Arg) {}
}

fn main() {
    let num = 0u8;
    let mut foo = Foo {
        borrowed: &num,
        owned_one: vec![0],
        owned_two: vec![1],
        output: 0
    };
    foo.do_stuff();
    let _out = &foo.output;
}

但是，现在编译失败，并显示：

error[E0502]: cannot borrow `foo.output` as immutable because it is also borrowed as mutable
  --> src/lib.rs:27:16
   |
26 |     foo.do_stuff();
   |     -------------- mutable borrow occurs here
27 |     let _out = &foo.output;
   |                ^^^^^^^^^^^
   |                |
   |                immutable borrow occurs here
   |                mutable borrow later used here

为什么在do_stuff的参数列表中指定self的生存期意味着我以后突然不能使用foo的不可变引用;我能做些什么呢