Rust 中的生命周期：borrow 的 mut 数据

以下代码报告错误: borrowed value does not live long enough个

fn main() {
    let e = 100;
    let mut a = (1,);
    let mut b = (&e,);
    {
        let mut c = (&mut a, &mut b);
        let f = &mut c;
        f.1 .0 = &(f.0 .0);
    }
    println!("{:?}", b);
}

但以下代码是正确的:

fn main() {
    let e = 100;
    let a = (1,);
    let mut b = (&e,);
    {
        let mut c = (&a, &mut b);
        let f = &mut c;
        f.1 .0 = &(f.0 .0);
    }
    println!("{:?}", b);
}

为什么？！！

第一个代码的完整错误消息:

error[E0597]: `c` does not live long enough
  --> src/main.rs:7:17
   |
6  |         let mut c = (&mut a, &mut b);
   |             ----- binding `c` declared here
7  |         let f = &mut c;
   |                 ^^^^^^ borrowed value does not live long enough
8  |         f.1 .0 = &f.0 .0;
9  |     }
   |     - `c` dropped here while still borrowed
10 |     println!("{:?}", b);
   |                      - borrow later used here

For more information about this error, try `rustc --explain E0597`.