Rust 交换引用时的生命周期

我正在学习Rust and Play通过一些场景来看看它是如何运行的.当将可变引用传递给函数，然后将另一个可变引用赋给它时，Rust编译器似乎希望引用的变量与最初引用的变量一样长.

但是，因为我实际上是将引用交换到其他地方的point，所以它永远不会影响调用函数的引用.它仍然会指向Foo，它是完全活着的.那么为什么foo2要和foo一样长寿呢？这看起来像是在期望替换原始引用中的引用...

struct Foo {
    x: u32,
    y: u32
}

fn do_something(mut foo: &mut Foo) {
    foo.x = 4;
    foo.y = 5;
    println!("foo.x = {}\nfoo.y = {}", foo.x, foo.y);
    let mut foo2 = Foo{x: 10, y: 20};
    foo = &mut foo2;
    println!("foo.x = {}\nfoo.y = {}", foo.x, foo.y);
}

fn main() {
    let mut foo = Foo{x: 1, y: 2};
    do_something(&mut foo);
    println!("foo.x = {}\nfoo.y = {}", foo.x, foo.y);
}

以下是编译器所说的:

error[E0597]: `foo2` does not live long enough
  --> main.rs:11:11
   |
6  | fn do_something(mut foo: &mut Foo) {
   |                          - let's call the lifetime of this reference `'1`
...
10 |     let mut foo2 = Foo{x: 10, y: 20};
   |         -------- binding `foo2` declared here
11 |     foo = &mut foo2;
   |     ------^^^^^^^^^
   |     |     |
   |     |     borrowed value does not live long enough
   |     assignment requires that `foo2` is borrowed for `'1`
12 |     println!("foo.x = {}\nfoo.y = {}", foo.x, foo.y);
13 | }
   | - `foo2` dropped here while still borrowed

error: aborting due to previous error

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