NLL 非词法生命周期

  源码:
      1 | fn main() {
      2 |     let mut v = vec![1, 2, 3];
      3 |     let first = &v[0];
      4 |     println!("first = {first}");      // first 的最后一次使用
      5 |     v.push(4);                         // 旧的「按作用域」规则会拒绝这一行;NLL 允许
      6 |     println!("{:?}", v);
      7 |     let r = &mut v;
      8 |     r.push(5);
      9 |     v.push(6);                         // r 已经不再使用
     10 |     println!("{:?}", v);
     11 | }
  编译:成功
  运行(退出码 0):
    first = 1
    [1, 2, 3, 4]
    [1, 2, 3, 4, 5, 6]

  源码:
      1 | fn main() {
      2 |     let mut s = String::from("a");
      3 |     let r = &s;
      4 |     let flag = std::env::args().count() > 5;
      5 |     if flag { println!("{r}"); }       // 只有这条路径用 r
      6 |     else { s.push('b'); }             // 另一条路径里 r 已经死了
      7 |     println!("{s}");
      8 | }
  编译:成功
  运行(退出码 0):
    ab

  源码:
      1 | fn main() {
      2 |     let mut v = vec![1, 2, 3];
      3 |     v.push(v.len());
      4 |     println!("{:?}", v);
      5 | }
  编译:成功
  运行(退出码 0):
    [1, 2, 3, 3]
  源码:
      1 | fn main() {
      2 |     let mut v = vec![1, 2, 3];
      3 |     let r = &mut v;
      4 |     r.push(v.len());     // 先取走了 &mut,再读 v:冲突
      5 | }
  编译:失败
    error[E0502]: cannot borrow `v` as immutable because it is also borrowed as mutable
     --> <实验目录>/src/u22d.rs:4:12
      |
    3 |     let r = &mut v;
      |             ------ mutable borrow occurs here
    4 |     r.push(v.len());     // 先取走了 &mut,再读 v:冲突
      |       ---- ^ immutable borrow occurs here
      |       |
      |       mutable borrow later used by call
    
    error: aborting due to 1 previous error
    
    For more information about this error, try `rustc --explain E0502`.

  源码:
      1 | struct Guard<'a>(&'a mut Vec<i32>);
      2 | impl Drop for Guard<'_> { fn drop(&mut self) { println!("Guard 释放,vec 长度 {}", self.0.len()); } }
      3 | fn main() {
      4 |     let mut v = vec![1];
      5 |     let g = Guard(&mut v);
      6 |     v.push(2);
      7 | }
  编译:失败
    error[E0499]: cannot borrow `v` as mutable more than once at a time
     --> <实验目录>/src/u22e.rs:6:5
      |
    5 |     let g = Guard(&mut v);
      |                   ------ first mutable borrow occurs here
    6 |     v.push(2);
      |     ^ second mutable borrow occurs here
    7 | }
      | - first borrow might be used here, when `g` is dropped and runs the `Drop` code for type `Guard`
    
    warning: unused variable: `g`
     --> <实验目录>/src/u22e.rs:5:9
      |
    5 |     let g = Guard(&mut v);
      |         ^ help: if this is intentional, prefix it with an underscore: `_g`
      |
      = note: `#[warn(unused_variables)]` (part of `#[warn(unused)]`) on by default
    
    error: aborting due to 1 previous error; 1 warning emitted
    
    For more information about this error, try `rustc --explain E0499`.
  源码:
      1 | struct Guard<'a>(&'a mut Vec<i32>);
      2 | impl Drop for Guard<'_> { fn drop(&mut self) { println!("Guard 释放,vec 长度 {}", self.0.len()); } }
      3 | fn main() {
      4 |     let mut v = vec![1];
      5 |     { let _g = Guard(&mut v); }      // 缩小 Guard 的作用域
      6 |     v.push(2);
      7 |     println!("{:?}", v);
      8 | }
  编译:成功
  运行(退出码 0):
    Guard 释放,vec 长度 1
    [1, 2]

  生成的 NLL 相关 MIR 转储文件: ['u22g.main.-------.nll.0.mir', 'u22g.main.-------.nll.0.regioncx.all.dot', 'u22g.main.-------.nll.0.regioncx.scc.dot'] (共 3 个)
  NLL 区域信息(借用区域 '?N 覆盖的 MIR 位置)节选:
    | '?0 | Global | ['?0, '?1]
    | '?1 | Local | ['?1]
    | '?0 | U0 | {bb0[0..=3], bb1[0..=2], bb2[0..=1], bb3[0..=6], bb4[0..=16], bb5[0..=12], bb6[0..=4], bb7[0..=9], bb8[0..=5], bb9[0..=1], bb10[0], bb11[
    | '?1 | U0 | {bb0[0..=3], bb1[0..=2], bb2[0..=1], bb3[0..=6], bb4[0..=16], bb5[0..=12], bb6[0..=4], bb7[0..=9], bb8[0..=5], bb9[0..=1], bb10[0], bb11[
    | '?2 | U0 | {bb3[5..=6], bb4[0..=16], bb5[0..=12], bb6[0..=4]}
    | '?3 | U0 | {bb4[1..=16], bb5[0..=12], bb6[0..=4]}
    | '?4 | U0 | {bb4[8..=16], bb5[0..=12], bb6[0..=4]}
    | '?5 | U0 | {bb4[15..=16], bb5[0..=12], bb6[0..=4]}