[Practice] 该段代码为什么会通过？ #3

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1. ZhangHanDong

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   Questions-Practice

   on Oct 28, 2017
2. ZhangHanDong

   changed the title [-]该段代码为什么会通过？[/-] [+][Practice] 该段代码为什么会通过？[/+] on Oct 28, 2017
3. ZhangHanDong commented on Oct 28, 2017

   

   ZhangHanDong

   on Oct 28, 2017

   ContributorAuthor

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   相关issue： rust-lang/rust#45587
4. huangjj27 commented on Feb 6, 2018

   

   huangjj27

   on Feb 6, 2018

   Last edited by huangjj27

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   当去掉let语句之后产生了如下的语义:

   1. rustc推断在main中声明的map的K, V是具有'static lifetime, 理由是foo函数接受了map作为参数;
   2. foo返回的Some是to_owned的关系, 所以变量baz会被move到闭包之中, 于是其析构也会提前到闭包调用结束, baz的生命周期为整个闭包调用

   因此baz的声明周期没有'static长, 所以报错:

    error[E0597]: `baz` does not live long enough
     --> test.rs:12:29
      |
   12 |         map.insert("sign", &baz);
      |                             ^^^ borrowed value does not live long enough
   13 |         Some(())
   14 |     });
      |     - borrowed value only lives until here
      |
      = note: borrowed value must be valid for the static lifetime...
   

   而当我们保留了let语句是, 相当于把mapmove到了闭包中, 并且在let语句的时候重新推断了生命周期条件:

   // let dist_map: BTreeMap(&'static str, &'a str) = moved_map;
   let map = map;
   

   而这种推断是合理的, 因为'static生命周期要比任何生命周期'a'要长('static: 'a);

   接着, rust再推断出满足'a lifetime的最小lifetime 要比&baz的lifetime要长('a: 'baz), 所有生命周期参数得以满足, 于是通过了编译.