你如何结合生锈的生命？

Question

这段代码:

struct Foo<'a> {
  value: Option<&'a int>,
  parent: Option<&'a Foo<'a>>
}

impl<'a> Foo<'a> {
  fn bar<'a, 'b, 'c: 'a + 'b>(&'a self, other:&'b int) -> Foo<'c> {
    return Foo { value: Some(other), parent: Some(self) };
  }
}

fn main() {
  let e = 100i;
  {
    let f = Foo { value: None, parent: None };
    let g:Foo;
    {
       g = f.bar(&e);
    }
    // <--- g should be valid here
  }
  // 'a of f is now expired, so g should not be valid here.

  let f2 = Foo { value: None, parent: None };
  {
    let e2 = 100i;
    let g:Foo;
    {
       g = f2.bar(&e2);
    }
    // <--- g should be valid here
  }
  // 'b of e2 is now expired, so g should not be valid here.
}

无法编译错误:

<anon>:8:30: 8:35 error: cannot infer an appropriate lifetime due to conflicting requirements
<anon>:8     return Foo { value: Some(other), parent: Some(self) };
                                      ^~~~~
<anon>:7:3: 9:4 note: consider using an explicit lifetime parameter as shown: fn bar<'a, 'b>(&'a self, other: &'b int) -> Foo<'b>
<anon>:7   fn bar<'a, 'b, 'c: 'a + 'b>(&'a self, other:&'b int) -> Foo<'c> {
<anon>:8     return Foo { value: Some(other), parent: Some(self) };
<anon>:9   }

(围栏:http://is.gd/vAvNFi)

这显然是一个人为的例子,但我偶尔也想这样做.

所以...

1)你如何结合一生？(即返回一个寿命至少为'a或'b的Foo,它的寿命更短)

2)有没有办法将测试写入资产生命周期编译失败？(例如,尝试编译以错误方式使用该函数的#[test],并以生命周期错误失败)

Answer 1

边界'c: 'a + 'b意味着'c至少与长度'a一样长'b.但是,在这种情况下,该Foo值对于恰好最短的'a和有效'b:一旦引用后面的数据超出范围,整个Foo必须无效.(这是说,数据的有效期为'c在有效工会的'a和'b.)

更具体地说,比方说'b = 'static,那'c: 'a + 'static意味着'c必须也是'static,所以返回值就是Foo<'static>.这显然是不对的,因为它将有限的'a self引用"升级"为永久持续的引用.

这里的正确行为是生命周期的交集:Foo只有在两个函数参数都有效时才有效.交集操作只是用相同的名称标记引用:

fn bar<'a>(&'a self, other: &'a int) -> Foo<'a>