生命周期限制
原文跟踪lifetime-mismatch.md Commit: d870b6788ba078ba398f020305ef9210f7cbd740
给出以下代码:
struct Foo;
impl Foo {
fn mutate_and_share(&mut self) -> &Self { &*self }
fn share(&self) {}
}
fn main() {
let mut foo = Foo;
let loan = foo.mutate_and_share();
foo.share();
}
人们可能期望它能够编译。我们称之为mutate_and_share临时借用 foo,但后来只返回共享引用。因此,我们期望foo.share()成功,因为foo不应该可变地借用。
但是当我们尝试编译它时:
error[E0502]: cannot borrow `foo` as immutable because it is also borrowed as mutable
--> src/lib.rs:11:5
|
10 | let loan = foo.mutate_and_share();
| --- mutable borrow occurs here
11 | foo.share();
| ^^^ immutable borrow occurs here
12 | }
| - mutable borrow ends here
发生了什么?好吧,我们得到以下内容:
struct Foo;
impl Foo {
fn mutate_and_share<'a>(&'a mut self) -> &'a Self { &'a *self }
fn share<'a>(&'a self) {}
}
fn main() {
'b: {
let mut foo: Foo = Foo;
'c: {
let loan: &'c Foo = Foo::mutate_and_share::<'c>(&'c mut foo);
'd: {
Foo::share::<'d>(&'d foo);
}
}
}
}
由于loan
的生命周期和mutate_and_share
的签名, 生命周期系统强制扩展&mut foo
的生命周期为'c
,然后当我们试着调用share
的时候,它说我们正试图别名&'c mut foo
!
根据我们实际关注的引用语义,这个程序显然是正确的,但是生命周期系统太粗糙而无法处理。