为什么我不能从 &'a T 返回 fmt::Arguments<'a>？

Question

根据我对生命周期的理解，如果函数的调用者在参数上指定了生命周期，我可以返回具有该生命周期的类型。

这有效，即使有省略：

pub fn substr(s: &str) -> &str {
    &s[0..1]
}

pub fn substr_ex<'a>(s: &'a str) -> &'a str {
    &s[0..1]
}

但这不是：

use std::fmt::Arguments;

pub fn as_format_arg<'a, T: 'a + ?Sized + Debug>(t: &'a T) -> Arguments<'a> {
    format_args!("{:?}", t)
}

error: borrowed value does not live long enough
  --> <anon>:16:18
   |
16 |     format_args!("{:?}", t)
   |                  ^^^^^^ does not live long enough
17 | }
   | - temporary value only lives until here
   |
   = note: borrowed value must be valid for the lifetime 'a as defined on unknown free region bounded by scope CodeExtent(38/CallSiteScope { fn_id: NodeId(42), body_id: NodeId(92) })...

error: `t` does not live long enough
  --> <anon>:16:26
   |
16 |     format_args!("{:?}", t)
   |                          ^ does not live long enough
17 | }
   | - borrowed value only lives until here
   |
   = note: borrowed value must be valid for the lifetime 'a as defined on unknown free region bounded by scope CodeExtent(38/CallSiteScope { fn_id: NodeId(42), body_id: NodeId(92) })...

这是一个错误吗？还是我误解了生命周期？

游戏围栏：https://play.rust-lang.org/?gist=5a7cb4c917b38e012f20c771893f8b3b&version=nightly

Answer 1

要了解发生了什么，让我们看看macro-expanded version：

fn as_format_arg<'a, T: 'a + ?Sized + Debug>(t: &'a T) -> Arguments<'a> {
    ::std::fmt::Arguments::new_v1(
        {
            static __STATIC_FMTSTR: &'static [&'static str] = &[""];
            __STATIC_FMTSTR
        },
        &match (&t,) {
            (__arg0,) => [::std::fmt::ArgumentV1::new(__arg0, ::std::fmt::Debug::fmt)],
        },
    )
}

这有助于解释第一个错误：

error: borrowed value does not live long enough
  --> src/main.rs:9:36
   |
9  |                                   &match (&t,) {
   |                                    ^ temporary value created here
...
15 | }
   | - temporary value only lives until here
   |
note: borrowed value must be valid for the lifetime 'a as defined on the block at 4:72...
  --> src/main.rs:4:73
   |
4  | fn as_format_arg<'a, T: 'a + ?Sized + Debug>(t: &'a T) -> Arguments<'a> {
   |                                                                         ^

具体来说，在堆栈上创建了一个ArgumentV1，并对其进行了引用。您不能从函数中返回该引用。

第二个错误：

error: `t` does not live long enough
  --> src/main.rs:9:44
   |
9  |                                   &match (&t,) {
   |                                            ^ does not live long enough
...
15 | }
   | - borrowed value only lives until here
   |
note: borrowed value must be valid for the lifetime 'a as defined on the block at 4:72...
  --> src/main.rs:4:73
   |
4  | fn as_format_arg<'a, T: 'a + ?Sized + Debug>(t: &'a T) -> Arguments<'a> {
   |                                                                         ^

注意format! family of macros doesn't take their arguments by value;他们会自动插入参考。你不希望println! 拥有你的价值！

这意味着打印的值实际上是一个&&'a T - 引用 到堆栈分配的t 值！同样，您不能返回对堆栈上分配的内容的引用。

如果函数的调用者在参数上指定了生命周期，我可以返回具有该生命周期的类型。

这是对的。您可以只返回返回该输入参数的一部分。您不能创建一个全新的值并在该生命周期内返回它。