创建返回实现serde :: Deserialize的值的函数时，发生生命周期错误

问题描述

我正在使用serde和serde_json 1.0从base64字符串中解码数据:

I'm using serde and serde_json 1.0 to decode data from a base64 string:

fn from_base64_str<T: Deserialize>(string: &str) -> T {
    let slice = decode_config(string, URL_SAFE).unwrap();
    serde_json::from_slice(&slice).unwrap()
}

当我编译时，我得到了:

When I compile, I got this:

error[E0106]: missing lifetime specifier
 --> src/main.rs:6:23
  |
6 | fn from_base64_str<T: Deserialize>(string: &str) -> T {
  |                       ^^^^^^^^^^^ expected lifetime parameter

检查Serde文档，Deserialize定义为:

Checking the serde doc, Deserialize is defined as:

pub trait Deserialize<'de>: Sized {

所以我增加了寿命:

fn from_base64_str<'de, T: Deserialize<'de>>(string: &str) -> T {
    let slice = decode_config(string, URL_SAFE).unwrap();
    serde_json::from_slice(&slice).unwrap()
}

然后编译器告诉我:

error: `slice` does not live long enough
  --> src/main.rs:11:29
   |
11 |     serde_json::from_slice(&slice).unwrap()
   |                             ^^^^^ does not live long enough
12 | }
   | - borrowed value only lives until here
   |
note: borrowed value must be valid for the lifetime 'de as defined on the body at 9:64...
  --> src/main.rs:9:65
   |
9  |   fn from_base64_str<'de, T: Deserialize<'de>>(string: &str) -> T {
   |  _________________________________________________________________^ starting here...
10 | |     let slice = decode_config(string, URL_SAFE).unwrap();
11 | |     serde_json::from_slice(&slice).unwrap()
12 | | }
   | |_^ ...ending here

我只知道Rust中生命周期的最基本知识，所以我对trait Deserialize中的'de感到非常困惑.

I only know the very basics of lifetimes in Rust, so I'm very confused by the 'de in trait Deserialize.

如何解决此类功能中的生命周期错误?我正在使用Rust 1.18.0-nightly(452bf0852 2017-04-19)

How can I fix the lifetime error in such function? I'm using Rust 1.18.0-nightly (452bf0852 2017-04-19)

推荐答案

我从 Serde问题891 :我应该使用DeserializeOwned而不是Deserialize.

I found an answer from Serde issue 891: I should use DeserializeOwned instead of Deserialize.

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