rust存取一个含有borrowed域的结构体
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2021-07-28 10:31:32
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直接存取不可以:
use std::io;use serde::{Serialize, Deserialize};extern crate bincode;#[derive(Debug, Serialize, Deserialize)] struct A <'a > { a: &'a [u8 ] } fn main () { let a : &[u8 ] = &[2 , 3 , 5 , 7 , 11 , 13 ]; let b = A{a}; let mut v = Vec ::new (); bincode::serialize_into (&mut v, &b).unwrap (); let mut reader = io::Cursor::new (&v); let c = bincode::deserialize_from::<_, A>(&mut reader).unwrap (); println! ("{:?}" , c); }
会报错:
error: implementation of `Deserialize` is not general enough --> src/main.rs:19:13 | 19 | let c = bincode::deserialize_from::<_, A>(&mut reader).unwrap(); | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ implementation of `Deserialize` is not general enough | = note: `A<'_>` must implement `Deserialize<'0>`, for any lifetime `'0`... = note: ...but `A<'_>` actually implements `Deserialize<'1>`, for some specific lifetime `'1` error: could not compile `serde-borrowed` due to previous error
这是因为BufWriter里的缓存的所有权不是调用者的,可能会被释放,这样构造含有borrowed域的结构体时就不能安全地借用读出来的数据了。
可以用std::borrow::Cow
实现在serialize的时候borrow,但是在deserialize的时候create,从而解决这个问题:
use std::{io, borrow::Cow};use serde::{Serialize, Deserialize};extern crate bincode;#[derive(Debug, Serialize, Deserialize)] struct A <'a > { a: Cow<'a , [u8 ]>, } fn main () { let a : &[u8 ] = &[2 , 3 , 5 , 7 , 11 , 13 ]; let b = A{a: Cow::Borrowed (a)}; let mut v = Vec ::new (); bincode::serialize_into (&mut v, &b).unwrap (); let mut reader = io::Cursor::new (&v); let c = bincode::deserialize_from::<_, A>(&mut reader).unwrap (); if let Cow ::Borrowed (_) = c.a { panic! (); } println! ("{:?}" , c); }
此外,bincode有forward_read_*
系列函数,好像也可以通过转移所有权解决这个问题:
https://docs.rs/bincode/1.3.1/bincode/de/read/trait.BincodeRead.html
但是我不会。
¶ 参考文献
https://stackoverflow.com/questions/60801133/how-do-i-use-serde-to-deserialize-structs-with-references-from-a-reader
https://stackoverflow.com/questions/52733047/how-to-borrow-a-field-for-serialization-but-create-it-during-deserialization
How to
deserialize maybe-borrowed maybe-copied data?