Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
873 views
in Technique[技术] by (71.8m points)

rust - Why does the usage of by_ref().take() differ between the Iterator and Read traits?

Here are two functions:

fn foo<I>(iter: &mut I)
where
    I: std::iter::Iterator<Item = u8>,
{
    let x = iter.by_ref();
    let y = x.take(2);
}

fn bar<I>(iter: &mut I)
where
    I: std::io::Read,
{
    let x = iter.by_ref();
    let y = x.take(2);
}

While the first compiles fine, the second gives the compilation error:

error[E0507]: cannot move out of borrowed content
  --> src/lib.rs:14:13
   |
14 |     let y = x.take(2);
   |             ^ cannot move out of borrowed content

The signatures of by_ref and take are almost identical in std::iter::Iterator and std::io::Read traits, so I supposed that if the first one compiles, the second will compile too. Where am I mistaken?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

impl<'a, I: Iterator + ?Sized> Iterator for &'a mut I is the reason why the first function compiles. It implements Iterator for all mutable references to iterators.

The Read trait has the equivalent, but, unlike Iterator, the Read trait isn't in the prelude, so you'll need to use std::io::Read to use this impl:

use std::io::Read; // remove this to get "cannot move out of borrowed content" err

fn foo<I, T>(iter: &mut I)
where
    I: std::iter::Iterator<Item = T>,
{
    let _y = iter.take(2);
}

fn bar<I>(iter: &mut I)
where
    I: std::io::Read,
{
    let _y = iter.take(2);
}

Playground


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...