Vec< struct>之间有什么区别？和&[struct]？ [英] What is the difference between Vec<struct> and &[struct]?

问题描述

我经常发现自己得到这样的错误：

 不匹配的类型：expected`collections :: vec :: Vec< u8&``，发现`& [u8]`（预期的结构集合:: vec :: Vec，found& -ptr）
  

据我所知，一个是可变的，一个不是但我不知道如何在类型之间，即采取一个& [u8] / code>，并使其成为 Vec< u8> ，反之亦然。

他们？是否与 String 和& str 相同

解决方案

与 String 和& str ？




是的。一个 Vec< T> 是& [T] 的所有变体。 & [T] 是一组 T 依次排列的参考在内存中（也称为切片）。它表示一个指向项目开头的项目和项目数量。引用是指您不拥有的内容，因此您可以使用的一组操作是有限的。有一个可变变体（& mut [T] ），它允许您突变切片中的项目。你不能改变片中有多少。换句话说，你不能改变片段本身。

采取& [u8] code>并使其成为 Vec




对于这个具体案例：

  let s：& [u8]; //设置在某处
 Vec :: from（s）; 
  

但是，这不得不在堆栈上分配内存，然后复制每个值进入该内存。它比其他方式贵，但对于某种情况可能是正确的。

反之亦然





  let v = vec！[1u8,2,3]; 
 let s = v.as_slice（）; 
  

这基本上是免费的，因为 v 仍然拥有数据，我们只是提供一个参考。这就是为什么许多API尝试在有意义的时候采取切片。

I often find myself getting an error like this:

mismatched types: expected `collections::vec::Vec<u8>`, found `&[u8]` (expected struct collections::vec::Vec, found &-ptr)

As far as I know, one is mutable and one isn't but I've no idea how to go between the types, i.e. take a &[u8] and make it a Vec<u8> or vice versa.

What's the different between them? Is it the same as String and &str?

解决方案

Is it the same as String and &str?

Yes. A Vec<T> is the owned variant of a &[T]. &[T] is a reference to a set of Ts laid out sequentially in memory (a.k.a. a slice). It represents a pointer to the beginning of the items and the number of items. A reference refers to something that you don't own, so the set of actions you can do with it are limited. There is a mutable variant (&mut [T]), which allows you to mutate the items in the slice. You can't change how many are in the slice though. Said another way, you can't mutate the slice itself.

take a &[u8] and make it a Vec

For this specific case:

let s: &[u8]; // Set this somewhere
Vec::from(s);

However, this has to allocate memory not on the stack, then copy each value into that memory. It's more expensive than the other way, but might be the correct thing for a given situation.

or vice versa

let v = vec![1u8, 2, 3];
let s = v.as_slice();

This is basically "free" as v still owns the data, we are just handing out a reference to it. That's why many APIs try to take slices when it makes sense.

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