问题描述
给出一个固定长度的 char 数组，例如：
Given a fixed-length char array such as:
let s: [char; 5] = ['h', 'e', 'l', 'l', 'o'];
如何获取& str ？
推荐答案
您不能没有分配，这意味着您最终会得到 String 。
You can't without some allocation, which means you will end up with a String.
let s2: String = s.iter().collect();
问题是Rust中的字符串不是<$ c的 not 集合$ c> char s，它们是UTF-8，这是一种编码，每个字符没有固定大小。
The problem is that strings in Rust are not collections of chars, they are UTF-8, which is an encoding without a fixed size per character.
例如，数组在这种情况下，总共需要20个字节才能占用5 x 32位。字符串的数据总共需要5个字节（尽管还有3个指针大小的值，所以在这种情况下，整个 String 会占用更多的内存）。
For example, the array in this case would take 5 x 32-bits for a total of 20 bytes. The data of the string would take 5 bytes total (although there's also 3 pointer-sized values, so the overall String takes more memory in this case).
我们从数组开始并调用，其产生的值类型为&字符。然后，我们使用到。这使用迭代器的 size_hint 来中预分配空间，从而减少了额外的分配。
We start with the array and call []::iter, which yields values of type &char. We then use Iterator::collect to convert the Iterator<Item = &char> into a String. This uses the iterator's size_hint to pre-allocate space in the String, reducing the need for extra allocations.
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