如何将 u8 的向量打印为字符串?

问题描述

这是我的代码:

let mut altbuf: Vec<u8> = Vec::new();

// Stuff here...

match stream.read_byte() {
    Ok(d) => altbuf.push(d),
    Err(e) => { println!("Error: {}", e); doneflag = true; }
}

for x in altbuf.iter() {
    println!("{}", x);
}

代码打印 u8 字节是正确的，但我终生不知道如何将纯 u8 字节向量转换为字符串?关于堆栈溢出的类似问题的唯一其他答案假定您正在使用类型为 &[u8] 的向量.

The code prints u8 bytes which are correct, but I can't figure out for the life of me how to convert a vector of pure u8 bytes into a string? The only other answer to a similar question on stack overflow assumes that you're working with a vector of type &[u8].

推荐答案

如果你看一下 String 文档，您可以使用一些方法.有 String::from_utf8 需要一个 Vec，还有 String::from_utf8_lossy 需要一个 &[u8].

If you look at the String documentation, there are a few methods you could use. There's String::from_utf8 that takes a Vec<u8>, and there's also String::from_utf8_lossy which takes a &[u8].

请注意，Vec 或多或少是 [T] 周围的一个拥有的、可调整大小的包装器.也就是说，如果你有一个 Vec，你可以把它变成一个 &[u8]，最简单的方法是重新借用它(ie &*some_vec).您还可以直接在 Vec 上调用任何定义在 &[T] 上的方法(通常，实现 Deref trait).

Note that a Vec<T> is more-or-less an owned, resizable wrapper around a [T]. That is, if you have a Vec<u8>, you can turn it into a &[u8], most easily by re-borrowing it (i.e. &*some_vec). You can also call any methods defined on &[T] directly on a Vec<T> (in general, this is true of things that implement the Deref trait).

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