string 将 str 转换为 &[u8]

Converting a str to a &[u8]

提问by ynimous

This seems trivial, but I cannot find a way to do it.

这似乎微不足道，但我找不到办法做到这一点。

For example,

例如，

fn f(s: &[u8]) {}

pub fn main() {
    let x = "a";
    f(x)
}

Fails to compile with:

无法编译：

error: mismatched types:
 expected `&[u8]`,
    found `&str`
(expected slice,
    found str) [E0308]

documentation, however, states that:

然而，文档指出：

The actual representation of strs have direct mappings to slices: &str is the same as &[u8].

strs 的实际表示直接映射到切片：&str 与 &[u8] 相同。

回答by fjh

You can use the as_bytesmethod:

您可以使用as_bytes方法：

fn f(s: &[u8]) {}

pub fn main() {
    let x = "a";
    f(x.as_bytes())
}

or, in your specific example, you could use a byte literal:

或者，在您的具体示例中，您可以使用字节文字：

let x = b"a";
f(x)