将&str转换为OsStr的正确方法是什么？

我正在使用防锈保险丝,它可以作为安装选项&[&std::ffi::os_str::OsStr].看来我应该拆分我的传入的逗号分隔选项字符串,我这样做:

mod fuse {
    use std::ffi::OsStr;

    pub fn mount(options: &[&OsStr]) {}
}

fn example(optstr: &str) {
    let mut options: &[&str] = &[];
    if optstr != "" {
        options = optstr.split(",").collect::>().as_slice();
    }
    fuse::mount(options)
}

这给出了以下错误:

error[E0308]: mismatched types
  --> src/main.rs:12:17
   |
12 |     fuse::mount(options)
   |                 ^^^^^^^ expected struct `std::ffi::OsStr`, found str
   |
   = note: expected type `&[&std::ffi::OsStr]`
              found type `&[&str]`

use std::ffi::OsStr;

fn main() {
    let a_string: &str = "Hello world";
    let an_os_str: &OsStr = OsStr::new(a_string);
    println!("{:?}", an_os_str);
}

let options: Vec<_> = optstr.split(",").map(OsStr::new).collect();
fuse::mount(&options)