Rust 在跨平台应用程序中使用std：：OS：：Linux和std：：OS：：Windows

根据目标操作系统使用#[cfg(…) attribute进行条件编译.这样编译器就不会try 编译针对不同操作系统的代码:

pub fn run(path: &str){
    let files = fs::read_dir(path).unwrap_or_else(|error: std::io::Error| {
        println!("Error: {error}");
        process::exit(1);
    });

    // The 'os' variable is obtained from std::env::consts::OS
    #[cfg (target_os = "windows")]
    {
        // Use std::os::windows in the following function
        use std::os::windows::*;
        printDir_win(files);
    }
    #[cfg (target_os = "linux")]
    {
        // Use std::os::linux in the following function
        use std::os::linux;
        printDir_lin(files);
    }
    #[cfg (not (any (target_os = "windows", target_os = "linux")))]
    {
        println!("{}OS not recognised: defaulting to std::fs methods{}","\x1b[31m","\x1b[0m");
        printDir(files);
    }
}

请注意，从组织的Angular 来看，将特定于平台的代码分组到一个模块中是很好的做法，该模块抽象了通用API背后的差异:

#[cfg (target_os = "windows")]
mod platform {
    use std::os::windows::*;
    pub fn print_dir (files: std::fs::ReadDir) {
        unimplemented!()
    }
}
#[cfg (target_os = "linux")]
mod platform {
    use std::os::linux::*;
    pub fn print_dir (files: std::fs::ReadDir) {
        unimplemented!()
    }
}
#[cfg (not (any (target_os = "windows", target_os = "linux")))]
mod platform {
    pub fn print_dir (files: std::fs::ReadDir) {
        unimplemented!()
    }
}

然后，您的代码的其余部分不需要判断每个地方的操作系统来知道要调用哪个函数:

pub fn run(path: &str){
    let files = std::fs::read_dir(path).unwrap_or_else(|error: std::io::Error| {
        println!("Error: {error}");
        std::process::exit(1);
    });
    platform::print_dir (files);
}

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