您如何写给pty master Rust

Question

我已经创建了一个简单的pty设置，但是我不确定一旦创建后如何实际写入主端或从端。我也不确定我的设置是否正确，因为在检查时，pty的子进程的Stdin，Stdout和Stderr全部为None，而不是被设置为从属文件描述符。任何人都可以澄清这是否正确，如果不正确，您对如何解决它有任何建议吗？

use libc::{self};
use nix::pty::openpty;
use std::os::unix::io::FromRawFd;
use std::process::{Child, Command, Stdio};

#[derive(Debug)]
pub struct Pty {
    process: Child,
    fd: i32,
}

fn create_pty(process: &str) -> Pty {
    let ends = openpty(None, None).expect("openpty failed");
    let master = ends.master;
    let slave = ends.slave;

    let mut builder = Command::new(process);
    builder.stdin(unsafe { Stdio::from_raw_fd(slave) });
    builder.stdout(unsafe { Stdio::from_raw_fd(slave) });
    builder.stderr(unsafe { Stdio::from_raw_fd(slave) });

    match builder.spawn() {
        Ok(process) => {
            let pty = Pty {
                process,
                fd: master,
            };

            pty
        }
        Err(e) => {
            panic!("Failed to create pty: {}", e);
        }
    }
}

fn main() {
    let shell = "/bin/bash";

    let pty = create_pty(shell);
    println!("{:?}", pty);

    println!("{}", pty.process.id());
}

Answer 1

这是预期的。未为原始文件句柄设置std::process::Child::stdin和朋友（因为Rust不知道它们是什么，所以构建器没有pty的master结尾）。

您可以自己为母版构建生锈的文件句柄：

fn main() {
    let shell = "/bin/bash";

    let pty = create_pty(shell);
    println!("{:?}", pty);

    let mut output = unsafe { File::from_raw_fd(pty.fd) };
    write!(output, "touch /tmp/itworks\n");
    output.flush();

    std::thread::sleep_ms(1000);

    println!("{}", pty.process.id());
}

您将看到它确实创建了文件“ / tmp / itworks”。

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