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。谁能澄清这是否正确,如果不正确,您是否有任何建议?
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());
}
答案 0 :(得分:3)
这是预期的。没有为std::process::Child::stdin
和朋友设置原始文件句柄(因为Rust不知道它们是什么,所以构建器没有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());
}
您会看到它确实创建了一个文件“ / tmp / itworks”。