How do I invoke a system command in Rust and capture its output?

Question

Is there a way to invoke a system command, like ls or fuser in Rust? How about capturing its output?

Answer 1

std::process::Command allows for that.

There are multiple ways to spawn a child process and execute an arbitrary command on the machine:

• spawn — runs the program and returns a value with details
• output — runs the program and returns the output
• status — runs the program and returns the exit code

One simple example from the docs:

use std::process::Command;

Command::new("ls")
        .arg("-l")
        .arg("-a")
        .spawn()
        .expect("ls command failed to start");

Answer 2

a very clear example from the docs:

use std::process::Command;
let output = Command::new("/bin/cat")
                     .arg("file.txt")
                     .output()
                     .expect("failed to execute process");

println!("status: {}", output.status);
println!("stdout: {}", String::from_utf8_lossy(&output.stdout));
println!("stderr: {}", String::from_utf8_lossy(&output.stderr));

assert!(output.status.success());

Answer 3

It is indeed possible! The relevant module is std::run.

let mut options = std::run::ProcessOptions::new();
let process = std::run::Process::new("ls", &[your, arguments], options);

ProcessOptions’ standard file descriptors default to None (create a new pipe), so you can just use process.output() (for example) to read from its output.

If you want to run the command and get all its output after it’s done, there’s wait_with_output for that.

Process::new, as of yesterday, returns an Option<Process> instead of a Process, by the way.

Answer 4

Or you can try this crate cmd_lib, it is a wrapper around std::process, to write shell-script like tasks in a clean, natural and rusty way:

let all_files = run_fun!("ls -a .")?;
// pipe commands are also supported
run_cmd!("du -ah . | sort -hr | head -n 10");