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

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");

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.


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());

Tags:

Rust