How can I wait for a Rust `Child` process whose stdout has been piped to another? - process

Question

I want to implement yes | head -n 1 in Rust, properly
connecting pipes and checking exit statuses: i.e., I want to be able to
determine that yes exits due to SIGPIPE and that head completes
normally. The pipe functionality is easy (Rust Playground):
use std::process;
fn a() -> std::io::Result<()> {
let child_1 = process::Command::new("yes")
.arg("abracadabra")
.stdout(process::Stdio::piped())
.spawn()?;
let pipe: process::ChildStdout = child_1.stdout.unwrap();
let child_2 = process::Command::new("head")
.args(&["-n", "1"])
.stdin(pipe)
.stdout(process::Stdio::piped())
.spawn()?;
let output = child_2.wait_with_output()?;
let result = String::from_utf8_lossy(&output.stdout);
assert_eq!(result, "abracadabra\n");
println!("Good from 'a'.");
Ok(())
}
But while we can wait on child_2 at any point, the declaration of
pipe moves child_1, and so it’s not clear how to wait on child_1.
If we were to simply add child_1.wait()? before the assert_eq!, we’d
hit a compile-time error:
error[E0382]: borrow of moved value: `child_1`
--> src/main.rs:15:5
|
8 | let pipe: process::ChildStdout = child_1.stdout.unwrap();
| -------------- value moved here
...
15 | child_1.wait()?;
| ^^^^^^^ value borrowed here after partial move
|
= note: move occurs because `child_1.stdout` has type `std::option::Option<std::process::ChildStdout>`, which does not implement the `Copy` trait
We can manage to wait with unsafe and platform-specific
functionality (Rust Playground):
use std::process;
fn b() -> std::io::Result<()> {
let mut child_1 = process::Command::new("yes")
.arg("abracadabra")
.stdout(process::Stdio::piped())
.spawn()?;
use std::os::unix::io::{AsRawFd, FromRawFd};
let pipe: process::Stdio =
unsafe { FromRawFd::from_raw_fd(child_1.stdout.as_ref().unwrap().as_raw_fd()) };
let mut child_2 = process::Command::new("head")
.args(&["-n", "1"])
.stdin(pipe)
.stdout(process::Stdio::piped())
.spawn()?;
println!("child_1 exited with: {:?}", child_1.wait().unwrap());
println!("child_2 exited with: {:?}", child_2.wait().unwrap());
let mut result_bytes: Vec<u8> = Vec::new();
std::io::Read::read_to_end(child_2.stdout.as_mut().unwrap(), &mut result_bytes)?;
let result = String::from_utf8_lossy(&result_bytes);
assert_eq!(result, "abracadabra\n");
println!("Good from 'b'.");
Ok(())
}
This prints:
child_1 exited with: ExitStatus(ExitStatus(13))
child_2 exited with: ExitStatus(ExitStatus(0))
Good from 'b'.
This is good enough for the purposes of this question, but surely there
must be a safe and portable way to do this.
For comparison, here is how I would approach the task in C (without
bothering to capture child_2’s output):
#include <stdio.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
#define FAILIF(e, msg) do { if (e) { perror(msg); return 1; } } while (0)
void describe_child(const char *name, int status) {
if (WIFEXITED(status)) {
fprintf(stderr, "%s exited %d\n", name, WEXITSTATUS(status));
} else if (WIFSIGNALED(status)) {
fprintf(stderr, "%s signalled %d\n", name, WTERMSIG(status));
} else {
fprintf(stderr, "%s fate unknown\n", name);
}
}
int main(int argc, char **argv) {
int pipefd[2];
FAILIF(pipe(pipefd), "pipe");
pid_t pid_1 = fork();
FAILIF(pid_1 < 0, "child_1: fork");
if (!pid_1) {
FAILIF(dup2(pipefd[1], 1) == -1, "child_1: dup2");
FAILIF(close(pipefd[0]), "child_1: close pipefd");
execlp("yes", "yes", "abracadabra", NULL);
FAILIF(1, "child_1: execlp");
}
pid_t pid_2 = fork();
FAILIF(pid_2 < 0, "child_2: fork");
if (!pid_2) {
FAILIF(dup2(pipefd[0], 0) == -1, "child_2: dup2");
FAILIF(close(pipefd[1]), "child_2: close pipefd");
execlp("head", "head", "-1", NULL);
FAILIF(1, "child_2: execlp");
}
FAILIF(close(pipefd[0]), "close pipefd[0]");
FAILIF(close(pipefd[1]), "close pipefd[1]");
int status_1;
int status_2;
FAILIF(waitpid(pid_1, &status_1, 0) == -1, "waitpid(child_1)");
FAILIF(waitpid(pid_2, &status_2, 0) == -1, "waitpid(child_2)");
describe_child("child_1", status_1);
describe_child("child_2", status_2);
return 0;
}
Save to test.c and run with make test && ./test:
abracadabra
child_1 signalled 13
child_2 exited 0

Answer 1

Use Option::take:
let pipe = child_1.stdout.take().unwrap();
let child_2 = process::Command::new("head")
.args(&["-n", "1"])
.stdin(pipe)
.stdout(process::Stdio::piped())
.spawn()?;
let output = child_2.wait_with_output()?;
child_1.wait()?;
See also:
How do I move out of a struct field that is an Option?
Unable to pipe to or from spawned child process more than once
How do I read the output of a child process without blocking in Rust?
Reading and writing to a long-running std::process::Child
Cannot move out of borrowed content when unwrapping