golang program terminates after calling linux shell command

Question

I am executing the following golang program (snippet) with root privilege:

  binary, lookErr := exec.LookPath("auditctl")
  if lookErr != nil {
    panic(lookErr)
  }
  env := os.Environ()

  args :=  []string{"auditctl", "-D"}
  execErr := syscall.Exec(binary, args, env)
  if execErr != nil {
    fmt.Println("error")
    panic(execErr)
  }

  fmt.Println("no error")

Because I dont have any auditctl rules in the system, the command print the following in the terminal. This is all normal just like when I type in the shell directly.

No rules

Except that neither "error" nor "no error" are printed. Which means the golang program terminates right after syscall.Exec. How did this happen and how can I continue execution after syscall.Exec because I have other things to run within the same program.

Answer 1

syscall.Exec invokes execve(2), which on linux does not return execution to the caller but replaces the current (Go) process with the process called.

As David Budworth, and mkopriva suggest.. if you actually want to spawn a separate process and return to your Go code after spawning; consider using exec.Command

Answer 2

As @mkopriva mentioned, you probably want exec.Cmd

here's an example:

cmd := exec.Command("auditctl", "-D")
cmd.Stdout = os.Stdout
cmd.Stderr = os.Stderr
cmd.Env = os.Environ()
if err := cmd.Run(); err != nil {
    log.Println("Failed to run auditctl:",err)
}