Python: fork, pipe and exec

Question

I want to execute a program in a python application, it will run in the background but eventually come to the foreground.

A GUI is used to interact with it. But con

Answer 1

First, the python subprocess module is the correct answer.

As an subprocess example:

import subprocess
x = subprocess.check_output(["echo","one","two","three"])

Where x will be the output (python3 bytes class: x.decode('utf-8') for string)

Note that this will NOT duplicate stderr. If you need stderr as well, you can do something like:

x = subprocess.check_output(["bash","-c", 'echo foo; echo bar >&2'],stderr=subprocess.STDOUT)

Of course, there are many other ways of capturing stderr, including to a different output variable.

Using direct control

However, if you are doing something tricky and need to have direct control, @Beginner's answer is flawed in a few ways: os._exit(0) was included which immediately causes the child to exit, rendering everything else pointless. No os.execve() rendering the primary goal of the question pointless. No way to access the child's stdout/stderr as another question goal.

import os
rside, wside = os.pipe()
if not os.fork():
    # Child

    os.close(rside)
    # Make stdout go to parent
    os.dup2(wside, 1)
    # Make stderr go to parent
    os.dup2(wside, 2)
    # Optionally make stdin come from nowhere
    devnull = os.open("/dev/null", os.O_RDONLY)
    os.dup2(devnull, 0)
    # Execute the desired program
    os.execve("/bin/bash",["/bin/bash","-c","echo stdout; echo stderr >&2"],os.environ)
    print("Failed to exec program!")
    sys.exit(1)

# Parent
os.close(wside)
pyrside = os.fdopen(rside)

for line in pyrside:
   print("Child (stdout or stderr) said: <%s>"%line)

# Prevent zombies!  Reap the child after exit
pid, status = os.waitpid(-1, 0)
print("Child exited: pid %d returned %d"%(pid,status))