Reading program return value in shell using python

Question

From the python document http://docs.python.org/2/library/subprocess.html

If I type the following in python

>>> subprocess.call(["ls", "-l"])

I will get a 0.

If I type the following in python,

>>> subprocess.call("exit 1", shell=True)

I will get a 1. However, if I type

>>> subprocess.call("exit 1")

It will show me an error

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/subprocess.py", line 493, in call
    return Popen(*popenargs, **kwargs).wait()
  File "/usr/lib/python2.7/subprocess.py", line 679, in __init__
    errread, errwrite)
  File "/usr/lib/python2.7/subprocess.py", line 1249, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

Why does this happen?

Second question, if ls crashes, will I get a non-zero return value by using the following command?

>>> subprocess.call(["ls", "-l"])

Answer 1

exit is a shell routine, not a real program. You get an error because the call can not find a program named exit . If you don't set the shell argument to True this makes no sense to executing it.

To answer your second question, yes you will get an non-zero returning value. Try listing a directory in which you don't have read rights.

Answer 2

This is because if shell=True the command will be executed through a shell. You can pass a command line like in a shell. If you omit that param, the first param is threatened as a file name, the file name of the command. and there is no command 'exit 1'

Here comes another example using the ls -al command:

import subprocess

subprocess.call("ls -al", shell=True) # works
subprocess.call("ls -al") # fails

Also note the answer of @ibi0tux regarding exit