I'm trying to understand -c option for bash better. The man page says:
-c: If the -c option is present, then commands are read from the first non-option argument command_string. If there are arguments after the command_string, they are assigned to the positional parameters, starting with $0.
I'm having trouble understanding what this means.
If I do the following command with and without bash -c, I get the same result (example from http://www.tldp.org/LDP/abs/html/abs-guide.html):
$ set w x y z; IFS=":-;"; echo "$*"
w:x:y:z
$ bash -c 'set w x y z; IFS=":-;"; echo "$*"'
w:x:y:z
bash -c isn't as interesting when you're already running bash. Consider, on the other hand, the case when you want to run bash code from a Python script:
#!/usr/bin/env python
import subprocess
fileOne='hello'
fileTwo='world'
p = subprocess.Popen(['bash', '-c', 'diff <(sort "$1") <(sort "$2")',
'_', # this is $0 inside the bash script above
fileOne, # this is $1
fileTwo, # and this is $2
])
print p.communicate() # run that bash interpreter, and print its stdout and stderr
Here, because we're using bash-only syntax (<(...)), you couldn't run this with anything that used POSIX sh by default, which is the case for subprocess.Popen(..., shell=True) or os.system(...); using bash -c thus provides access to capabilities that wouldn't otherwise be available without playing with FIFOs yourself.
Incidentally, this isn't the only way to do that: One could also use bash -s, and pass code in on stdin. Below, that's being done not from Python but POSIX sh (/bin/sh, which likewise is not guaranteed to have <(...) available):
#!/bin/sh
# ...this is POSIX sh code, not bash code; you can't use <() here
# ...so, if we want to do that, one way is as follows:
fileOne=hello
fileTwo=world
bash -s "$fileOne" "$fileTwo" <<'EOF'
# the inside of this heredoc is bash code, not POSIX sh code
diff <(sort "$1") <(sort "$2")
EOF