python: sys.argv[0] meaning in official documentation

Question

Quoting from docs.python.org:

"sys.argv The list of command line arguments passed to a Python script. argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string."

Am I missing something, or sys.argv[0] always returns the script name, and to get '-c' I'd have to use sys.argv[1]?

I'm testing with Python 3.2 on GNU/Linux.

Mathes

Eheh I stupidly misinterpreted -c as just an example option, instead of a real option for python, nevermind... – Mathes

Answer 1

No, if you invoke Python with -c to run commands from the command line, your sys.argv[0] will be -c:

C:\Python27>python.exe -c "import sys; print sys.argv[0]"
-c

Carburize

Answer 2

When Python is invoked as python script.py then sys.argv[0] == 'script.py'. When you invoke python -c 'import sys; print sys.argv' then sys.argv[0] == '-c' indicating the script body was passed as a string on the command line.

Prolonge

Answer 3

python -c executes a command passed on the command line, rather than a script from a file. sys.argv[0] will be set to "-c".

If you run a script with a -c flag, then yes, sys.argv[1] will be set to "-c" and sys.argv[0] will be set to the name of the script.

Carlo

I just thought of -c as an example, like in script -c, and didn't think of it as the real python option... ^^' – Mathes