Barak, Ron wrote:
Hi,
I wanted to make a python byte-code file executable, expecting to be able to run it
without specifying "python" on the (Linux bash) command line.
So, I wrote the following:
[r...@vmlinux1 python]# cat test_pyc.py
#!/usr/bin/env python
print "hello"
[r...@vmlinux1 python]#
and made its pyc file executable:
[r...@vmlinux1 python]# ls -ls test_pyc.pyc
4 -rwxr-xr-x 1 root root 106 Jul 29 14:22 test_pyc.pyc
[r...@vmlinux1 python]#
So, I see:
[r...@vmlinux1 python]# file test_pyc.py*
test_pyc.py: a python script text executable
test_pyc.pyc: python 2.3 byte-compiled
[r...@vmlinux1 python]#
If I try to do the following, no problem:
[r...@vmlinux1 python]# python test_pyc.pyc
hello
[r...@vmlinux1 python]#
However, the following fails:
[r...@vmlinux1 python]# ./test_pyc.pyc
-bash: ./test_pyc.pyc: cannot execute binary file
[r...@vmlinux1 python]#
Is there a way to run a pyc file without specifying the python path ?
Bye,
Ron.
I don't currently run Unix, but I think I know the problem.
In a text file, the shell examines the first line, and if it begins #!
it's assumed to point to the executable of an interpreter for that text
file. Presumably the same trick doesn't work for a .pyc file.
Why not write a trivial wrapper.py file, don't compile it, and let that
invoke the main code in the .pyc file?
Then make wrapper.py executable, and you're ready to go.
DaveA
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