Developer Forum

What I am after is Python sys.argv[0] equivalent which gives the full name
(including path) of the currently running script. I have unzip.exe binary in
the same directory as the main scheme program and need to run this unzip.exe
from scheme. Using the latest DrScheme environment...

in
unzip.exe

The following will give the path of the script in which this code appears:

(require (lib "etc.ss"))
(this-expression-source-directory)

I would have suggests (current-path), but I didn't knew the clever
this-expression-source-directory:

< http://www.yqcomputer.com/ #node_sec_11.3.3>

--
Jens Axel Saard

Thank you, it works...




appears:
< http://www.yqcomputer.com/ #n
ode_sec_11.3.3>