Question

How do I know the difference in the ‘execution env’ of a process in two different contexts?

To articulate the question properly, I have plan9port installed in /opt/plan9/ and when I run the fortune program from /opt/plan9/bin/fortune it works fine. (reads the list of fortunes from /opt/plan9/lib/fortune and /opt/plan9/lib/fortune.index ). When I call it from inside of a c code (test.c) with

char* opts[] = {"fortune"};
execve("/opt/plan9/bin/fortune", opts, NULL);

It doesn’t read the fortune list. I used strace to see what is the difference when I call these two binaries.

strace -f -eopen ./test

open("/usr/local/plan9/lib/fortunes", O_RDONLY) = -1 ENOENT (No such file or directory)
Misfortune!
+++ exited with 1 +++

Gives out the default message “Misfortune”.

strace -f -eopen fortune

open("/opt/plan9/lib/fortunes", O_RDONLY) = 3
Snob intellectual bachelors can't have fun in San Antonio.  -Ted Nelson
+++ exited with 0 +++

which works perfectly fine.

How do I change ./test read fortunes file. It must have something to do with the exec environment, from where the binary reads the libraries from.

Answer 1

When you call execve(), you are explicitly setting up a NULL environment. So the fortune program is probably depending on some environment variable to find /opt/plan9/.... Type env at a shell prompt to find out which environment variables are set.