Retrieving binary file name from a process

This is a somewhat complicated question.

Is there anyway for a currently running process to identify the full path to the binary file that launched it?

Basically, I have this binary file which can be installed anywhere in the UNIX file system, and it creates a running process. I need to be able to find the binary file that created the process. Because it can be anywhere, I can't hard code the path. Because this is a C program, I can't use a shell command like "find". So it has to be a C API call or something like that. Does anyone know of such a function?

Well if you can trust your shell, then look in argv[0].

That's a very good idea actually. I totally forgot about that. I'm so used to always using argv[1] and so on just to get the arguments I need that I totally forgot that argv[0] actually has something occasionally useful. Thanks for the help.

I realized that I misstated my question.

1) A binary executable file in a random location

2) A shared library (akin to a DLL in Windows) in a random location

3) A process that is currently running, launched from the binary executable that invokes routines in the shared library

I need the routine in the shared library to know where the location of it's shared library is within the UNIX file system. argv[0] will not work because that will give me the EXE and not the DLL.

/etc/ld.so.conf contains (indirectly) the list of places to search (in addition to /usr/lib)

If you do
man ld
This lists other search strategies as well.

The name of the current executable can be found at "/proc/self/exe". Use readlink() to the get path.

Use dladdr() to get the path to a shared library. Try "man dladdr".