Executable Unix script cannot be found by which command

I have a Unix bash script written by a former teammate that must be in my PATH, though I can’t find it by manual inspection (see below). I can however execute it anywhere by just typing

$ my_script

I would like to view and edit this script. However, when I try to find it via the which command, I get a blank response. Return code indicates an error:

$ which my_script
$ echo $?
1

And yet, I can run the script. I manually combed my PATH and could not find it. Honestly, I have not encountered anything like this in 20+ years. Is there any other command besides which, and/or ways such a script could be hidden?

Answer

To determine the type of any command callable from the shell, use the type builtin.

In your case, since the presumptive script turned to be a shell function, you would have seen (assuming bash):

$ type my_script  # Bash: option -t would output just 'function`
my_script is a function
my_script () 
{                
...               # The function's definition

From there – as you did – you can examine the profile (e.g., ~/.bash_profile) and / or initialization scripts (e.g., ~/.bashrc) to determine where the function was sourced.

Caveat: The function signature output by type is normalized to the <name> () form – even if you defined the function as function <name>

In bash you can even find out directly where a given function was defined (tip of the hat to this superuser.com answer and Charles Duffy for inspiring me to find it and coming up with the most succinct form):

$ (shopt -s extdebug; declare -F my_func)
my_func 112 /Users/jdoe/.bashrc   # sample output

112 is the line number inside the script indicated.