Re: Good placement of bash export functions

[uzibalqa]

------- Original Message -------
On Tuesday, October 11th, 2022 at 8:52 PM, Greg Wooledge <greg@wooledge.org> 
wrote:


> On Tue, Oct 11, 2022 at 05:51:57PM +0000, uzibalqa wrote:
> 
> > I have a number of bash script files that perform a specific tasks. After 
> > defining the appropriate set of functions, I included an additional utility 
> > function that allows one to export the function names.
> 
> 
> Why are you exporting the functions at all? Do you run a complicated
> find command, with -exec bash -c '....' which uses one of the functions?
> 
> You call these "script files". A script is normally a command implemented
> in bash. It's got a shebang (#!/bin/bash or equivalent), and it lives in
> a directory in your $PATH, and you run it as a command, by typing its
> name in a shell.
> 
> When it runs, it does so as a separate child process. Therefore,
> exporting a function has no effect on the calling shell.
> 
> > What would be easier to use, an export function in every file, or having a 
> > separate file containing all the necessary export functions for the entire 
> > package?
> 
> 
> ... "package"?

A have a number of bash script files, each with its own set of functions that a 
user can call.
Each files focuses on specific tasks (for instance. comint.rc handles parallel 
versions of 
search for both find and grep).  So I call the whole collection a terminal 
shell "package".
 
> What are you trying to do here?
> 
> How do you actually run these "script files"?

I source the files.  Then users can just call the functions they want, without 
having to worry
about specifying any paths.
 
> What's the intent behind exporting the functions?

It is a facility, normally one does not call the export functions.  But one can 
call them if one wants.