Shell Programming and Scripting

Hi,

Suppose I have a User define function get_abc in which I am using $0 to get the name of function. But when I call that function in any script, $0 will give the script name, not the function name.

For example:
Function: get_abc
-------------------
get_abc( ){
echo $0
}

Script: call_get_abc.ksh
---------------------- ---
echo "Calling get_abc"
get_abc
exit 0

Output:
Calling get_abc
call_get_abc.ksh <---- Here I want function name (get_abc)


Regards,
Yeheya

All the documentation I could find says that $0 will always give the name of the script - not the function - no matter how deep inside a function it lies.. and yet when I copied your script and ran it in ksh, $0 inside a function returned the name of the function! However, when I ran it in bash, it returned the name of the script like all the documentation says it should..

Code:

get_abc( ){
 echo $0
}

echo "Calling get_abc"
get_abc
exit 0

In POSIX, $0 is the name of the script and not the function. However, if you define the functions as Korn functions using the Korn syntax:

function funcname { list; }

then $0 will be set to the function name, not the script name.

But my problem is not still solved . This is not the generic and reliable way of doing things.

Thanks & Regards,
Yeheya

I havent tried this so it still may print out the name of the script, but what about using typeset -fu to define the function in a separate function library. In the script, set the FPATH variable to the path to search for the function. Call the function. See what happens

function library script

my_function () {
echo "\$0 is $0"
}


your-script.ksh

#defines my_function as an undefined function. FPATH is the
#search path to auto-load the function.

FPATH=/path/to/function-library-script
typeset -fu my_function

my_function #call the function

end your-script.ksh