SPUG: Finding name for sub ref

[SPUG-list-owner]

> I've created a subroutine to time the execution of another subroutine; so I pass in a reference to the subroutine that is to be timed.
> 
> Zub timeThis
> {
>       my ($sub) = shift;
>       my $startTime = time();
>       &{$sub}(@_);
>       my $endTime = time();
>       # if possible, I'd like to get the subroutine name from the reference right here,
>       # then I'd be able to print it out in the following print statement.
>       print " took " . ($endTime - $startTime) . " seconds\n";
> }
> 
> timeThis(\&createTheFiles, $path, $numberOfFiles, $fileSize, $dataType);
> 
> 
> Is there a way to extract the name of the subroutine from the reference to the subroutine?
> 
> Thanks...
> 
> -Keith

I didn't know of any easy way to do this, so I asked "The Damian", while we were
on a break in our Adv. OO Perl class this morning.  And he says, 1) first of all,
if it's an anonymous sub, it won't have a name, and 2) if it is named, you'd
have to walk through the symbol tables for each of your packages looking
for its address, to retrieve the name.

Maybe you should ask the sub's caller to provide the name as an additional arguent!

-Tim

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