Print the perl script's path without resolving symlink -

how print script's directory path without resolving symlink?

edit: realized below comment not possible using findbin, there way around script's directory ?

use findbin; use cwd; use cwd 'abs_path'; use file::spec::functions qw( rel2abs ); begin {    push @inc, "$findbin::bin"; } print "rel2abs:"; print rel2abs("$findbin::bin"); print "\n";  print "abs_path:"; print abs_path("$findbin::bin"); print "\n"; 

this script prints:

/tmp/one/two$ perl 3/symlink_to_four/test.pl  rel2abs:/tmp/one/two/3/four abs_path:/tmp/one/two/3/four 

i want print /tmp/one/two/3/symlink_to_four

 /tmp/one/two/3$ ls -l  4  symlink_to_four -> 4 

why need this: samba configured in linux box perl script going run, if replace first portion of linux path \\linux-machine-name , rest of / \, windows path, save in file , consumed other script, instance, in above example, trying convert /tmp/one/two/3/symlink_to_four \\linux-machine-name\one\two\3\symlink_to_four, know how figure-out linux-machine-name, don't want resolved path, because cannot figure out windows equivalent path resolved path, \\linux-machine-name\one\two\3\symlink_to_four valid, not \\linux-machine-name\one\two\3\four, it's how samba configured.

if correctly understand you're trying achieve, can path::tiny:

#!/usr/bin/perl  use strict; use warnings;  use path::tiny;   $path =  path(__file__);  print $path->absolute->parent, "\n"; 

path::tiny's realpath method resolve symlinks, absolute method not.

__file__ special token resolves name of perl source file it's contained in (either main script or module loads). use $0 name of script being executed. either of these absolute or relative depending on whether script invoked relative or absolute path.

if scripts invoked absolute path use $0 , not bother module.