Re: How to get the unique canonical path of a given path?

[tybalt89]

Are these answers correct?

#!/usr/bin/perl

use strict; # https://perlmonks.org/?node_id=11145493
use warnings;

for my $path ( qw(
  /a/b/c/d/../../../e
  /a/../b/./c//d
  ../invalid_do_not_change
  ./tmp
  /tmp/../../../tmp
  A//B A/B/ A/./B A/foo/../B
  ) )
  {
  local $_ = $path;
  1 while s{
    /+(?=/) |                    # multiple ///
    ^/..(?=/) |                  # stay at root
    /\z |                        # remove trailing /
    (?<=/)(?!\.\./)[^/]+/\.\./ | # remove 'name/../'
    (?<![^/])\./                 # remove ./
    }{}x;
  printf "%30s  ->  %s\n", $path, $_;
  }

Outputs:

           /a/b/c/d/../../../e  ->  /a/e
                /a/../b/./c//d  ->  /b/c/d
      ../invalid_do_not_change  ->  ../invalid_do_not_change
                         ./tmp  ->  tmp
             /tmp/../../../tmp  ->  /tmp
                          A//B  ->  A/B
                          A/B/  ->  A/B
                         A/./B  ->  A/B
                    A/foo/../B  ->  A/B

[pryrt]

<Reveal this spoiler or all in this thread>

Nevermind; ignore my objection from the spoiler. I just tried the cd /tmp/../../../tmp and found it did go to the /tmp directory. And perl -le 'use autodie; open my $fh, ">", "/tmp/../../../tmp/worked.txt"; print {$fh} "it worked";' works as expected as well, so apparently that weird notation is a perfectly-valid syntax. Sorry.

[tybalt89]

It works because on a *nix system the root directory is pretty much defined by " .. is the same as . "

[ikegami]

[Nothing to see here]