Re^2: file comparison using file open in binary mode.

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Re^3: file comparison using file open in binary mode. by gmargo (Hermit) on Nov 27, 2009 at 18:15 UTC
That is not the case. Only the content matters. `gmargo@tesla 1368$ ls -l perl10.pl -rw-r--r-- 1 gmargo gmargo 1150 Oct 16 21:43 perl10.pl gmargo@tesla 1369$ cp perl10.pl perl10a.pl gmargo@tesla 1370$ md5sum -b perl10.pl perl10a.pl 976ed3393d1e967b2d8b4432c92b1397 perl10.pl 976ed3393d1e967b2d8b4432c92b1397 perl10a.pl gmargo@tesla 1371$ sha1sum -b perl10.pl perl10a.pl 1f62d2bcc8dcdf3f9ef0f3728f9f99c85eb21d81 perl10.pl 1f62d2bcc8dcdf3f9ef0f3728f9f99c85eb21d81 perl10a.pl` [download]	[reply] [d/l]
Re^4: file comparison using file open in binary mode. by Karger78 (Beadle) on Nov 27, 2009 at 18:19 UTC
ok, let me give md5 a shot again. perhaps i did somthing wrong.	[reply]
Re^4: file comparison using file open in binary mode. by Karger78 (Beadle) on Nov 27, 2009 at 20:06 UTC
Thanks look's like the MD5 works nicely. one last question, how do i do the comparioson with each value in the hash? This is what i go thus far, but doesn't seem to work. `while ( my ($key, $value) = each(%hash1) ) { while ( my ($key1, $value1) = each(%hash2) ) { if ($value eq $value1) { print "\n$key => $value\n is equal with\n"; print "$key1 => $value1\n" } } print "$key => $value\n"; }` [download]	[reply] [d/l]
Re^5: file comparison using file open in binary mode. by zwon (Abbot) on Nov 27, 2009 at 20:35 UTC
It'll be funnier if you would use hash values as keys and file names as values. Then in order to find if there's same file in another tree you could use something like: `for (keys %hash1) { if (exists $hash2{$_}) { say "Files $hash1{$_} and $hash2{$_} are equal"; } }` [download] (Not tested)	[reply] [d/l]