See mkdir and umask. In short, the mkdir function honors $ENV{UMASK}. If you really intend to make the directory world-writable, I guess the following code will work:

my $old_umask = umask;
umask 0777;
mkdir $dump_dir
    or die "Couldn't create '$dump_dir': $!";
umask $old_umask;
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The umask is an attribute of the process (like its current working directory or nice level) and is manipulated by the system call of the same name. It's not related to the process' environment.</pedant>

(Well, it is part of the process' environment, but it's not contained in %ENV or char *env or what have you . . . :)

Update: And I just noticed, umask returns the previous value when you call it so you could just do my $old_umask = umask 077; in one swell foop.</nit>

my $old_umask = umask;
umask 0777;
mkdir $dump_dir
    or die "Couldn't create '$dump_dir': $!";
umask $old_umask;
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This actually does the opposite of what the OP asked for. A permissive umask is a smaller number (IE: less bits turned on) than a non-permissive mask. Setting the umask to 0777 ensures that no permissions are assigned, regardless of what mode you set for mkdir (or if you leave it as the default, as above). The number in the umask is applied to the mode bits of a created file with the following logic:

MODE & ~MASK

So, for the mode bits 0777 and a umask of 0777:

perl -e'printf "%o\n", 0777 & ~0777' => 0 (or 0000)

For a umask of 0000:

perl -e'printf "%o\n", 0777 & ~0000' => 777 (or 0777)

By default (at least it is here), the umask is set to 022. If we use the logic above with this knowledge, we can see why the OP's mode argument to mkdir didn't take affect:

$ perl -e'printf "umask: %04o\n", umask; printf "mode: %04o\n", 0777; 
+printf "masked mode
: %04o\n", 0777 & ~umask'
umask: 0022
mode: 0777
masked mode: 0755
$
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In short, the answer to the OP's question is to set the umask to 0. Note the following:

$ perl -e'umask 0; printf "umask: %04o\n", umask; printf "mode: %04o\n
+", 0777; printf "ma
sked mode: %04o\n", 0777 & ~umask'
umask: 0000
mode: 0777
masked mode: 0777
$ perl -e'umask 0; mkdir "FOO", 0777'
$ ls -dl FOO
drwxrwxrwx  2 matt matt 6 Apr 13 20:07 FOO
$
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umask is slightly counter-intuitive, but I found that when I saw the boolean math it helped me to understand it much better. I hope this helps.

Best Regards

m.att

Thank you, this really helped me!

try chmod the dir after creating it using :

`chmod 777 $dump_dir`
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HTH

chmod 0777, $dump_dir is preferable to `chmod 777 $dump_dir`.  No backticks needed; chmod is a Perl builtin.

(This is just a side-note.  umask is the correct solution.)

I am currently writing a script to dump data from a mysql database

I am trying to create a directory which is writable to by everyone
so the user which the mysqld is running under has access to write
the file

this is the type of command I am creating the directory with


my $dump_dir = $output_dir."/".$dbname;
if(! -e $dump_dir){
mkdir($dump_dir, 0777) or throw("Couldn't make ".$dump_dir." $! ");
}


but this is the privileges i get for the directory

drwxr-xr-x 2 lec ensembl 8192 Jun 4 15:15 briggsae_new_2

does anyone know what I am doing wrong 
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Sure, you've changed the group name from "lec ensembl" to "lec sieter", but the question has almost the same wording, the code is identical, and you even kept time ("15:15") and size (8192) of the file the same.

So I repeat, madtoperl, "Why are you plagiarizing other people's questions?!"