I think that the best way to understand umask is to say that umask is used to revoke permissions, not to set permissions.
umask sets which permissions must be removed from the system default when you create a file or a directory.
For example, a mask 0022 means that you don't want group and others modify the file.
default 0666 rw-.rw-.rw-
umask 0022 ---.-w-.-w-
Final 0644 rw-.r--.r--
That means that any file from now on will have 0644 permissions.
It is important to understand that umask revokes, deletes permissions from system default, so it can´t grant permissions the system default hasn't. In the example above, with the 666 system default, there is no way you can use umask to create a file with execute permission. If you want to grant more permissions, use chmod.
Be aware that system default permissions are not related to PHP (they depends upon server configuration). PHP has a default umask that is applied after system default base permissions. And there are different system default base permissions for files and directories.
Usually, system default permissions for files are 666 and for directories 0777. And usually, default PHP umask is 0022
umask
(PHP 4, PHP 5, PHP 7, PHP 8)
umask — 現在の umask を変更する
説明
umask() は PHP の umask を
mask & 0777 にセットし、
元の umask 値を返します。
PHP がサーバー・モジュールとして動作中の場合、
各リクエストが終了するたびに umask は元の値に戻されます。
パラメータ
-
mask -
新しい umask。
戻り値
mask が null の場合、
umask()は単に現在の umask 値を返します。
そうでない場合、元の umask 値を返します。
変更履歴
| バージョン | 説明 |
|---|---|
| 8.0.0 |
mask は、nullable になりました。
|
例
例1 umask() の例
<?php
$old = umask(0);
chmod("/path/some_dir/some_file.txt", 0755);
umask($old);
// チェック
if ($old != umask()) {
die('An error occurred while changing back the umask');
}
?>注意
注意:
マルチスレッドな Web サーバーでこの関数を使用することは避けてください。 ファイルを生成後、chmod() を使用してファイル権限を 変更するのがより良い方法です。全て同じ umask が使用されるので、 umask() の使用は、スクリプトを同時に実行する場合や Web サーバー自身の予期しない動作を引き起こす原因になる可能性があります。
add a note
User Contributed Notes 14 notes
librodot at ciberpiula dot net ¶
15 years ago
sean at awesomeplay dot com ¶
17 years ago
"It is better to change the file permissions with chmod() after creating the file."
The usual lacking of security knowledge within the PHP team rears its head once again. You *always* want to have the file created with the proper permission. Let me illustrate why:
(a) you create new file with read permissions
(b) an attacking script opens the file
(c) you chmod the file to remove read permissions
(d) you write sensitive data to the file
Now, you might think that the changes of an attacking script getting to open the file before you chmod them are low. And you're right. But low changes are never low enough - you want zero chance.
When creating a file that needs increased permissions, you always need to create the file with the proper permissions, and also create it with O_EXCL set. If you don't do an exclusive create, you end up with this scenario:
(a) attacker creates the file, makes it writable to everyone
(b) you open the file with restricted permissions, but since it already exists, the file is merely opened and the permissions left alone
(c) you write sensitive data into the insecure file
Detecting the latter scenario is possible, but it requires a bit of work. You have to check that the file's owner and group match the script's (that is, posix_geteuid(), not myuid()) and check the permissions - if any of those are incorrect, then the file is insecure - you can attempt to unlink() it and try again while logging a warning, of course.
The only time when it is reasonable or safe to chmod() a file after creating it is when you want to grant extra permissions instead of removing them. For example, it is completely safe to set the umask to 0077 and then chmoding the files you create afterward.
Doing truly secure programming in PHP is difficult as is, and advice like this in the documentation just makes things worse. Remember, kids, anything that applies to security in the C or UNIX worlds is 100% applicable to PHP. The best thing you can possibly do for yourself as a PHP programmer is to learn and understand secure C and UNIX programming techniques.
Anonymous ¶
18 years ago
Using (cmask - umask) is a wrong way to calculate the new mask:
0022 - 0700 = 0656 WRONG
0700 & ~0022 = 0700 CORRECT
Correct php code:
<?php
$rmask = ($cmask & ~$umask);
?>
neon at neonjs dot com ¶
13 years ago
In case you don't understand why you need to "Avoid using this function in multithreaded webservers":
It's because this function changes the umask at the process level, rather than only for PHP or for the current script. If there are multiple simultaneous threads running in the process in which your PHP script is running, the change will apply to all of those threads at the same time hence why this is not safe for multithreaded use.
I understand that if you are using the PHP module and Apache's prefork MPM, which is not multi-threaded, then you at least won't get race-condition problems such as this. However, it is still worth noting that the umask setting, if not re-set, will persist for the life of that process even if the process is re-used to serve future PHP or non-PHP requests.
jphansen at uga dot edu ¶
16 years ago
On most UNIX environments the recommended default umask for files, defined in /home/user/.profile or /etc/profile, is 022 (chmod: 644). On trusted systems it is 002. Exercise caution when applying more liberal settings.
malcolm.murphy ¶
8 years ago
Clarification Of "mask & 0777":
The manual's comment "umask() sets PHP's umask to mask & 0777 [...]" is merely implying that the method only affects file permissions, but not special modes such as the setuid, setgid or sticky bits. Curiously, PHP does not actually perform the bitwise operation itself, but instead assumes it will be done by a system call of the same name. On some systems such as OS X, umask effectively sets the umask as mask & 07777, but the extraneous bits are not applicable to subsequent PHP calls like mkdir(). Linux's umask does use 0777. Its manual entry has a comment similar to the PHP one, but with a parenthetical statement that helps explain what it means:
"umask() sets the calling process's file mode creation mask (umask) to mask & 0777 (i.e., only the file permission bits of mask are used) [...]"
The fact that permissions can be determined by inverting a mask using the operation $mask & ~0777 is irrelevant, despite its similar appearance to $mask & 0777. The latter operation instead truncates $mask to the first nine low-order bits (i.e., the three rightmost octal digits [and note that the leading zero for octal notation is not itself a digit]). It does not change the remaining bits.
For example, all of the following calls have the same effect: umask(0022), umask(07022), umask(261650) (decimal value of 0777022), and umask(0b111000010010) (binary notation for 07022).
Richard Snell ¶
9 years ago
It is important to note that the mask parameter will accept values other than octal and that this can cause unexpected results.
Setting umask(22) could be expected to reduce a file with default 0666 permission to 0644 by applying a mask of 0022 but as the parameter is being supplied as a decimal it will be converted to octal silently and actually apply a mask of 0026 resulting in a final file permission of 0642.
Similarly the value returned by umask is in decimal format. If you correctly apply a mask using umask(0022) and then query the new setting with umask() it will return a value of 18 (0022 octal is 18 decimal).
In short, when applying permissions it is best to pad the supplied value with zeros to create an octal value (22 becomes 0022) and if you want to analyze the returned value remember to convert it to octal for ease of interpretation.
ahmad dot mayahi at gmail dot com ¶
7 years ago
Simply umask means the default permissions for new files/directories:
<?php
umask(022);
?>
This sets the default permissions for user, groups, and others respectively:
• 0 - read, write and execute
• 1 - read and write
• 2 - read and execute
• 3 - read only
• 4 - write and execute
• 5 - write only
• 6 - execute only
• 7 - no permissions
webmaster at iacomputing dot co dot uk ¶
14 years ago
You can use umask to solve the PHP session bug that appears in several PHP versions.
<?php
umask(0022);
session_start();
?>
This will prevent sessions being created with inadequate permissions.
bishop ¶
14 years ago
"It is better to change the file permissions with chmod() after creating the file."
If you take that advice seriously, consider setting your umask so that files are created private to your user, then use chmod to open them up.
<?php
// files will create as -rw-------
umask(0077);
// create a file, eg fopen()
// give access: -rw-r--r--
chmod('/path/to/file', 0644);
?>
Whenever reasonable, default to shut and open as needed (like above) instead of default to open and shut as needed. The above still has a race condition, but the race condition will deny appropriate access instead of granting inappropriate access.
andi<at>splitbrain.org ¶
19 years ago
To play around with umasks and permissions use this little fragment:
<?
$umask = 0012;
$perm = 0777;
printf("umask: %04o perm: %04o result: %04o\n",
$umask,$perm,$perm & (0777 - $umask));
?>
maulwuff at gmx dot de ¶
17 years ago
umask takes away the given values from the standard mask 777.
A graphical view shows this better:
standard:
rwxrwxrwx = 777
will get with umask 002:
rwxrwxr-x = 775
or will get with umask 077:
rwx------ = 700
and so on.
sam at totallydigital dot co dot nz ¶
22 years ago
The first comment perhaps didn't quite make clear what's on with your umask and the permissions.
The permission passed to a command is first bitwise ANDed with the _INVERSE_ of the current umask, then applied to the file.
For example, umask = 0011 and permission = 0775
The inverse of 0011 = 0766
0775 AND 0766
= 111.111.101 AND 111.110.110
= 111.110.100
= 0764
ShaD@TW ¶
18 years ago
Notice that directory(s) and file(s) sometimes have different results.
<?php
umask(0670); //- set umask
$handle = fopen('file', 'w'); //- 0006
mkdir("/path/dir"); //- 0107
?>
calculate the result:
<?php
$umask = 0670;
umask($umask);
//- if you are creating a new directory, $permission = 0777;
//- if you are creating a new file, $permission = 0666.
printf( "result: %04o", $permission & ( 0777 - $umask) );
?>
BTW, as the manual said, the form of umask() is "int umask ( [int mask] )", so if you want to print/echo any umask, don't forget to convert it from DEC (because it returns a "int") to OCT.
<?php
$umask = umask(); //- returns the current umask, which is a "int"
$umask = decoct($umask); //- Now, $umask is a "string"
echo $umask;
?>
Don't forget that the argument(parameter) is a "int", too.
<?php
umask(777); //- WRONG! Even though you maybe use "umask 777" in some OS.
umask(0777); //- OK
?>
If there was any mistake, please correct my statement.
