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Sinoosh
Asked: 2016-09-26 22:52:51 +0800 CST

Diffrence between "perm -mode" &"perm /mode" in find command

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I read man page for findbut it is not clear for me.

find -perm -mode ------>at least these bit(s) must be set for a file to match

For example: find -perm -754 finds 754,755,757,774,777

What about? find -perm /754 please explain to me by examples

command-line

2 Answers

  1. Best Answer

    It's basically the difference between all three bits (-mode) and any single bit (/mode) permission (-perm) subset test.

    • find -perm -mode:

      In this case the permission bits mentioned must be present for the file. For example, if you do find -perm -666 and if a file has 776, it will be matched. Similarly 666, 777 etc will be matched too, but 665 won't be matched. In summary, the mentioned (three) bits must be a subset of the permission bits.

    • find -perm /mode:

      Here any one bit of subset would do. For example, if we do find -perm /666, and if a file has 644, the file will be matched because the user permission bit is 6, and we are looking for a single bit subset. Similarly, 700, 060, 006 etc will be matched, but not e.g. 444, as no bit contains any subset of the required permission bits.

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  2. The other answer correctly explains the find -perm -mode part. However, this answer corrects an incorrect claim about the find -perm /mode part.

    Here any one bit of subset would do. For example, if we do find -perm /666, and if a file has 644, the file will be matched because the user permission bit is 6, and we are looking for a single bit subset. Similarly, 700, 060, 006 etc will be matched, but not e.g. 444, as no bit contains any subset of the required permission bits.

    The first sentence is true, however, the part claiming 444 would not be matched is not.

    -perm /mode matches if any permission bit matches, not a whole number.

    When passing /666, we are asking the find command to look for files that have any of the following bits rw-rw-rw (6 = 4+2 which means rw-). So it will match any file that is readable, or writable, or both by any type of owner (user, group or other). That means the only files that will not be matched are files with the following permissions: 000, 001, 010, 011, 100, 101, 110, and 111. Any other permission will be matched by -perm /666 as it would have either the read or write flags assigned to it.

    For more clarification, check the following example.

    for h in {0..7}; \
do for i in {0..7}; \
do for j in {0..7}; \
do for k in {0..7}; \
do \
touch $h$i$j$k; \
chmod $h$i$j$k $h$i$j$k; \
done; \
done; \
done; \
done 

    find -perm /666 -printf '%04m %M\n' | grep 444
7444 -r-Sr-Sr-T
6444 -r-Sr-Sr--
5444 -r-Sr--r-T
4447 -r-Sr--rwx
4446 -r-Sr--rw-
4445 -r-Sr--r-x
4444 -r-Sr--r--
4443 -r-Sr---wx
4442 -r-Sr---w-
4441 -r-Sr----x
4440 -r-Sr-----
3444 -r--r-Sr-T
2444 -r--r-Sr--
1444 -r--r--r-T
0444 -r--r--r--

    So the only thing it does not match are the files that have permissions not containing 2 or 4: i.e. all files that contain only 1 or 0:

    # remove special permissions (since it was 0 every file that matches and only differs in the set of special permissions is found again)
find \! -perm /666 -printf '%04m %M\n' | grep '^0'| sed 's/ .../ /g'
0111 x--x--x
0110 x--x---
0101 x-----x
0100 x------
0011 ---x--x
0010 ---x---
0001 ------x
0000 -------

    maybe to clarify a bit:

    given the following permissions

    • WXYZ: file permissions (e.g. 0755 W=0, Z=5)
    • wxyz: search permissions
    • w&W is > 0 if at least one bit matches

    / is an OR operation: return w&W || x&X|| y&Y || z&Z

    i.e. return if file contains any permission (W-Z) that matches (at least) the searched permissions (w-z).

    while - is an AND operation: return w&W && x&X && y&Y && z&Z

    i.e. return if file contains all permissions (W-Z) match (at least) the searched permissions (w-z).

    In both cases that can mean that W has set more than w bits (6 sets 2 and 4, but not 3 or 1)

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