UNIX / Linux: File Permissions in Octal

To set file permissions using the chmod command, we can specify the permissions in either numeric mode or symbolic mode. But when we want to get the file permissions, most often with the ls -l command, we are normally only given a printout using the symbolic mode.

There are a couple of ways to print permissions in octal mode. The first method is by using the stat command.

$ stat -c "%a" /path/to/file

1	$ stat -c "%a" /path/to/file

The above will printout the octal permissions of a file. Noting more, nothing less. But if we try to use the command and pass a wildcard argument, the information would be meaningless.

$ stat -c "%a" *
755
644

$ stat -c "%a" *

755

644

You will not know which permissions are for which file. To include the filenames alongside with the permissions, make a slight modification to the stat command.

$ stat -c "%a %n" *
755 foo
644 foo.c

$ stat -c "%a %n" *

755 foo

644 foo.c

Now we know that file foo has a permission of 755 and file foo.c has a permission of 644.

The octal permissions of the stat command also takes care of setuid, setgid and sticky bits. We can test this by printing the permissions of su – a standard setuid command.

$ stat -c "%a %A %n" $(which su)
4755 -rwsr-xr-x /bin/su

1 2	$ stat -c "%a %A %n" $(which su) 4755 -rwsr-xr-x /bin/su

I have written a small shell script that will make use of the ls and awk commands to print the octal permissions.

#!/bin/sh

ls -l "$@" | awk '{
        i=0;
        for(j=0;j<=8;j++) {
            p=substr($1,j+2,1);
            i+=((p~/[rwxst]/)*2^(8-j));    
            if(j==2&&p~/[sS]/) i+=04000;    
            if(j==5&&p~/[sS]/) i+=02000;
            if(j==8&&p~/[tT]/) i+=01000;
        }
        if(i)printf("%04o ",i);
        print
    }'

#!/bin/sh

ls -l "$@" | awk '{

i=0;

for(j=0;j<=8;j++) {

p=substr($1,j+2,1);

i+=((p~/[rwxst]/)*2^(8-j));

if(j==2&&p~/[sS]/) i+=04000;

if(j==5&&p~/[sS]/) i+=02000;

if(j==8&&p~/[tT]/) i+=01000;

}

if(i)printf("%04o ",i);

Copy the script to a file called lso (ls octal), make it executable and run it like this:

$ lso /path/to/file

1	$ lso /path/to/file

Alternatively, if you are not interested in the shell script, you can opt to copy the below function to your shell initialisation file ( eg. .bashrc or .bash_profile) instead.

function lso  () {
ls -l "$@" | awk '{
        i=0;
        for(j=0;j<=8;j++) {
            p=substr($1,j+2,1);
            i+=((p~/[rwxst]/)*2^(8-j));    
            if(j==2&&p~/[sS]/) i+=04000;    
            if(j==5&&p~/[sS]/) i+=02000;
            if(j==8&&p~/[tT]/) i+=01000;
        }
        if(i)printf("%04o ",i);
        print
    }'
}

function lso () {

ls -l "$@" | awk '{

i=0;

for(j=0;j<=8;j++) {

p=substr($1,j+2,1);

i+=((p~/[rwxst]/)*2^(8-j));

if(j==2&&p~/[sS]/) i+=04000;

if(j==5&&p~/[sS]/) i+=02000;

if(j==8&&p~/[tT]/) i+=01000;

}

if(i)printf("%04o ",i);

}

Run the function by calling the function name from the command line.

$ lso -d /tmp ; 
1777 drwxrwxrwt 5 root root 4096 Jan 11 21:51 /tmp

$ stat -c "%a %n" /tmp
1777 /tmp

$ lso -d $(which passwd) ; stat -c "%a %n" $(which passwd) 
4755 -rwsr-xr-x 1 root root 47032 Jan 27  2016 /usr/bin/passwd
4755 /usr/bin/passwd

$ lso -d $(which su) ; stat -c "%a %n" $(which su)             
4755 -rwsr-xr-x 1 root root 36936 Jan 27  2016 /bin/su
4755 /bin/su

$ lso -d /tmp ;

1777 drwxrwxrwt 5 root root 4096 Jan 11 21:51 /tmp

$ stat -c "%a %n" /tmp

1777 /tmp

$ lso -d $(which passwd) ; stat -c "%a %n" $(which passwd)

4755 -rwsr-xr-x 1 root root 47032 Jan 27 2016 /usr/bin/passwd

4755 /usr/bin/passwd

$ lso -d $(which su) ; stat -c "%a %n" $(which su)

4755 -rwsr-xr-x 1 root root 36936 Jan 27 2016 /bin/su

4755 /bin/su

Update (11-Jan-2017): Script updated to handle setuid, setgidand sticky bits.

Mohamed Ibrahim

ibrahim = { interested_in(unix, linux, android, open_source, reverse_engineering); coding(c, shell, php, python, java, javascript, nodejs, react); plays_on(xbox, ps4); linux_desktop_user(true); }