reset permissions back to default

Question

How can I reset permission to default according to the mask, so they will have permissions set as the file was just created

Example of what I want to achieve: umask is set to 0022 so

touch file
mkdir directory

file's permissions now are rw-r--r--

directory's permissons now are rwxr-xr-x

chmod 777 file
chmod 777 directory

file's permissions now are rwxrwxrwx

directory's permissions now are rwxrwxrwx

is there a way to reset perms back to default so file will be rw-r--r-- and directory rwxr-xr-x using chmod?

Answer 1

Using subtraction is only correct for some umask values - it's better to use mode & ~umask, like in PSkocik's answer, or chmod =rw (or chmod =rwx for directories) which applies the umask as specified by the standard, as noted by ilkkachu in the comments.

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In some cases, you can just subtract the umask from 0666 for files and 0777 for directories to get default permissions:

$ printf "%04d\n" "$((0777 - $(umask)))"
0755
$ printf "%04d\n" "$((0666 - $(umask)))"
0644

Accordingly, you can apply chmod:

chmod $((0666 - $(umask))) file
chmod $((0777 - $(umask))) directory

In bash, you have to use printf to force the output in octal:

$ printf "%04d\n" "$((0777 - $(umask)))"
0493
$ printf "%04o\n" "$((0777 - $(umask)))"
0755

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Another way would be to create a new file and directory, and use them as reference:

touch file2
mkdir directory2 
chmod --reference=directory2 directory
chmod --reference=file2 file

Answer 2

The masks are applied via bitwise AND with the bitwise negated mask so if you want to create your own final permission mode, you can do:

$((mode & ~umask))

You'll need to print that back in octal so that you can pass it to chmod:

$ chmod `printf '%o' $((0777 & ~$(umask)))` directory
$ chmod `printf '%o' $((0777 & ~0111 & ~$(umask)))` file
#^additional implicit mask of 0111 for files

where 0777 is the permission mode you want to apply the mask to (you can obtain it with stat -c %a file or stat -c %a directory).

You can echo the above to see what the process substitution will evaluate to (for an umask of 0022, you will get 755 and 644).

You could make a generic function out of it:

#takes a umask as first param and applies it to each folowing param (files)
maskMode(){
  local mask="$1" dmask mode a
  dmask="$((mask & ~0111))"; shift
  for a; do
    mode=0`stat -c "%a" "$a"`
    chmod `printf "%0.4o\n" $(($mode & ~mask))` "$a"
  done
}

For your particular use, chmod reference files are another option.

Answer 3

If you want to do it recursively through the current directory -

chmod -R $((777 - `umask`)) .
chmod -R -x+X .

You can replace `umask` with $(umask).

Answer 4

Recursive files and directories

For anyone looking to reset all files and directories recursively within a parent directory:

sudo find . -type f -print0 | xargs -0 sudo chmod $((0666 - $(umask)))
sudo find . -type d -print0 | xargs -0 sudo chmod $((0777 - $(umask)))

Explanation

The other answers showed how to reset individual files and directories (or apply one rule to both recursively) with chmod -R $((0666 - $(umask))) and chmod -R $((0777 - $(umask))) respectively. (Great answers, thanks!)

I then found this answer showing how to apply permissions selectively using find throughout the directory:

find . -type f -print0 | xargs -0 chmod 644

This will change files recursively, and therefore

find . -type d -print0 | xargs -0 chmod 755

will apply to directories.

Stringing these answers together provides the above solution. Note that this requires the GNU version of find and xargs for the -print0/-0 options, respectively.

Hope it helps others!