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Show file and directory permissons from umask value

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

0

I have homeword about use umask value to make default file and directory permissons, but I got a problem in the following:

If I have umask = 0541

• 0666 - 0541 = 0125, File permisson: ---x-w-r-x
  


• 0777 - 0541 = 0236, Directory permisson: d-w--wxrw-
  

But the actual result on linux terminal:

• 
  d-w--wxrw for the directory


• 
  --w--w-rw- for the file
  Please explain to me!

linux command

share|improve this question

edited 2 days ago

Kusalananda♦

140k17261436

asked 2 days ago

jason momoajason momoa

32

New contributor

jason momoa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

0

I have homeword about use umask value to make default file and directory permissons, but I got a problem in the following:

If I have umask = 0541

• 0666 - 0541 = 0125, File permisson: ---x-w-r-x
  


• 0777 - 0541 = 0236, Directory permisson: d-w--wxrw-
  

But the actual result on linux terminal:

• 
  d-w--wxrw for the directory


• 
  --w--w-rw- for the file
  Please explain to me!

linux command

share|improve this question

edited 2 days ago

Kusalananda♦

140k17261436

asked 2 days ago

jason momoajason momoa

32

New contributor

jason momoa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

I have homeword about use umask value to make default file and directory permissons, but I got a problem in the following:

If I have umask = 0541

• 0666 - 0541 = 0125, File permisson: ---x-w-r-x
  


• 0777 - 0541 = 0236, Directory permisson: d-w--wxrw-
  

But the actual result on linux terminal:

• 
  d-w--wxrw for the directory


• 
  --w--w-rw- for the file
  Please explain to me!

linux command

share|improve this question

edited 2 days ago

Kusalananda♦

140k17261436

asked 2 days ago

jason momoajason momoa

32

New contributor

jason momoa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 2 days ago

Kusalananda♦

140k17261436

asked 2 days ago

jason momoajason momoa

32

New contributor

jason momoa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 2 days ago

Kusalananda♦

140k17261436

asked 2 days ago

jason momoajason momoa

32

New contributor

jason momoa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

Kusalananda♦

140k17261436

edited 2 days ago

Kusalananda♦

140k17261436

asked 2 days ago

jason momoajason momoa

32

New contributor

jason momoa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

jason momoajason momoa

32

2 Answers
2

active

oldest

votes

1

Note that umask's are not a subtraction, but a logical AND operation based on the umask.

Think of permissions in binary:

So a permission of 0666 would be

110 110 110

Now a umask of 0541 would be

101 100 001

To calculate the results we negate the umask:

010 011 110

and then AND that with the permission.

So

110 110 110 AND
010 011 110
=== === ===
010 010 110

And that maps to -w- -w- rw-

share|improve this answer

answered 2 days ago

Stephen HarrisStephen Harris

27.4k35383

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks you so much @Stephen ^^!, you saved my time.
  
  – jason momoa
  2 days ago
  

  
  
  
  

add a comment | 

0

Consider it like this, if you have default permissions 666 means that you have only r+w for user, group and others, and by umask 541 you want to remove permissions r+x,r,x.

For user:

By default you can read+write and by umask you want to remove read+executable 
permissions, but it will never happen because initially file has no executable 
permissions. So only read permission will be removed.

For group:

read permission will removed

For others:

There is no executable permission in the file, so there will be no change.

Same concept applies for directories.

share|improve this answer

answered 2 days ago

Prvt_YadvPrvt_Yadv

3,29631430

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for your answer ^^, I found a trick and I think it's right:
  
  – jason momoa
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please add it as answer, if it solves your problem.
  
  – Prvt_Yadv
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  default file=0666 => -rw-rw-rw- here Umask = 0541 => -r-xr----x substract them => --w--w-rw- for the file
  
  – jason momoa
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I was checked any value and it's always return a true value
  
  – jason momoa
  2 days ago
  

  
  
  
  

add a comment | 

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1

Note that umask's are not a subtraction, but a logical AND operation based on the umask.

Think of permissions in binary:

So a permission of 0666 would be

110 110 110

Now a umask of 0541 would be

101 100 001

To calculate the results we negate the umask:

010 011 110

and then AND that with the permission.

So

110 110 110 AND
010 011 110
=== === ===
010 010 110

And that maps to -w- -w- rw-

share|improve this answer

answered 2 days ago

Stephen HarrisStephen Harris

27.4k35383

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks you so much @Stephen ^^!, you saved my time.
  
  – jason momoa
  2 days ago
  

  
  
  
  

add a comment | 

1

Note that umask's are not a subtraction, but a logical AND operation based on the umask.

Think of permissions in binary:

So a permission of 0666 would be

110 110 110

Now a umask of 0541 would be

101 100 001

To calculate the results we negate the umask:

010 011 110

and then AND that with the permission.

So

110 110 110 AND
010 011 110
=== === ===
010 010 110

And that maps to -w- -w- rw-

share|improve this answer

answered 2 days ago

Stephen HarrisStephen Harris

27.4k35383

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks you so much @Stephen ^^!, you saved my time.
  
  – jason momoa
  2 days ago
  

  
  
  
  

add a comment | 

Note that umask's are not a subtraction, but a logical AND operation based on the umask.

Think of permissions in binary:

So a permission of 0666 would be

110 110 110

Now a umask of 0541 would be

101 100 001

To calculate the results we negate the umask:

010 011 110

and then AND that with the permission.

So

110 110 110 AND
010 011 110
=== === ===
010 010 110

And that maps to -w- -w- rw-

share|improve this answer

answered 2 days ago

Stephen HarrisStephen Harris

27.4k35383

110 110 110

101 100 001

010 011 110

110 110 110 AND
010 011 110
=== === ===
010 010 110

share|improve this answer

answered 2 days ago

Stephen HarrisStephen Harris

27.4k35383

answered 2 days ago

Stephen HarrisStephen Harris

27.4k35383

answered 2 days ago

Stephen HarrisStephen Harris

27.4k35383

0

Consider it like this, if you have default permissions 666 means that you have only r+w for user, group and others, and by umask 541 you want to remove permissions r+x,r,x.

For user:

By default you can read+write and by umask you want to remove read+executable 
permissions, but it will never happen because initially file has no executable 
permissions. So only read permission will be removed.

For group:

read permission will removed

For others:

There is no executable permission in the file, so there will be no change.

Same concept applies for directories.

share|improve this answer

answered 2 days ago

Prvt_YadvPrvt_Yadv

3,29631430

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for your answer ^^, I found a trick and I think it's right:
  
  – jason momoa
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please add it as answer, if it solves your problem.
  
  – Prvt_Yadv
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  default file=0666 => -rw-rw-rw- here Umask = 0541 => -r-xr----x substract them => --w--w-rw- for the file
  
  – jason momoa
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I was checked any value and it's always return a true value
  
  – jason momoa
  2 days ago
  

  
  
  
  

add a comment | 

0

Consider it like this, if you have default permissions 666 means that you have only r+w for user, group and others, and by umask 541 you want to remove permissions r+x,r,x.

For user:

By default you can read+write and by umask you want to remove read+executable 
permissions, but it will never happen because initially file has no executable 
permissions. So only read permission will be removed.

For group:

read permission will removed

For others:

There is no executable permission in the file, so there will be no change.

Same concept applies for directories.

share|improve this answer

answered 2 days ago

Prvt_YadvPrvt_Yadv

3,29631430

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks for your answer ^^, I found a trick and I think it's right:
  
  – jason momoa
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Please add it as answer, if it solves your problem.
  
  – Prvt_Yadv
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  default file=0666 => -rw-rw-rw- here Umask = 0541 => -r-xr----x substract them => --w--w-rw- for the file
  
  – jason momoa
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I was checked any value and it's always return a true value
  
  – jason momoa
  2 days ago
  

  
  
  
  

add a comment | 

Consider it like this, if you have default permissions 666 means that you have only r+w for user, group and others, and by umask 541 you want to remove permissions r+x,r,x.

For user:

By default you can read+write and by umask you want to remove read+executable 
permissions, but it will never happen because initially file has no executable 
permissions. So only read permission will be removed.

For group:

read permission will removed

For others:

There is no executable permission in the file, so there will be no change.

Same concept applies for directories.

share|improve this answer

answered 2 days ago

Prvt_YadvPrvt_Yadv

3,29631430

By default you can read+write and by umask you want to remove read+executable 
permissions, but it will never happen because initially file has no executable 
permissions. So only read permission will be removed.

read permission will removed

There is no executable permission in the file, so there will be no change.

share|improve this answer

answered 2 days ago

Prvt_YadvPrvt_Yadv

3,29631430

answered 2 days ago

Prvt_YadvPrvt_Yadv

3,29631430

answered 2 days ago

Prvt_YadvPrvt_Yadv

3,29631430