Timeline for What are the final 3 bits in the UNIX permission mode bits?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2018 at 4:35 | history | tweeted | twitter.com/StackUnix/status/1002770780546007041 | ||
| Jun 1, 2018 at 21:47 | vote | accept | winnie99 | ||
| Jun 1, 2018 at 15:45 | answer | added | ilkkachu | timeline score: 7 | |
| Jun 1, 2018 at 15:28 | answer | added | Stephen Kitt | timeline score: 3 | |
| Jun 1, 2018 at 15:27 | comment | added | Kevin Kruse | The 0 (first digit) in your example (0777) references the suid, sgid and "sticky" bit per Stephen's link. The first 7 (second digit) is the 3 bits (RWX) of the owner's permissions. The second 7 (third digit) is the 3 bits of the group's permissions. The final 7 (fourth digit) is the 3 bits of the others' permissions. You're correct in understanding that 3 bits together can represent a number from 0 to 7. | |
| Jun 1, 2018 at 15:09 | comment | added | winnie99 | Thanks Stephen. As a matter of fact I have read the answer before but I'm a little dense at this so I didn't get a clue about the last 3 bits.. The closest thing I can infer from that answer is 3 bits can represent 0 to 7, and there are 4 values that can set permissions apparently (like 0777), and each number in each position represent some permission? | |
| Jun 1, 2018 at 14:50 | review | First posts | |||
| Jun 1, 2018 at 15:07 | |||||
| Jun 1, 2018 at 14:46 | comment | added | Stephen Kitt | This answer should cover the first part of your first question. | |
| Jun 1, 2018 at 14:44 | history | asked | winnie99 | CC BY-SA 4.0 |