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I'm saying there are 8 different umasks due to image problems.
I am 666-640 = 026
Thought that was the answer,

What kind of reasoning will umask come out in 8 ways?
Why is 027 the same result?
If i am familiar with Linux, thank you.

  • Answer # 1

    As explained,0666is the default for files, and then aumaskmask is drawn. Therefore, whether or not the bits for the execution bits that were not originally set up are inumask, it does not matter how much the file is created.

  • Answer # 2

    There are the following 8 umask values ​​that can generate 0666 to 0640.

    0137
    0136
    0127
    0126
    0037
    0036
    0027
    0026

    If you can help.

  • Answer # 3

      

    What kind of reasoning will umask come out in 8 ways?

    Since it is a

    file, the execution bit is not attached by default at the time of creation. Therefore, if the execution bit is ignored, the value of the one-digit umask and the permissions correspond as follows.

    0 and 1 =>rw *
    2 and 3 =>r- *
    4 and 5 =>-w *
    6 and 7 =>-*

    rw- r-- ---, so 8 is probably the umask below.

    0026, 0027, 0036, 0037, 0126, 0127, 0136, 0137