Talk:Umask
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This doesn't look like a bitwise AND to me; more like a subtraction. 777 & 177 makes 177, not 600, wherease 777-177 does make 600. The unix man pages for umask(1) also refer to subtraction rather than ANDing.
- That sounds right to me. With my umask set to 002, if I create a new file or directory it has its permissions set to 0775, which of course is 777-002. On the other hand, the man page for the C function (umask(2)) that I have says "umask sets the umask to mask & 0777." I don't know enough on this subject to make an edit, though. Does anyone have an idea of how that bit should be written? It's somewhat confusing as it is. --Dirk Gently 03:48, 28 July 2005 (UTC)
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- Doesn't the current notation (^ ~) indicate the XOR of the complement of the argument? I think that's incorrect, and at least, could be simplified greatly to say simply the complement, (~) of the argument (although it should probably be explicitly stated as such for clarity). It comes out like the following:
- ~1778 = 6008
- --Max Magee
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- This is correct. ^ = XOR. I'm the editor who did those major changes back in 17 August 2005. Note it was a simple AND before, and totally wrong. The XOR with the argument itself (not with the complement!) also works. Prove me wrong. A ^ B in this case will give the same result as in A AND ( ~ B). -andy 80.129.114.15 04:05, 14 January 2007 (UTC)
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- 666 AND NOT 174 = 602 = WTF?
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It is correct that 666 AND NOT 174 = 602. It is also NOT true in general that A ^ B = A AND ~B.
