How can I change leading zero in umask 0022? Can it be changed? I searched the usual methods for information and my local system was of little help. $ man umask   No manual entry for umask    $ umask --help bash: umask: --: invalid option umask: usage: umask [-p] [-S] [mode]  I found an  with the following:  umask [-p] [-S] [mode] mode ... File creation mask -S  ........  Print the mask in symbolic format -p  ........  Output in a form that can be reused as input  Based on this, I tried a few basic umask commands: $ umask 0022  $ umask -p umask 0022  $ umask -S u=rwx,g=rx,o=rx  I was surprised to find that I could not run umask with sudo $ sudo umask   [sudo] password :   sudo: umask: command not found    I changed the regular umask as follows: $ umask 021 $ umask 0021  $ umask 0022 $ umask 0022    Finally, I hit an error while trying to change the leading zero $ umask 1022 -sh: umask: 1022: octal number out of range    So, back to my original question, how does one change the leading zero (0)?  Is it for the super user (root)? Can it be updated, and if so, to what? I reviewed a similar question  and found:  I don't know that you would unmask the first bit with umask, but technically you could ...  and  I don't believe you can actually set the umask to allow you to enable any of these extra bits by default  So, these uncertain and contradictory statements do not answer this question.
So I know umask can restrict privileged users, using this format umask ugo.  I understand that the read = 4, write = 2, and exec = 1. However, when I type umask, it returns 4 digits which is 0022 or 0073. I have no understanding of how does this work now because there is an extra digit. What is that extra digit and what does 0022 mean?
PHP - Get rid of notation Let $u:\mathbb{C}\rightarrow\mathbb{R}$ be a harmonic function such that $0\leq u(z)$ for all $ z \in \mathbb{C}$. Prove that $u$ is constant.  I think i should use Liouville's theorem, but how can i do that? Help! Custom command does not reference a label created with a previous custom command Why is my camera so forgiving for overblown exposure when shooting in RAW? What updates should be done to older monsters to bring them in line with MM3 era monsters? I know that to better balance paragon and epic level players against similar level monsters, WotC changed the math used for monsters for MM3, and I believe that they changed some of the types of powers they gave monsters.   If I want to update monsters from MM1 or MM2 to be in line with MM3 era monsters, what changes should I make? 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My feeling is such stages-wise charging &amp; control/monitor circuitry is already within the phone set, i just want to reconfirm. When I try to save a new file in TextEdit, I always get a popup saying I can't save with ".txt" at the end of the file name.  I want to save a plaintext file, but TextEdit insists on saving as .rtf instead. It also lists .doc, .odt, .html, and some others.  How can I save a TextEdit document as plain text? Is there any term used for multiple hat tricks? If a bowler takes more than three sequential wickets (say six wickets) in one over, would this be called as two hat-tricks or does this feat, however unlikely, has a different name? How to receive sms from our android app? Why do we use "have" with does and not "has"? You can’t buy your kids clever. What’s more, if they’re merely above average, by sending them to some hideous Holland Park hothouse, you’re probably buying them miserable.  — Alex Proud,   ""   As far as I know, clever and miserable are both adjectives, not nouns. So why were they used with to buy? Is this an idiomatic way of saying You can't buy your kids intelligence and you're probably buying them (moral) misery? If so, what is the idiom used here? I'm not familiar with it. "We plan" versus "we are planning" Is there a semantic difference between the following two sentences?     In the future, we are planning to migrate our tool to the Z3 solver.         In the future, we plan to migrate our tool to the Z3 solver. Tool for straightening a bent metal bar of the gate? Someone's car hit our metal bar gate, so one of the rods is bent as shown in the photo. What kind of a tool can I use to bend this back, so it's straight? Thank you! What potions actually grant fame or glory, or put a stopper in death? Are there any? ‘You are here to learn the subtle science and exact art of potionmaking,’ [Snape] began. He spoke in barely more than a whisper, but they caught every word – like Professor McGonagall, Snape had the gift of keeping a class silent without effort. ‘As there is little foolish wand-waving here, many of you will hardly believe this is magic. I don’t expect you will really understand the beauty of the softly simmering cauldron with its shimmering fumes, the delicate power of liquids that creep through human veins, bewitching the mind, ensnaring the senses ... I can teach you how to bottle fame, brew glory, even stopper death – if you aren’t as big a bunch of dunderheads as I usually have to teach.’&#185;   Philosopher's Stone - page 102 - Bloomsbury - chapter 8, The Potions Master   Are there any canon potions that actually grant fame or glory, or stopper death? I'm looking for specific canon-based answers from sources such as the Harry Potter books, interviews with J.K. Rowling, or information from Pottermore. Maybe because it's early I'm overlooking the obvious, but offhand I can't think of any canon potions that fit this criteria exactly.   &#185;I edited in the entire paragraph that I can teach you how to bottle fame, brew glory, even stopper death comes from for clarity. Hopefully this will be helpful and provide a better context for the comment in question. Showing that $\lnot Q \lor (\lnot Q \land R) = \lnot Q$ without a truth table I've done a truth table after reducing it to this and it seems to be equal to $\neg Q$:  $$\lnot Q \lor (\lnot Q \land R) = \lnot Q$$  But when I try to show it without a truth table (with just transformations), I end up in a loop between that and:  $$\lnot Q \land (\lnot Q \lor R)$$  Is there a way to show this is true without using a truth table? What am I missing?!  Thanks in advance! I have 3987MB of RAM on my laptop. Task (System) Manager says that I am using 700MB of this RAM, whilst the command "free -m" says I am using nearly 1500MB. Which one is correct and why do they show two different values?   Thanks for any help. Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$ Proving at least 99 pairwise sums of the reals are non-negative Suppose, one hundred real numbers are given and their sum is 0.Then how can I prove that at least 99 of the pairwise sums of these hundred numbers are non-negative?  I tried this: Let the real numbers be $a_i$,$1\leq i\leq100$. By the given condition, $a_1+a_2+\dots +a_{100}=0$.Suppose for the sake of contradiction, that at most 98 of the pairwise sums of these given reals are non-negative.I just tried to examine the case when exactly 98 of them are non-negative. I labelled the sums $S_1$,$S_2,\dots S_{4950}$,WLOG assume that sums $S_1,S_2,\dots S_{4852}$ are negative, the rest are non-negative. $$S_1+S_2+\dots +S_{4950}=99(\sum_{i=1}^na_i)=0$$ Case I: $S_{4853}+a_{4854}+\dots +a_{4950}=0$.That simply means that $S_1+S_2+\dots +S_{4852}=0$.But, by the given condition, $S_1+S_2+\dots +S_{4852}&lt;0$,a contradiction.After that, I am not able to make any progress.  Can anyone please suggest a more efficient method to solve this problem.I feel this approach will involve enormous case work. Getting a standard form quadratic from a set of points ($3$ points) I came home from school today, pulled out my homework, now I'm stumped.  I don't want the answer, I just want to know how to do it.  Here is the question that I'm reading:  Determine a quadratic function in standard form for the set of points $(x_1,y_2)(x_2,y_2)(x_3,y_3)$  Where the $x$'s and $y$'s represent the points.  Here are the points I am working with (remember, I don't want the answer, I just want to know how to do it, there are lots of questions on this so I want to be able to know how to do it).     $(-1, 2), (0, 1), (1, -4)$   Thanks! What is the best way to add options to a select from a JavaScript object with jQuery? What is the best method for adding options to a &lt;select&gt; from a JavaScript object using jQuery?  I'm looking for something that I don't need a plugin to do, but I would also be interested in the plugins that are out there.  This is what I did:  selectValues = { "1": "test 1", "2": "test 2" };  for (key in selectValues) {   if (typeof (selectValues[key] == 'string') {     $('#mySelect').append('&lt;option value="' + key + '"&gt;' + selectValues[key] + '&lt;/option&gt;');   } }   A clean/simple solution:  This is a cleaned up and simplified :  $.each(selectValues, function(key, value) {      $('#mySelect')           .append($('&lt;option&gt;', { value : key })           .text(value)); });   Changes from matdumsa's: (1) removed the close tag for the option inside append() and (2) moved the properties/attributes into an map as the second parameter of append().