On boot my nas mounts are timing out, but when I manually run sudo mount -a after login they all mount just fine. The problem occurs with and without vers=3.0. $ dmesg [    5.521690] FS-Cache: Loaded [    5.530037] FS-Cache: Netfs 'cifs' registered for caching [    5.532069] Key type cifs.spnego registered [    5.532071] Key type cifs.idmap registered [    5.532305] CIFS: Attempting to mount //192.168.1.3/Documents [    5.532307] CIFS: Attempting to mount //192.168.1.3/Projects [    5.535674] CIFS: Attempting to mount //192.168.1.3/Media [    9.467263] e1000e: enp0s31f6 NIC Link is Up 1000 Mbps Full Duplex, Flow Control: Rx/Tx [   13.671407] CIFS VFS: Error connecting to socket. Aborting operation. [   13.671412] CIFS VFS: Error connecting to socket. Aborting operation. [   13.671415] CIFS VFS: Error connecting to socket. Aborting operation. [   13.671431] CIFS VFS: cifs_mount failed w/return code = -2 [   13.673594] CIFS VFS: cifs_mount failed w/return code = -2 [   13.675588] CIFS VFS: cifs_mount failed w/return code = -2 [   23.155328] snd_hda_codec_hdmi hdaudioC1D0: HDMI: invalid ELD data byte 2 [   23.227324] snd_hda_codec_hdmi hdaudioC1D0: HDMI: invalid ELD data byte 2 [   57.613993] CIFS: Attempting to mount //192.168.1.3/Projects [   57.906356] CIFS: Attempting to mount //192.168.1.3/Media [   57.941905] CIFS: Attempting to mount //192.168.1.3/Documents  $ cat /etc/fstab //192.168.1.3/Projects /home/ehammer/.Projects/ cifs credentials=/home/ehammer/.smbcredentials,vers=3.0,iocharset=utf8,uid=1000,gid=1000 0 0 //192.168.1.3/Media /home/ehammer/.Media/ cifs credentials=/home/ehammer/.smbcredentials,vers=3.0,iocharset=utf8,uid=1000,gid=1000 0 0 //192.168.1.3/Documents /home/ehammer/.Doc-Share/ cifs credentials=/home/ehammer/.smbcredentials,vers=3.0,iocharset=utf8,uid=1000,gid=1000 0 0
right now   I have setup /etc/fstab like this:  //192.168.5.167/H /mnt/ssd cifs credentials=/root/.smbreds,noauto,x- systemd.automount 0 0   I want to set it up like first network should be connected and then fstab mount should get mounted.
In case of shell-builtins (eg type itself):  $ type type type is a shell builtin  $ which type &lt;Doesn't return anything since it's a shell builtin, silently exits&gt;   In case of commands (normally) (eg python):  $ type python python is /usr/bin/python  $ which python /usr/bin/python   In case of which (which is a command located at /usr/bin/which)  $ type which which is hashed (/usr/bin/which) $ which which /usr/bin/which   Why does type which say that which is hashed? What is the significance of which being hashed and what does it actually mean? Is drinking water during or after a meal better? When did you wake up?      I woke up at 10.   But then if I say, “When did you sleep?”     I slept at 10. – seems difficult to digest!   That's because sleep is a process that includes duration.     I slept for 10 hours. – this way, it's digestible!   But when I mention the point of time, is it okay?   Strange thing is, if I'm describing my habit/routine, it's perfectly fine..     I sleep at/by 10     I request not to take this question 'logically'. It's about English. Nobody is perfect in timing. Please don't argue that you cannot have 'perfect time' of sleeping because you don't know when you slept. I'm talking about the sentence structure and not precise timings.   By the way, “I slept at 10” is very common in India. Distance between surfaces Drawing wavy, wiggly lines in QGIS? Is there a QGIS function or plugin to draw wiggly line?   I have used Spline Tool to manually draw some waves, but it is time consuming.  If possible, I would like to draw something like:    Inkscape Function Plotter ( sin(x) curve). I have a hook_form_alter in a module, and it should be firing, but it appears it's not. I've cleared caches and rebuilt the registry. Is there someplace in the database or somewhere else where I can see if my hook is registed? FWIW here is my hook:  /**  * Implements hook_form_alter  */ function my_module_form_alter(&amp;$form, &amp;$form_state, $form_id) {   drupal_set_message($form_id); } How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number? It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well.  On the other hand, clearly $\sqrt{n^2} = n$ for any positive integer $n$.  It seems that any positive integer has a square root that is either an integer or irrational number.        How do we prove that if $a \in  \mathbb N$, then $\sqrt a$ is an integer or an irrational number?      I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ are all irrational.  This suggests we can extend the previous result to other radicals.        Can we extend 1? That is, can we show that for any $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational? Isn't $HOME supposed to be the location of the user's home directory? I'm having trouble with a web application I'm writing: The web application, which runs as user nobody, needs to launch an inkscape process to manipulate some SVG files. Because this runs as user nobody, there is no home directory. What I am doing is creating a temporary directory under /tmp and trying to set that as the home directory via export HOME=/tmp/someUniqueId. I then set a few things in that "temporary" home directory, like a symlink .fonts to the folder for the font files to use.  I know this works for finding the ~/.fonts directory. But when I launch Inkscape, all I get is a message that Inkscape could not create the .inkscape directory:    Doesn't $HOME set the location of my home directory? Why is Inkscape still trying to access / as my home directory? Last night, I migrated my MacBook's operating system from 10.10 "Yosemite" to 10.11 "El Capitan". Everything seemed to go smoothly. Happily, regarding my TeX-related work, I am able to use the GUI front ends "TeXworks" and "TeXshop" just fine since completing the OS upgrade. I use MacTeX2015, with all the latest updates installed.  Oddly, though, since completing the upgrade I can no longer run any MacTeX command-line utilities such as texdoc and kpsewhich. (Actually, as I just discovered, I can no longer run pdflatex either from a command line. However, I never do run pdflatex from a command line prompt, so that's not an urgent problem for me -- at least not for now.)  The suggestions given in the posting  do not seem to be applicable to this case, as they all seem to revolve around setting up /usr/texbin correctly -- /usr/texbin is no longer allowed under MacOSX 10.11. The document  by Herb Schulz deals in depth with adjusting settings in the GUI front ends, but it appears not to touch on what to do about /usr/texbin, which appears to be the now-missing link to TeX's command line utilities.   Pointers and solutions most welcome! If my question gets migrated to another site, is it possible somehow to reverse it back to the original site? Ignore .DS_Store and Icon files in a folder with Cocoa NSFIleManager It's a series of fantasy books in which the inhabitants of the world, from the time when they're children, get to somehow have an accompanying small animal that represents their soul or their personality. Bash tab completion fails inside of command substitution I've been using kill $(pidof ) to terminate programs for a while, and since last night, every time I try to use tab completion inside the $() it throws an error,   bash: unexpected EOF while looking for matching `)' bash: syntax error: unexpected end of file   Tab-completion works fine outside of command substitution.  I guess my question is where are the configuration files for tab-completion located on an Ubuntu system? If anyone has had this specific problem, or can see plainly what I cannot, how to fix this, and/or why this would occur? (I always want to know why.)  and if not that, even where and how I should begin my investigation into this problem, would be greatly appreciated.  UPDATE  tab completion following $(pidof is no longer throwing an error, but is now not working at all. I've never really messed with any bash internals before, but am not afraid to get my hands dirty. Any guidance greatly appreciated. cat files with directory Is there a command to show the directory/file name when cat files?  For example: assume two files f1.txt and f2.txt are under ./tmp  ./tmp/f1.txt ./tmp/f2.txt    Then when I do cat ./tmp/*.txt, only the content of files will be shown. But how to firstly show the file name, then the content?, e.g.:   (The needed command):  ./tmp/f1.txt:    This is from f1.txt  and so on  ./tmp/f1.txt:  This is from f2.txt ...   Is there a command to do it? (There seems to be no option for 'cat' to show the file names) How could I go about detecting (returning true/false) whether an ArrayList contains more than one of the same element in Java?  Many thanks, Terry  Edit Forgot to mention that I am not looking to compare "Blocks" with each other but their integer values. Each "block" has an int and this is what makes them different. I find the int of a particular Block by calling a method named "getNum" (e.g. table1[0][2].getNum(); How to Convert Persian Digits in variable to English Digits Using Culture? When do I pronounce a non-existent “r” between adjacent vowel sounds? If I say two words consecutively, with the first ending in a vowel sound and the second starting with one, when is it correct to include a non-existent r between those two words?  Examples from phrases I've heard:        I saw (r)a movie   They saw (r)us   Law (r)and order   That's the idea, (r)anyway.      I heard the last one in the movie Edge of Tomorrow yesterday.  Is this grammatically correct? If so, is the omission of the r grammatically incorrect? Are there any rules regarding this? typeof is an operator and a function Meaning of the word "axiom" Thus, axioms are propositions we assume to be true and we use them in an axiomatic theory as premises to infer conclusions, which are called "theorems" of this theory.  For example, we can use the  to prove theorems of arithmetic.  This is one meaning of the word "axiom". But I recognized that the word "axiom" is also used in quite different contexts.  For example, a group is defined to be an algebraic structure consisting of a set $G$, an operation $G\times G\to G: (a, b)\mapsto ab$, an element $1\in G$ and a mapping $G\to G: a\mapsto a^{-1}$ such that the following conditions, the so-called group axioms, are satisfied:   $\forall a, b, c\in G.\ (ab)c=a(bc)$, $\forall a\in G.\ 1a=a=a1$ and $\forall a\in G.\ aa^{-1}=1=a^{-1}a$.      Why are these conditions (that an algebraic structure has to satisfy to be called a group) called axioms? What have these conditions to do with the word "axiom" in the sense specified above? I am really asking about this modern use of the word "axiom" in mathematical jargon. It would be very interesting to see how the modern use of the word "axiom" historically developed from the original meaning.   Now, let me give more details why it appears to me that the word is being used in two different meanings:  As peter.petrov did, one can argue that group theory is about the conclusions one can draw from the group axioms just as arithmetic is about the conclusions one can draw from the Peano axioms. But in my opinion there is a big difference: while arithmetic is really about natural numbers, the successor operation, addition, multiplication and the "less than" relation, group theory is not just about group elements, the group operation, the identity element and the inverse function. Group theory is rather about models of the group axioms. Thus: The axioms of group theory are not the group axioms, the axioms of group theory are the axioms of set theory.  Theorems of arithmetic can be formalized as sentences over the signature (a. k. a. language) $\{0, s, +, \cdot\}$, while theorems of group theory cannot always be formalized as sentences over the signature $\{\cdot, 1, ^{-1}\}$. Let me give an example: A typical theorem of arithmetic is the case $n=4$ of Fermat's last theorem. It can be formalized as follows over the signature $\{0, s, +, \cdot\}$: $$\neg\exists x\exists y\exists z(x\not = 0\land y\not = 0\land z\not = 0\quad\land\quad x\cdot x\cdot x\cdot x + y\cdot y\cdot y\cdot y = z\cdot z\cdot z\cdot z).$$ A typical theorem of group theory is  which states that for any finite group G, the order of every subgroup H of G divides the order of G. I think that one cannot formalize this theorem as a sentence over the "group theoretic" signature $\{\cdot, 1, ^{-1}\}$; or can one? so there are some idiosyncrasies with the method they use to invert that part of the image which I can't figure out. It could be that they just invert the luminosity but I don't know how i would do that in a dynamic way.