I am trying to understand the following           var obj1 = {             value: "first value"         };          var obj2 = {             value: "second value"         };          var  obj3 = obj2;          function change(obj1, obj2) {             obj1.value = "asdasd";             obj2 = "lol";         }          change(obj1, obj2);          console.log("1",obj1.value);//"asdasd"         console.log("2",obj2.value);//"secondvalue"   why only obj1.value is changed ??
The primitive types (number, string, etc.) are passed by value, but objects are unknown, because they can be both passed-by-value (in case we consider that a variable holding an object is in fact a reference to the object) and passed-by-reference (when we consider that the variable to the object holds the object itself).  Although it doesn't really matter at the end, I want to know what is the correct way to present the arguments passing conventions. Is there an excerpt from JavaScript specification, which defines what should be the semantics regarding this?
Does an ability or spell grant me the bonus action with which to use it? The  description of casting times says that spells with a casting time of &quot;bonus action&quot; require a bonus action to cast (p. 79, emphasis mine):  Most spells require a single action to cast, but some spells require a bonus action, a reaction, or much more time to cast. Bonus Action A spell cast with a bonus action is especially swift. You must use a bonus action on your turn to cast the spell, provided that you haven't already taken a bonus action this turn. You can't cast another spell during the same turn, except for a cantrip with a casting time of 1 action.  It's pretty clear that bonus actions aren't automatic (p. 69):  You can take a bonus action only when a special ability, spell, or other feature of the game states that you can do something as a bonus action. You otherwise don't have a bonus action to take.  So to cast a spell like Healing Word (p. 93), do I  need to be granted a bonus action that I can use, or do I have a bonus action I can use?  I guess I'm unclear on whether I need to have a bonus action before I can spend it on something, or whether having an ability, spell, or whatnot that can be done as a bonus action grants me the bonus action I need to use it. Currently, Stack Exchange’s Markdown parser only allows four-space indents to represent code blocks:      // some code     // another line of code    and other Markdown implementations allow for an alternative syntax, that doesn’t require indenting each code line:  ``` // some code // another line of code ```   This is much more convenient to type out. It would be super useful if Stack Exchange could support this syntax.    By extension, this syntax also allow you to specify the source language right after the opening ```:  ```js // some code // another line of code ```   …which would then enable syntax highlighting for that specific language. Although it’s interesting metadata, I don’t think this feature is needed on Stack Overflow, as the syntax highlighting library works pretty well for various languages.  So, even if you would allow this syntax but ignore the ```language, this would greatly improve my productivity on Stack Exchange.  What do you think?  The moderators are currently collecting feedback regarding this feature, so please post a comment with your thoughts. Good idea? Bad idea? Don’t really care? What do you think are the benefits/drawbacks? Experiences? Let them know! Thanks! package makeidx: print first letter for each group of keys I've created index with makeidx package. It works. What really need is to print the first letter above every group of keys (i.e. referring to the image I want to print the red letter, that I have manually inserted). What do you call a person who is never content with anything that is given to him? I am looking for an adjective for a person who, no matter what you give him/her (materially or otherwise), is never content and always expects more. MWE for externalize + automatic recompilation after tikz figure modification Moderncv has a line of empty space between address and phone/email, like this:   MWE:  \documentclass[12pt]{moderncv} \moderncvstyle{banking} \name{Sherlock}{Holmes} \address{221B Baker Street}{London}{U.K.} % Your current address \phone[mobile] {+44~(0)71234 00000}    % Your mobile phone number \email{superdetective@mail.com}  \begin{document} \makecvtitle \end{document}   How do I remove/decrease that space by shifting the phone and email line up? How to get over "device or resource busy"? Computational complexity of Fibonacci Sequence How do I use the Chinese remainder theorem to find all the square roots of 11 in $\mathbb Z_{35}$? I execute quite a lot of sudo commands.  I noticed that if I execute one and then another one within few minutes, the 2nd time I don't get the message:     [sudo] password for my_username:   but I do get it when there is more time between the execution of the two commands.  What's the period of time in which the message is not appearing? How can we check it / update it? I'd say Microsoft have a way of bending the rules and I know that McLaren have won the championship. While this sounds strange, I believe it is correct English (sorry, I'm not native).   But when it's a small company, would you still use it this way? Is a company always plural, or are small companies singular? I.e., would you say Bakery Johnson makes fine bread or Bakery Johnson make fine bread? Is it My book seller, Woody's, have moved or is it has moved? When & How often do 4e Monster powers recharge? What is the 64-bit Firefox Beta PPA? I recently discovered that my computer is 64-bit. I have backed up my Home folder, and reinstalled Ubuntu. The reinstall wasn't nearly as painful as I thought. There is one thing that I can't quite seem to figure out: how do I get the 64-bit Firefox Beta build? I always get the Beta builds, but I want to take advantage of the 64-bit architecture of my computer.  says that Mozilla has come out with a 64-bit version of Firefox, but I can't seem to find it.  I do understand  of using a 64-bit browser, but I've decided to jump right in and do it anyway. (Flash and Java are already 64-bit, and who cares about Silverlight, since it's not for Linux anyway?)  There's only one issue, and it's a big one: I can't find the 64-bit Beta PPA!!! (I really hate using .tar.gz files, but I'd be willing to do that as long as I could still access Firefox via the Launcher. Oh, speaking of which, I don't understand .tar.gz files. Once, I managed to run one (the Dropbox Beta build,) but I have no idea whatsoever on how to install them: as in, click on the icon and go.) How do I remove a configuration object from the active configuration? I'm writing a custom module for Drupal 8. It defines a field storage configuration object, in a file called modules/mymod/config/install/field.storage.bloom_stem.bs_twitter_ac.yml.  The problem is that this is not removed when I uninstall the module; it is apparently left behind because, when I re-install the module, I get a crash.   drush -y pm-uninstall mymod drush cr drush -y en mymod      exception 'Drupal\Core\Config\PreExistingConfigException' with message 'Configuration objects (field.storage.bloom_stem.bs_twitter_ac) provided by mymod already exist in active configuration'   So it seems I need to some how manually remove this configuration object. Do I need to provide some more code that deletes it when the module is uninstalled? Given $E[Y]=1$ , $E[Y^2]=2$ and $E[Y^3]=5$, Y non negative integer random variable, find min of $P[Y=0]$ Divisibility Rule for 9 I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic.    Question:  If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9.    Proof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\mod9$  Now consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\mod9$   therefor  $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)$  $9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\mod9)(d_1+d_2+...d_n)$  Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself.    The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\mod 9$.  I think this is fair game here but not 100% confident.  Thanks. Few days back I have installed ecryptfs, created a private directory using it. Now my Ubuntu login became super slow. I have tried removing ecryptfs from Synaptic. I found It, it can not be deleted that way. Somewhere I have read about this command  ecryptfs-setup-private --undo   But this command displays a rather confusing list of commands on terminal. I don't want to mess up my installation, neither want to loose my administrator account.  Now my question is, how can I safely remove /home/user/Private (.Private) and remove ecryptfs encryption ? What is a decoupling capacitor and how do I know if I need one? What is a decoupling capacitor (or smoothing capacitor as referred to in the link below)?  How do I know if I need one and if so, what size and where it needs to go?    mentions many chips needing one between VCC and GND; how do I know if a specific chip is one?   Would an  4-bit parallel access shift register used with an Arduino need one? (To use my current project as an example) Why or why not?  I feel like I'm starting to understand the basics of resistors and some places they're used, what values should be used in said places, etc, and I'd like to understand capacitors at the basic level as well. I normally use the iPhone timer to sleep the device after I've dozed off to some music or podcast. Unfortunately this means that when I ask Siri to set a timer for some cooking, Siri waits the allotted time and turns off the phone rather than notify me that the time is up. Any solutions to this? Is the following proof correct, or are certain steps illegal?  Suppose $f:G_1\to G_2$ is a surjective group isomorphism, $N_1\lhd G_1, N_2\lhd G_2$ normal subgroups, with $N_1=f^{-1}(N_2)$. Prove that $(G_1/N_1)\cong (G_2/N_2)$.  Proof:  As $f$ is surjective, we an use the first isomorphism theorem to get $G_1/\ker(f)\cong G_2$, which implies that $$(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)$$  Using the first isomorphism theorem we also know that $$N_1/\ker (f)\cong f(N_1)=f(f^{-1}(N_2))=N_2$$  Combining this we can see that $(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)\cong G_2/N_2$.  Using the third isomorphism theorem we have that: $$ (G_1/\ker f)/(N_1/\ker f)\cong G_1/N_1$$  And from the last two lines we have $G_1/N_1\cong G_2/N_2$ $\tag*{$\Box$}$