Confusing about passing by refrenece in js object
Is JavaScript a pass-by-reference or pass-by-value language?
Difference between shadowing and overriding in C#? What's difference between shadowing and overriding a method in C#? The outlet for my range hood was misplaced. It needs to be 4 inches to the side. I can make the junction and move the wiring to the new location but I really don't want to have the empty plate exposed.   The wire is not long enough to reach the required location, so I will make a junction in the existing box, and add a new box at the correct location.  Is there any acceptable circumstance where I could tile over the "old" in-wall junction box?  I'm guessing not, but I figured I'd ask.  EDIT: I should clarify, based on the answers below... I know for a fact that this outlet is the only line on this entire circuit, all the way back to the panel. I am asking explicitly if there is any way to acceptably splice the wiring without exposing a junction box. Evaluating $\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$ for $x\in\mathbb{R}^{+}$ I am trying to evaluate the following sum. Here $x\in\mathbb{R^{+}}$. $$f(x)=\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$$ At present I do not know how to proceed. Any hints are appreciated. Thanks. Should first and last name be formatted? How does the /proc//exe symlink differ from ordinary symlinks? If I start a process and then delete the binary of it, I can still recover it from /proc/&lt;pid&gt;/exe:  $ cp `which sleep` . $ ./sleep 10m &amp; [1] 13728 $ rm sleep $ readlink /proc/13728/exe                            /tmp/sleep (deleted) $ cp /proc/13728/exe ./sleep-copy $ diff sleep-copy `which sleep` &amp;&amp; echo not different not different $ stat /proc/13728/exe    File: ‘/proc/13728/exe’ -&gt; ‘/tmp/sleep (deleted)’   Size: 0           Blocks: 0          IO Block: 1024   symbolic link   On the other hand, if I make a symbolic link myself, delete the target and attempt to copy:  cp: cannot stat ‘sleep’: No such file or directory   /proc is an interface to the kernel. So does this symbolic link actually point to the copy loaded in memory, but with a more useful name? How does the exe link work, exactly? I'm a 16-year-old male, and I have been interested in tabletop RPGs for a long time, especially D&amp;D. I sorta dove into D&amp;D really fast and hard by buying the Red box, Dungeon Master Kit, Monsters Vault, Players Handbook, Rules book, and even the Shadowfell Campaign not quite knowing what I'd get myself into. I've done tooonnsss of research and even watched live games on Youtube.   For the most part, I'm basically on my own, My friends are somewhat interested in it and want to play, but expect me to Teach/Dungeon Master the game for them, I'd really wish to do this, but need more insight and expertise. I've played the solo Campaign in the Red box and even thought about joining a local group from a popular gaming store in Phoenix, AZ called Imperial Outpost. But I'm not sure if they would take a newbie like myself. Any advice? Logistic regression vs. LDA as two-class classifiers Pixel Perfect 2D tiles I'm trying to make an 64x64 tilemap using Super Tilemap Editor &amp; Unity but my tiles looks terrible. I did my research about pixel perfect 2D and all of the solutions were for pixel art, they didn't work for me.  I have the following tileset I made in Adobe Illustrator as an example    Tiles looks like this when placed in game (1x zoom)    I'm pretty sure they're not 64x64 in game, they also look pixelated probably because they're smaller than they actually are on camera.  So my question is, how can I calculate the correct camera size for 64x64 tiles and do I have to make any changes in order to make sprites look pixel perfect? I'm currently using 1080/64/2=8.4375 (vertical resolution / PPU / 2) as camera size. Does That's very sweet of you  in the US sound like That's very sweef you? the of is only speak f and link with swee How to print the full NumPy array, without truncation? I currently have an ec2 &quot;t2.nano&quot; instance up and running. I would like to switch to a &quot;t2.micro&quot;. In order to do that, I need to stop the instance. But I would lose my IPv4 address in the process. And I do not want that (because I have several ssh tunnels up and running from rasps in the wild that are currently hard coded to connect to this IPv4 address). I decided to create an Elastic IP address. But I understand that by creating an Elastic IP address and assigning it to my current ec2 instance, I would lose my current IPv4 address too. Is it possible to create an Elastic IP address from the current IPv4 address of my ec2 instance? If so, how can I do this operation? Short story about a man stranded on a red planet with a sentient hand as his friend I remember reading a short story 10 to 15 years ago about a man stranded alone on a red planet, where nothing would be green, and so he would miss that color. There was a sentient hand to keep him company, and it talked to him. I don't remember if the hand was a native life form or his own hallucination.  The story was not very old at that time, so it may have been written in the late 90s or 00s. Let's fix some terminology first. A category $\mathcal{C}$ is preabelian if:  1) $Hom_{\mathcal{C}}(A,B)$ is an abelian group for every $A,B$ such that composition is biadditive,  2) $\mathcal{C}$ has a zero object,  3) $\mathcal{C}$ has binary products,  4) $\mathcal{C}$ has kernels and cokernels.  A category $\mathcal{C}$ is abelian if it is preabelian and satisfies:  5) every monomorphism is a kernel and every epimorphism is a cokernel.  Define the coimage of a map to be the cokernel of its kernel, and the image to be the kernel of its cokernel. We have the following commutative diagram:    where $\overline{f}$ is the only existing map (because of universality of kernel and cokernel).  I'm having trouble proving the following:     A preabelian category $\mathcal{C}$ is abelian iff $\overline{f}$ is an isomorphism.   The converse is easily shown, I'm having trouble proving $\Rightarrow$... How can I accept the Microsoft EULA agreement for ttf-mscorefonts-installer? After a recent update, ttf-mscorefonts-installer prompted me to accept its license agreement.     ┌─────────────────┤ Configuring ttf-mscorefonts-installer ├─────────────────┐  │                                                                           │   │ TrueType core fonts for the Web EULA                                          │                                                                               │ END-USER LICENSE AGREEMENT FOR MICROSOFT SOFTWARE                             │                                                                               │ IMPORTANT-READ CAREFULLY: This Microsoft End-User License Agreement           │ ("EULA") is a legal agreement between you (either an individual or a          │ single entity) and Microsoft Corporation for the Microsoft software           │ accompanying this EULA, which includes computer software and may include      │ associated media, printed materials, and "on-line" or electronic              │ documentation ("SOFTWARE PRODUCT" or "SOFTWARE"). By exercising your          │ rights to make and use copies of the SOFTWARE PRODUCT, you agree to be        │ bound by the terms of this EULA. If you do not agree to the terms of          │ this EULA, you may not use the SOFTWARE PRODUCT.                              │                                                                               │                                  &lt;Ok&gt;                                         │                                                                           │   └───────────────────────────────────────────────────────────────────────────┘    Screenshot:   For some reason my terminal will not allow me to accept, or for some reason I am pressing the wrong hotkey...  I've tried every letter on the keyboard and Enter among others...  I'm sure there is a very simple and obvious solution to this.  I've also just tried to remove the package completely however the terminal states that due to the package not being correctly installed, I should reinstall the package before removing it.  Very frustrating!  Essentially, because I cannot successfully install this package, I can't really ever upgrade my system because I always have to end up terminating the terminal with the license agreement (thus the upgrade fails). The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$ So my solution went like this:  $$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$. Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).  Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$ So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning? Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution  I realised my mistake after some hours of pondering over it. (In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$. The correct way, indeed, would be through the link supplied Starting iPhone app development in Linux? I've heard that you need to get a Mac if you want to develop iPhone apps. Is this true?   Is it possible to develop iPhone apps using Linux? If yes, what do I need and where do I download the necessary tools? Speed $v$ of a wave on a tightrope Quantum teleportation - the alternative to destroying atoms Can't you just disassemble, not destroy our atoms and transport them to another teleporter via networking or telecoms?  This is my thinking of how to keep the same person, not an exact copy of him/her, alive when teleporting.  Please answer to tell me if it's possible or not. I can't create an onCheckedChanged listener for my RadioGroup Why is removing instationarities a good thing when trying to forecast a time series?