How to get current timestamp
How do you get the current time in milliseconds (long), in VB.Net?
What are the most widely used C++ vector/matrix math/linear algebra libraries, and their cost and benefit tradeoffs? java.net.SocketException: Connection reset I am getting the following error trying to read from a socket. I'm doing a readInt() on that InputStream, and I am getting this error. Perusing the documentation this suggests that the client part of the connection closed the connection. In this scenario, I am the server.  I have access to the client log files and it is not closing the connection, and in fact its log files suggest I am closing the connection. So does anybody have an idea why this is happening? What else to check for? Does this arise when there are local resources that are perhaps reaching thresholds?    I do note that I have the following line:  socket.setSoTimeout(10000);   just prior to the readInt(). There is a reason for this (long story), but just curious, are there circumstances under which this might lead to the indicated error? I have the server running in my IDE, and I happened to leave my IDE stuck on a breakpoint, and I then noticed the exact same errors begin appearing in my own logs in my IDE.  Anyway, just mentioning it, hopefully not a red herring. :-( I would like to get the path to the execution directory of a Windows Forms application. (That is, the directory in which the executable is located.)  Does anyone know of a built-in method in .NET to do this? Can anonymous class implement interface? Is it possible to have an anonymous type implement an interface?  I've got a piece of code that I would like to work, but don't know how to do this.  I've had a couple of answers that either say no, or create a class that implements the interface construct new instances of that. This isn't really ideal, but I'm wondering if there is a mechanism to create a thin dynamic class on top of an interface which would make this simple.  public interface DummyInterface {     string A { get; }     string B { get; } }  public class DummySource {     public string A { get; set; }     public string C { get; set; }     public string D { get; set; } }  public class Test {     public void WillThisWork()     {         var source = new DummySource[0];         var values = from value in source                      select new                      {                          A = value.A,                          B = value.C + "_" + value.D                      };          DoSomethingWithDummyInterface(values);      }      public void DoSomethingWithDummyInterface(IEnumerable&lt;DummyInterface&gt; values)     {         foreach (var value in values)         {             Console.WriteLine("A = '{0}', B = '{1}'", value.A, value.B);         }     } }   I've found an article  that describes one approach. Is this the best way of doing this? iOS app stalls phone when on weak wifi sign I've recently noticed a bug on the Stack Exchange IOS app. When I open the Stack Exchange app on weak wifi, it hangs on the blue load screen as it launches and either crashes the app or the entire phone lags and didn't respond for about a minute.  I'm running iOS 8.4.1, I have the most up to date version of the app, and by 'weak' wifi I mean two out of four bars of strength. Evaluate $$\sum\limits_{k=1}^{n} \frac{k}{2^k}$$ proof that on a compact manifold a vector field is complete I would like to know if my proof of the following statement is correct     If $M$ is a compact manifold, then every vector field $X$ over $M$ is complete.   Proof:  I take $p\in M$ and $(\gamma_p,I_p)$ a maximal integral curve through $p$. $I_p$ is open and non empty so if $I_p$ is closed, it is equal to $\mathbb{R}$.  So take $s\in \text{Adh}(I_p)$. There exists $t_i\rightarrow s, t_i\in I_p$. Define $q$ as $\text{lim}(\gamma_p(t_i))$ and take a maximal integral curve passing by $q$ as $(\gamma_q,I_q)$. Then since the vector fields are $C^\infty$, $\gamma_q(I_q)\cap\gamma_p(I_p)$ is nonempty (intuitively, this is clear but don't really know how to show it).  On the intersection, $\gamma_p$ and $\gamma_q$ coincide. So we can extend the maximal curve $\gamma_p$ to include $\gamma_q$. This implies that $q\in\gamma_p(I_p)$ and is actually equal to $\gamma_p(s)$  I think the main idea is there even though it probably lacks a bit of rigour. Copy & Pasting from Photoshop to Illustrator loses transparency I often run into this issue. I'm creating a file in Photoshop with masked layers etc. I want to make the final output available for some mock-ups in Illustrator, so I rasterize the layers and marquee select the entire area, copy and move to Illustrator.  Pasting the area in Illustrator flattens all transparencies, which is quite annoying. I'm left with my object and a flat white background.  Is there a reason for this? $\left|x_{n+1} - x_n\right| < C\left|x_n - x_{n-1}\right|$ prove this is Cauchy $R/Ra$ is an injective module over itself Let $R$ be a PID, $a\in R$ be a nonzero nonunit in $R$. Prove that $R/Ra$ is an injective module over itself.   If $R$ is a PID, every $R$- divisible module is injective, but the question concerns with $R/Ra$-module, so I have no idea to solve this problem. Help me. I was working on a problem in group theory, which asks about the automorphism group of a direct product of groups.  Okay, so I know that if $G,H$ are two groups whose orders are relatively prime, then $\operatorname{Aut}(G\times H)\cong\operatorname{Aut}(G)\times \operatorname{Aut}(H)$.  I also know that if $G,H$ are abelian and simple and isomorphic, then $\operatorname{Aut}(G\times H)\cong \operatorname{GL}_2(\mathbf{Z}_p)$ where $p$ is the order of $G$.  What I don't know is the following: what if $G,H$ are simple, yet not abelian? Let us for now focus on a more restricted case where $G\cong H$, in particular, $G=H$. What can we say about the group $\operatorname{Aut}(G\times H)$? (I heard something about this group being related to what is called "wreath product", which I do not know...)  Thanks bunch in advance! Windows installer for Ubuntu 14.04 LTS onwards Is there any Windows installer (wubi) for Ubuntu 14.04 LTS?  When I mount the .iso file with Daemon Tools and run wubi.exe, it just asks me if I want to reboot (I think that's useful only when the .iso  is burned on CD), so is there no "real" installation? Libvirt: how to pass qemu command line args? I have the current libvirt 1.2.20 i want to pass to my machine create with virt-manager some qemu flags. I edited the file and add    &lt;/devices&gt;   &lt;qemu:commandline&gt;           &lt;qemu:arg value='-L pc-bios'/&gt;   &lt;/qemu:commandline&gt; &lt;/domain&gt;   But doesn't work i also tried to edit from virsh  virsh edit domain   and put   &lt;qemu:commandline&gt;           &lt;qemu:arg value='-L pc-bios'/&gt;   &lt;/qemu:commandline&gt;   But give me this error  errore: XML document failed to validate against schema: Unable to validate doc against /usr/share/libvirt/schemas/domain.rng Element domain has extra content: qemu:commandline  Failed. Try again? [y,n,i,f,?]:    How to pass correctly qemu command line? Thanks How do I gunzip all files recursively in a target directory? I'd like to know what is the command to use to gunzip all files in a target directory recursively? I tried to use the unzip command but it didn't work.  I tried the command from The reciprocal of many prime numbers p in base 10 have a set of repeating digits p-1. Why? I've noticed that the reciprocal of many prime numbers have a curious number of digits. For example. 1/7 has 6 repeating digits, 1/17 has 16 repeating digits, and 1/47 has 46. There's a rule here that makes this pattern, and I haven't quite figured it out. Does anyone have some insight on this?  There are a few other numbers that follow the pattern subversively. For example, 1/3 has two repeating digits: 33. 1/11 has ten repeating digits: 0909090909. The number 1/13 has a set of six that repeats twice, which makes a set that's 12 digits long.  For reference, here is a website that lists reciprocals of numbers 2 through 70, including non-primes.   The ones that break the rule are 1/2 and 1/5, but they seem to be a matched pair, where the 2 and 5 switch places in the equation 1/X = Y/10  What is the rule that makes this pattern? I'm very curious. Recently I've noticed more and more sign-in / sign-up workflows defaulting to the sign-up option first and providing a minimal "Sign-In / Login" action somewhere else on the page. What's the reason behind this practice?  Examples where I've seen this recently: In episode 10 of Full Metal Alchemist: Brotherhood, during the events in the Research Lab 5, Edward Elric is severely hurt. Later in the hospital, he writes down everything he had come to know, and he discusses with Alphonse, Hughes, and Armstrong about what to do with the information to continue the investigation.  At this point, King Bradley arrives with lemon into the room. He finds out what they are doing but acts very calm and chilled out and allows them to continue investigation while being careful.  Why did he not get Hughes and Armstrong killed at that point? Or at least tell them to not investigate at all? Using the fact that $n^3 + 5n$ is divisible by $6$, infer that $n^4 - 2n^3 + 11n^2 + 14n$ is always divisible by $24$ The problem is as stated in the title. I've tried factorization, but it doesn't seem to yield anything. Any help is greatly appreciated. Also please don't use modular arithmetic, thank you. skype 4.3 cannot connect to account on ubuntu 14.04 LTS I have being using Skype for a long time on my Ubuntu 14.04 LTS. From this morning on, Skype simple cannot connect to my account anymore.  That is my Skype version:     Skype 4.3.0.37   © 2014 Skype and/or Microsoft   My Ubuntu system:  LSB Version:    core-2.0-amd64:core-2.0-noarch:core-3.0-amd64:core-3.0-noarch:core-3.1-amd64:core-3.1-noarch:core-3.2-amd64:core-3.2-noarch:core-4.0-amd64:core-4.0-noarch:core-4.1-amd64:core-4.1-noarch:cxx-3.0-amd64:cxx-3.0-noarch:cxx-3.1-amd64:cxx-3.1-noarch:cxx-3.2-amd64:cxx-3.2-noarch:cxx-4.0-amd64:cxx-4.0-noarch:cxx-4.1-amd64:cxx-4.1-noarch:desktop-3.1-amd64:desktop-3.1-noarch:desktop-3.2-amd64:desktop-3.2-noarch:desktop-4.0-amd64:desktop-4.0-noarch:desktop-4.1-amd64:desktop-4.1-noarch:graphics-2.0-amd64:graphics-2.0-noarch:graphics-3.0-amd64:graphics-3.0-noarch:graphics-3.1-amd64:graphics-3.1-noarch:graphics-3.2-amd64:graphics-3.2-noarch:graphics-4.0-amd64:graphics-4.0-noarch:graphics-4.1-amd64:graphics-4.1-noarch:languages-3.2-amd64:languages-3.2-noarch:languages-4.0-amd64:languages-4.0-noarch:languages-4.1-amd64:languages-4.1-noarch:multimedia-3.2-amd64:multimedia-3.2-noarch:multimedia-4.0-amd64:multimedia-4.0-noarch:multimedia-4.1-amd64:multimedia-4.1-noarch:printing-3.2-amd64:printing-3.2-noarch:printing-4.0-amd64:printing-4.0-noarch:printing-4.1-amd64:printing-4.1-noarch:qt4-3.1-amd64:qt4-3.1-noarch:security-4.0-amd64:security-4.0-noarch:security-4.1-amd64:security-4.1-noarch Distributor ID: Ubuntu Description:    Ubuntu 14.04.3 LTS Release:    14.04 Codename:   trusty   Any suggestion why or how to fix it?  Thanks in advance. Can my spouse receive an emergency UK visa to visit a funeral? I'm a British national. My grandmother just passed. My wife is Chinese. In general it's a major pain to get my wife a visa to visit the UK, we have never been before.  Is there any way to get an emergency visa to attend the funeral?  What sort of documents would we need?