Could someone show how it is possible to check whether another instance of the program (e.g. test.exe) is running and if so stop the application from loading if there is an existing instance of it.
Using C# and WPF under .NET (rather than  or console), what is the correct way to create an application that can only be run as a single instance?  I know it has something to do with some mythical thing called a mutex, rarely can I find someone that bothers to stop and explain what one of these are.  The code needs to also inform the already-running instance that the user tried to start a second one, and maybe also pass any command-line arguments if any existed.
Error message "sudo: unable to resolve host (none)" When I run sudo the terminal is stuck for a few seconds and then outputs an error message. My terminal looks like this:  ubuntu@(none):~$ sudo true sudo: unable to resolve host (none)   What can I do to solve it? How much effect does the Bernoulli effect have on lift? I understand that the Bernoulli effect is a flawed explanation for the cause of lift, and does not cause much at all, but how much?   Is there any experimental data on the force caused by the Bernoulli effect? Maybe implicitly through data of the pressure difference between the top and underside of an aeroplane's wings. After that, I assume I could (crudely approximating the pressure to be acting perpendicularly to the flight direction) use $\Delta P A$ to work out the net force on the plane.    Perhaps there is another way to quantitatively analyse the extent to which the Bernoulli effect causes lift.  Edit: see  short cartoon (content similar to Mike Dunlavey's answer). Why doesn't the Related question list show that questions are closed? I saw a question on Meta that was obviously going to be a duplicate:        And indeed, looking at the "Related" questions list for that question showed a number of duplicates:    Being the good citizen I am, I tried to vote to close (technically Flag as duplicate as I've not got the rep here to vote to close) it as an Exact Duplicate only to find myself falling into a rabbit hole of Closed, Duplicate questions. Is there any way to flag a closed question as such in the Related list?  Mostly because the first one I picked was closed as a duplicate of the second one, which was closed as a duplicate of the one that is now listed as the duplicate/linked answer.   Edit to respond to comment:  I take the point about "not voting to close on title alone", and normally wouldn't, however as the questions had identical titles, and it's pretty clear what they should be asking, I clicked the first one:    which lead to:  which leads to:  - but I'd got bored following links and didn't check that one as the first two were identical, so assumed this was too and didn't want to see how deep the rabbit hole led.   In the related list, there's only one other question in there that is an exact duplicate, but hasn't (currently) been closed as such which is:      Reasoning for highlighting closed questions:  The "Questions with similar titles" does display the closed titles when asking the question, so it should be fairly possible to include.  If there are clearly duplicate titles, then as a reader/reviewer would be helpful to know which ones are already closed as duplicates so you can either:   Use the remaining open questions to focus on which of those is the canonical answer to close this with. Use the closed questions to find the canonical answer (which failed for me with my route of clicks). Should I buy a Canon 55-250mm lens or 70-300mm for my Rebel T3i? I have the basic 18-55mm lens that came with my camera but I need a next step up that's not too expensive.   What are the most common types of photos shot with either lenses? Or what is each lens best for? Prime spectrums bijection but not homeomorphism What is the difference between Ubuntu and its derivatives? A function $f : I \rightarrow \mathbb R$ on an interval $I$ is convex   if $f((1-t)x+ty)\le (1-t)f(x)+tf(y), \forall x,y \in I, t \in [0,1]$      Assume now that $I$ is an open interval.  Show that if $f$ is convex   then for each $c \in I$ there exists $m \in \mathbb R$ such that $m(x  − c) + f(c) \le f(x)$ for all $x \in I$ , and if in addition $f$ is   differentiable at $c$ then $f'(c)$ is the unique $m$ that works. In   general, must m be unique?   (we have previously shown that convex functions are continuous)  I think I'm missing the point of the question.  Surely, by the mean value theorem we can find some $y\in (x,c)$ such that $f'(y)=\frac{f(x)-f(c)}{x-c}$ setting $m=f'(y)$ we get $m=\frac{f(x)-f(c)}{x-c}, m(x-c)+f(c)=f(x)$ such an $m$ fits the inequality we're asked to show.  I also cannot see why $f'(c)$ would be the unique solution or why we would ever get a unique solution of the inequality at all for that matter.  Clearly I'm misunderstanding the question, so if someone could point out where I would appreciate that.  Thank you As per the title, what hardware do I need to be able to run various versions of Android? Line integration in complex analysis bash is not terminating I'm trying some things with my shell. This includes the following command:  kill $$  My expectation was that the shell would log me out and terminate itself. But instead the bash persists without change.   Any idea why?  I ran this command on different environments (archlinux, debian). The behavior was always the same. How to send AT commands to modem once the connection is established? I have a zte 3g modem. I use carrier provided dialer for connection establishment. Once the ppp connection is active, i would like to send some AT commands(for ex. Query signal strength, AT+CSQ). But the dialer i use locks the /dev/ttyUSB0 port, which is the command port to send AT commands for my modem. So is there any way, to send the commands, once the connection is active?    Edit: I also tried the additional port /dev/ttyUSB1. But the port is flowing with random data from the modem. A sample is given below.  T^PREFMODE??                                                                     ^PREFMODE:8                                                                       OK                                                                               TC                                                                               ^DSDORMANT:1                                                                      +CSQ:19, 99                                                                       OK                                                                               T^SYSINFO                                                                        ^SYSINFO:2,3,0,4,255                                                              OK                                                                               TT^SYSINFO?                                                                      ^SYSINFO:2,3,0,4,255     I tried adding my commands, i even got output. But the response is very poor. Most of the times, my AT commands went unnoticed. Can I nohup/screen an already-started process? I'm doing some test-runs of long-running data migration scripts, over SSH. Let's say I start running a script around 4 PM; now, 6 PM rolls around, and I'm cursing myself for not doing this all in screen.  Is there any way to "retroactively" nohup a process, or do I need to leave my computer online all night? If it's not possible to attach screen to/nohup a process I've already started, then why? Something to do with how parent/child proceses interact? (I won't accept a "no" answer that doesn't at least address the question of 'why' -- sorry ;) ) The reCAPTCHA has become too difficult for humans to solve The reCAPTCHAs have suddenly become almost illegible. I regularly have to change it four or five times before I find one I can read. I am tempted to just enter the known word (an easy guess), and leave the other. Can something be done? Indefinite articles used with plural nouns: It was AN amazing TWO DAYS I'm a committer for the .  It went to the private beta phase sometime in the last 12 hours, but yet to have been invited. When I visit  I get the standard "To log in, you must have commited to the Area 51 site proposal and received the invitation email." message.  However, when I navigate to the Pets site via the Android app, I'm able to view the site in it's entirety (questions, answers and comments). When I attempt to post, I get a message saying my user token isn't allowed to take action on this site.  Same behavior is observed when using the iOS app. UK visa: unsigned application form Can I use Named Credential merge fields as URL parameters on the endpoint? I have a named credential as follows:  Name   Test_Credential URL    https://www.somesite.com Identity Type    Named Principal Authentication Protocol    Password Authentication Username   foobar Password   mySomeSiteDeveloperKey Generate Authorization Header   FALSE Allow Merge Fields in HTTP Header    TRUE Allow Merge Fields in HTTP Body     TRUE   The service that I am trying to authenticate against uses URL parameters to pass in the developer key and username. Is there any way to use  to pass those in to the URL?   The Named Credential attributes seem not to get decoded until the request is actually sent, as debugging the endpoint, or headers if I set them per the linked doc, show the un-merged data. I can get the callout to go to the right place, and return the expected error because the auth information is not provided. Anybody know of a "show me my HTTP request" site that I could point to to get a better idea of how this works?  I've tried both of these:  req.setEndpoint('callout:Test_Credential/api/action?id={!$Credential.Password}&amp;account={!$Credential.UserName}');  req.setEndpoint('callout:Test_Credential/api/action' + '?id=' + '{!$Credential.Password}' + '&amp;account=' + '{!$Credential.UserName}');   (Edited to remove the part about $Credential vs. $Test_Credential as I was able to determine that it is the former, and see the merged creds by calling out to .) Why does `ls -d *[!e]*` display all files instead of omitting all files that contain an e? ls -d *e* prints out all files containing an "e" somewhere in it. But the opposite, ls -d *[!e]* does not seem to work. Why exactly is that?  is it because ls takes * as its own argument and expands all the files?  I understand I can do: ls -d !(*e*) (in Bash with extglob set, or in ksh). Why do the Localization of a Ring This question may be a bit vague, but neverthless, i would like to see an answer. Wikipedia tells me that:   In abstract algebra, localization is a systematic method of adding multiplicative inverses to a ring.   It is clear that for integral domains, we have the Field of Fractions, and we work on it. It's obvious that a Field has multiplicative inverses. Now if we consider any arbitrary ring $R$, by the definition of localization, it means that we are adding multiplicative inverses to $R$ thereby wanting $R$ to be a division ring or a field. My question, why is it so important to look at this concept of Localization. Or how would the theory look like if we never had the concept of Localization. Is there a standard way to do an fopen with a unicode string file path?