When calculating the size of a float array it prints a size of 2 instead of 41 in C
How to find the 'sizeof' (a pointer pointing to an array)?
Where should the period be put when an entire sentence is quoted at the end of a sentence? As in the following:     A common proverb is: “A rolling stone gathers no moss.”        A common proverb is: “A rolling stone gathers no moss”.        A common proverb is: “A rolling stone gathers no moss.”.   If the proverb weren't a full sentence, I would immediately put a single period after the quote, at the very end.  But if it is a sentence, which period should be removed so that there is only one period? Two periods look absolutely dorky. What are character special and block special files in a unix system? Can't execute .out files, getting permission denied I have written a C++ program and complied it to produce a.out file. However, whenever I try to run it, I get Permission Denied. I read that we can use sudo, but I can't quite get it to work. I use something like, sudo "./a.out" but that too doesn't work.  Edit:  Here is the message I get when I try "./a.out".  bash: ./a.out: Permission denied A short story I read long ago about the phone system 'waking up' Let $A$ be a closed and bounded subset of $\mathbb{R}^{n}$. Given that $f: A \to A$ satisfies $d(f(x),f(y)) &lt; d(x,y)$ for all $x,y \in A$, $x\neq y$ where $d$ is the usual Euclidean metric. Show that $f$ has a unique fixed point.    My observations:    Since $\mathbb{R^{n}}$ is complete, then $(A,d)$ is complete since $A$ is closed; Since $A$ is bounded, then $A \subset B_{R}$ for some $R &gt; 0$; Banach Fixed Point Theorem guarantees a unique fixed point if $f$ is contraction on $A$.   Please correct me if I am wrong. My quandry here is showing that $f$ is a contraction, i.e.  $$ d(f(x),f(y)) \leqslant k\cdot d(x,y), \hspace{3mm} k \in (0,1) $$  I'm looking for some hints or tips. Thank you. Is there a direct, elementary proof of $n = \sum_{k|n} \phi(k)$? How can I clear event subscriptions in C#? Take the following C# class:  c1 {  event EventHandler someEvent; }   If there are a lot of subscriptions to c1's someEvent event and I want to clear them all, what is the best way to achieve this? Also consider that subscriptions to this event could be/are lambdas/anonymous delegates.  Currently my solution is to add a ResetSubscriptions() method to c1 that sets someEvent to null. I don't know if this has any unseen consequences. I was wondering what is the purpose of piping the downloaded file by curl into sudo -E bash - in the following shell:   curl -sL https://deb.nodesource.com/setup_5.x | sudo -E bash - Hide processes from other users based on groups (under Linux)? Is it possible to configure process hiding for certain user groups under a linux system?  For example: Users from group X should not see processes owned by users from group Y in ps/top or under /proc.  Is it possible to configure such a setup with SELinux?  (I vaguely remember a similar feature in the funny grsecurity patch set - but IIRC, it was more generic - and besides, I want to configure a stock linux distro without having to maintain a custom kernel.)  Edit: For better illustration, . The example is not that generic, but one can configure that a user or some users can only see information of their own processes in ps etc. Log of products to sum of logs Can I access constants in settings.py from templates in Django? I have some stuff in settings.py that I'd like to be able to access from a template, but I can't figure out how to do it. I already tried  {{CONSTANT_NAME}}   but that doesn't seem to work. Is this possible? In which scenario do I use a particular STL container? I've been reading up on STL containers in my book on C++, specifically the section on the STL and its containers. Now I do understand each and every one of them have their own specific properties, and I'm close to memorizing all of them... But what I do not yet grasp is in which scenario each of them is used.  What is the explanation? Example code is much prefered. What is the alternative for Teamviewer? Is there any Ubuntu alternative for team-viewer software for Windows? Inner product of two continuous maps is continuous Let f,g be continuous maps of S in $\mathbb{R}^n$, to $\mathbb{R}^m$. Show the inner product h(x) = $\langle f(x), g(x)\rangle$ is continuous.  My attempt: Since f,g are continuous, their product is a continuous mapping, so for $\epsilon \gt 0\: t.e\:  \delta \gt 0$ such that $\parallel x - a\parallel \lt \delta $ implies $\parallel \langle f(x),g(x)\rangle - \langle f(a)g(a)\rangle \parallel \le\parallel f(x)g(x) - f(a)g(a)\parallel \lt \epsilon$ by schwartz inequality. Do gravitational lenses work on gravitational waves? Do gravitational lenses work on gravitational waves? Could we get an Einstein cross of gravity? Differentiation - simple case I'm planning on selling a USB external hard drive that currently contains an old Ubuntu installation with stored passwords and banking information.  How can I securely erase the drive before selling it? Show that $u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$ is in $W^{1,n}(U)$, where $U=B(0,1)\subset\mathbb{R}^n$. The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback! I know it is so weird, but last night I set an image on my desktop background and delete the image, then I right-click on the desktop and press the desktop properties to change the style of background (center, stretch and etc) then the desktop properties stopped something like hanging on Windows.   Then a black window alerted something and I press yes without reading it. after that with right-click on the desktop I just see this:    Even some system applications don't work now. for example, I cannot take a screenshot.  Also, my pcmanfm-qt is dead completely, with the command line or clicking on the icon nothing happens.  I don't know what happens. and I have no idea how to fix it. How to convert every pixel of raster picture to square objects in Illustrator?