If we give return statement like this in try, what will be the order of execution   try{ --- ---- ----- return a; }  catch{ } finally{ }   Here what will be order of execution if there is return in try. Please let me know
Considering this code, can I be absolutely sure that the finally block always executes, no matter what something() is?  try {       something();       return success;   }   catch (Exception e) {        return failure;   }   finally {       System.out.println("I don't know if this will get printed out"); }
Many times I see that URL extensions are hidden (e.g. .html, .php), but not all websites do this.  Why do webmasters hide the extension? Is it for security, for the URL to look cleaner, or for some other purpose? Sum of points and rank according to points 3D arrows with TikZ Asymptote does give 3D arrows when drawing three dimensional figures (they even have shade!). Could we fake it with TikZ? Here's an example, manually written, with just one arrow worked:  \documentclass{scrartcl} \usepackage{tikz} \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[-&gt;,scale=2]   \draw[&gt;={Triangle[width=3pt,length=5pt]}] (0,0,0) -- (1,0,0);   \draw[&gt;={Triangle[width=3pt,length=5pt]}] (0,0,0) -- (0,1,0);   \draw[&gt;={Circle[sep=-.9pt,width=3.08pt,length=2pt]Triangle[width=3pt,length=5pt]}] (0,0,0) -- (0,0,1); \end{tikzpicture} \end{document}     But that's (wrongly) manually calculated by trial and error (apart from the fact that it's not perfect, seeing how the other two axis —x and y— are oriented, it should have certain slant). It would be nice to have an interface like the other arrows  &gt;={3D[length=⟨real length⟩,width=⟨real width of the base⟩]}   Which would then calculate from those values (and, of course, also taking into account x=⟨x vect⟩,y=⟨y vect⟩,z=⟨z vect⟩)  Circle[sep=⟨value⟩,width=⟨value⟩,length=⟨value⟩,slant=⟨value⟩] Triangle[width=⟨value⟩,length=⟨value⟩] % And I'm not sure if `slant=⟨value⟩`                                        % is necessary in this Triangle case   Example  Using the code from the answer (thanks to Symbol1), for instance, this code (taking the idea from another question, link in the comments)  \begin{tikzpicture}[x={(0cm,1cm)},                     y={({cos(30)*1cm},{sin(30)*-1cm})},                     z={({cos(20)*1cm},{sin(20)*1cm})},                     thick,                     scale=3]   \pgfmathsetmacro{\ax}{.4}   \pgfmathsetmacro{\ay}{.6}   \pgfmathsetmacro{\az}{-.6}   \pgfmathsetmacro{\bx}{.4}   \pgfmathsetmacro{\by}{-.7}   \pgfmathsetmacro{\bz}{-.8}   \draw[densely dotted,thin] (\ax,\ay,\az) -- (\ax,0,\az);   \draw[densely dotted,thin] (\ax,\ay,\az) -- (0,\ay,\az);   \draw[densely dotted,thin] (\ax,0,0) -- (\ax,0,\az) -- (0,0,\az);   \draw[densely dotted,thin] (0,0,\az) -- (0,\ay,\az) -- (0,\ay,0);   \draw[densely dotted,thin] (\bx,\by,\bz) -- (\bx,0,\bz);   \draw[densely dotted,thin] (\bx,\by,\bz) -- (0,\by,\bz);   \draw[densely dotted,thin] (\bx,0,0) -- (\bx,0,\bz) -- (0,0,\bz);   \draw[densely dotted,thin] (0,0,\bz) -- (0,\by,\bz) -- (0,\by,0);   \draw[-{Cone3}] (0,0,0) -- (1,0,0) node[right] {$x$};   \draw[-{Cone3}] (0,0,0) -- (0,1,0) node[above right] {$y$};   \draw[-{Cone3}] (0,0,0) -- (0,0,1) node[below right] {$z$};   \draw[-{Cone3}] (0,0,0) -- (0,0,-1) node[above left] {$-z$};   \draw[thin] (0,0,0) -- (0,-.9,0);   \draw[-{Cone3}] (0,0,0) -- (\ax,\ay,\az) node[below left] {source};   \draw[-{Cone3}] (0,0,0) -- (\bx,\by,\bz) node[above right] {sim}; \end{tikzpicture}   gives this figure When the phrase is used as an object, why so many native speakers are saying "you and I" instead of "you and me"? I'm not a native speaker but I thought "you and me" is correct. Not sure if this falls into the same category, but "Just between you and me" sounds more natural than "Just between you and I". Prove that negative absolute temperatures are actually hotter than positive absolute temperatures Could someone provide me with a mathematical proof of why, a system with an absolute  (such that of a spin system) is  than any system with a positive temperature (in the sense that if a negative-temperature system and a positive-temperature system come in contact, heat will flow from the negative- to the positive-temperature system). What exactly happens to the signals hitting a common mode choke? I'm trying to better understand the principles behind the common mode choke. I made a few drawings to clarify.  &nbsp; Differential Mode Signals  Differential currents (driven by differential voltages) create equal but opposite magnetic fields B in the inductor core:    These magnetic fields cancel each other out, so the net flux in the core is zero. As such, these differential currents don't "feel" any impedance.  &nbsp; Common Mode Signals  In contrast, common mode currents generate equal and additive magnetic fields in the core. That's why they "feel" a high impedance, and cannot get through (or getting through means they are highly attenuated).    But what exactly happens? I have several theories, which I will describe below.  &nbsp; Common Mode Signals - Theory 1  My first thought would be that the common mode signal hits the choke and creates a magnetic flux inside. By doing this, lots of energy is lost (hysteresis and perhaps other effects) as heat. Only a small part gets through:      What kind of common mode choke would behave in this particular way? "Burning up" the voltage spike seems a very desirable effect to me.  &nbsp; Common Mode Signals - Theory 2  Perhaps the voltage spike doesn't really get the chance to build up much magnetic flux in the core, or maybe the core is simply not "lossy" enough. The voltage spike bounces off the core and turns back. Only small part gets through:    Although the system on the right side of the choke is protected, the system on the left has to deal with reflected signals. Nasty things like standing waves might appear.  &nbsp; My questions  I've got a few questions for you:   Do you think theory 1 or theory 2 is most plausible? Do you think certain types of common mode chokes tend to behave as described in theory 1, others like in theory 2? Perhaps both of my theories are just plain wrong. If so, what actually does happen?   Please enlighten me. How do I verify a PGP signature? What GUI (no command line) software or websites can I use to verify a PGP signature?  If I have a message like this  -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1  Because anyone can claim to be me. There's no validation of the user name or email address when someone posts a comment. While I do try to remove imposters, some may slip through. By signing my comments using this technique, anyone can independently verify that I was the author of the message by validating the signature. -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.6 (MingW32)  iD8DBQFFxqRFCMEe9B/8oqERAqA2AJ91Tx4RziVzY4eR4Ms4MFsKAMqOoQCgg7y6 e5AJIRuLUIUikjNWQIW63QE= =aAhr -----END PGP SIGNATURE-----   How can I verify the message against the public key to get the same signature back?  This  does not allow me to do that. Why is the node editor showing different kinds of nodes? I want to show without using Sylow theorem that     Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)   My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$ Then $H=\{e,a\}\leq G.$ Let $S$ be the set of all left cosets of $H$ in $G.$ Now $S$ forms a group w.r.t. the composition $g_1Hg_2H=(g_1g_2)H.$ The mapping $\sigma:S\to G:gH\mapsto g$ is an $1-1$ homomorphism. Then $S\simeq\sigma(S)\leq G.$ Thus $G$ has a subgroup of order $p.$  Please tell me whether I'm right!     I'm skecptical as I didn't use that $p$ is prime. Can you reuse marinade used with raw chicken? Does $A$ a UFD imply that  $A[T]$ is also a UFD? I'm trying to prove that $A$ a UFD implies that $A[T]$ is a UFD.   The only thing I am sure I could try to use is Gauss's lemma.  Also, how can we deduce that the polynomial rings $\mathbb{Z}[x_1,\ldots,x_n]$ and $k[x_1,\ldots,x_n]$ are UFDs? I don't understand why su - is preferred over su to login as root. Photoelectric effect intensity I understand the PE effect quite well but I'm failing to understand one thing. Intensity is the amount of energy per second incident to a given area.   So can you can increase the intensity by either increasing the electrical energy or making the area smaller? By increasing the electrical energy, what exactly is happening?  I assume that by increasing intensity you are increasing the number of photons incident to an area of the metal thus increasing the rate of photoelectric emission. But why doesn't increasing intensity via more electrical power lead to greater energy of the electrons, if by increasing intensity you are increasing the energy of the incident waves?  I understand the results but I'm trying to understand why the intensity works the way it is thanks How to make a bootable ISO image from a bootable DVD? I want to make a .iso image of a bootable OS DVD, to use in VirutalBox.    Nautilus can make a .iso, but only of data files, it seems. The .iso it created was not bootable.  What is available to make a bootable .iso from a bootable DVD? Automotive electronic circuit design, Voltage divider to uC ADC pin Automotive Circuit requirement:   If input pin is connected to Battery(T87 line), the output of the circuit shall provide a output voltage of 5V to ADC pin of microcontroller, if the input pin is connected to GND, the output of the circuit shall provide a output voltage of 0V to ADC pin of microcontroller and if the input pin is left floating, the output of the circuit shall provide a output voltage of 2.5V to ADC pin of microcontroller.   The battery voltage in load dump condition could be up to 60V, but in normal operation without any fault condition, it would be up to 32V.  Would a simple voltage divider (with a pull-up to 5V) with a blocking diode be enough? Any other ideas? What parameters of the ADC of the microcontroller should be considered? Injection current into the uC- ADC pin would be critical, right? How do I find out the voltage thresholds? Also, what are the parameters of the Schottky diode that I should look to select? The diode would decouple the input pin from the centre tap of the voltage divider, because in case the input pin is connected to battery, the injection current into the uC ADC pin would be too high.  I have used a switching diode, since it was already available in the Bill of Materials. I have recently noticed a huge (>3.5GB) file in the DCIM/.thumbnails folder. I have tried deleting it, but the next time I open the Camera app it rebuilds the file (and locks up the phone, sometimes displaying a "media scanning in progress" message, in the process.)  The total file size of all the (approximately 2,000) photos displayed in the Gallery is around 500MB. There are in addition around 35,000 images in folders which contain .nomedia files to instruct Android to ignore the media therein. The total file size of these is around 1.5GB. The Gallery app is correctly ignoring these images, but I wonder whether the Camera app is misbehaving and actually processing them.  I think this problem has emerged since the ICS upgrade ... either when the phone (Samsung Galaxy Note) was first updated from Gingerbread or at some later date.  Any ideas, please? Proving that a Hölder space is a Banach space I am trying to show that the Hölder space $C^{k,\gamma}(\bar{U})$ is a Banach space. To do this, I successfully proved that the mapping $\| \quad \| : C^{k,\gamma}(\bar{U}) \to [0,\infty)$ is a norm, by proving its properties.   But how do I show that the sequence $\{u_k\}_{k=1}^\infty \subset C^{k,\gamma}(\bar{U})$ converges to $u \in C^{k,\gamma}(\bar{U})$, that is, how do I show that $$\lim_{k \to \infty}\|u_k-u\|=0,$$ which would mean the normed linear space is complete, and hence a Banach space?  Here are the following taken from PDE Evans, 2nd edition, page 255:     Definition. The Hölder space $$C^{k,\gamma}(\bar{U})$$ consists of all functions $u \in C^k(\bar{U})$ for which the norm $$\|u\|_{C^{k,y}(\bar{U})}:= \sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})}$$ is finite.   Also from page 254,     Definitions. (i) If $u : U \to \mathbb{R}$ is bounded and continuous, we write $$\|u\|_{C(\bar{U})}:=\sup_{x\in U}|u(x)|.$$      (ii) The $\gamma$th-Hölder seminorm of $u : U \to \mathbb{R}$ is $$[u]_{C^{0,\gamma}(\bar{U})}:=\sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|u(x)-u(y)|}{|x-y|^\gamma} \right\},$$ and the $\gamma$th-Hölder norm is $$\|u\|_{C^{0,\gamma}(\bar{U})}:=\|u\|_{C(\bar{U})}+[u]_{C^{0,\gamma}(\bar{U})}.$$   This is all I got so far: \begin{align} \|u_k-u\|_{C^{k,\gamma}(\bar{U})}&amp;=\sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})} \\ &amp;= \sum_{|\alpha|\le k} \sup_{x\in U} |u_k(x)-u(x)|+ \sum_{|\alpha|=k} \sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|[u_k(x)-u(x)]-[u_k(y)-u(y)]|}{|x-y|^\gamma} \right\}. \end{align} Now, where can I go from here, to show that $\lim_{k \to \infty}\|u_k-u\|_{C^{k,\gamma}(\bar{U})}=0$? The sequence is Cauchy, and I have to use that fact somehow. Last activity name still missing from migrated question Would you recommend using a  or a  field, and why (using MySQL)?   I'm working with PHP on the server side. Studying the equation $\ln(x)- 1+ x^2=0$