(Newbie to Java, old time C# guy.)  I have noticed a lot of the use of @Override in Android example code. I thought that all Java methods were by default "Virtual"?  What then does @Override do?  Example:   private class HelloWebViewClient extends WebViewClient {      @Override     public boolean shouldOverrideUrlLoading(WebView view, String url) {         view.loadUrl(url);         return true;     } }
What are the best practices for using Java's @Override annotation and why?     It seems like it would be overkill to mark every single overridden method with the @Override annotation.   Are there certain programming situations that call for using the @Override and others that should never use the @Override?
Can Raise Dead revive a character with zero maximum Hit Points? So, a character died from Draining Kiss of a succubus and his Max HP is 0. Can he be restored by Raise Dead, or does he need a higher level spell? Or a Greater Restoration cast on his lifeless body before Raise dead? I'm probably interested in RAW interpretation, since the character is in Adventurerer's league. Protected Question does not show Add an Answer button This question is protected. To answer it, I must have at least 10 reputation.        Ignoring the association bonus of 100, I should .  Yet, the Android app does not show the blue "Add an Answer" button at the bottom. The app also does not show the protection status.  If I view the question in the desktop or mobile browser (and I'm signed in), I can add an answer there. Finn knows nothing about his family because he was taken while he was very little.   But a black main character makes me think (not being racist guys). And when he used the lightsaber I thought that he had a Jedi parent. Master Windu is the only black Jedi we know so is there any canon explanation to that? Is it right? Divisor in $\mathbb{C}[X]$ $\implies$ divisor in $\mathbb{R}[X]$? let $P \in \mathbb{R}[X]$ be a real polynomial divisible by a polynomial $Q \in \mathbb{R}[X]$ in $\mathbb{C}[X]$. How can I easily show that $P$ is also divisible by $Q$ in $\mathbb{R}[X]$?  A simple argument without using higher algebraic theorems is desirable. If I could use instruments of higher algebra, the exercise I have to do in whole would be done in two lines. But I'm not allowed to use. I think there would be an easy argument which I can't see yet because of my mental fogginess that I have sometimes.  Thank you beforehand. Extrude and bevel an imported svg curve I exported an svg 1.1 file from Adobe Illustrator CS6. I joined all the shapes by pressing Ctrl+J. I set the curve shape to 2D. Then I extruded using the right hand side panel:    So far so good. But when I try to apply even a slight bevel, all hell breaks loose:    It does get less insane if I bring the bevel down by multiple factors of 10... But then I really don't have much of a bevel.  Am I doing something wrong? Am I using the wrong kind of object? Should I be doing this with meshes? Do my curves just need cleaning up? $t^3$ is never equal to Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is never equal to: (A). $4t+3$ (B). $8t+5$ (C). $10t+3$ (D). $6t+5$  We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I don't see how to continue since this is a cubic equation. Is there an easier way of finding a contradiction? remove files when name does NOT contain some words Hypernym for “import” and “export”? AppFabric / WF 4.0 instances stuck despite no bookmark and pending timer Jack probs with heaphones/earbuds So I already asked this question, but I couldn't respond to the answers. I can't hear out of the left ear of any headphones/earbuds because I have to push the jack towards the Earth to hear out of it. It seems you guys think the jack is the problem when I'm pretty sure it's the computer (my Acer laptop). I also forgot to mention the headphones (PX22 Turtle Beaches) work with both ears, but I have to plug in the usb extension cord in order for that to work (it's also the same cord that enables the mic on my headphones), plus I lost the cord. Thanks 4 hewps. What is involved in converting a PC video card to have boot screen on the Mac? I am working on the following problem from group theory:     If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.   That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$. Occasionally, I get part of an application lingering behind on my desktop:    This is part of a context menu for an application. It is always top-most, meaning that it is always on the screen, and I cannot click on it (it just clicks through it). It has happened before, sometimes with whole windows. It remains on screen even if the application it originated from is no longer running.  How do I get rid of it without rebooting? Any tools to generate an XSD schema from an XML instance document? I am looking for a tool which will take an XML instance document and output a corresponding XSD schema.  I certainly recognize that the generated XSD schema will be limited when compared to creating a schema by hand (it probably won't handle optional or repeating elements, or data constraints), but it could at least serve as a quick starting point. Looking for qt-at-spi for ubuntu 20.04 I'm trying to prove the problem below - which comes from Munkres' "Analysis on Manifolds" book in the section on the Inverse function theorem.  Since its in the chapter on the Inverse Function Theorem I figured I'd start by showing that $f$ satisfies the conditions of the theorem.  Writing the Jacobian shows that it's both $C^r$ and we get $\det f'(x,y)=4x^2+4y^2\neq0$ when $x&gt;0$.  So we can apply the theorem but I'm unsure of how to proceed to show that $f$ is $1$-$1$.  And I didn't see how to use the hint they provided.  Thanks!      Let $f\colon \mathbf R^2\to \mathbf R^2$ be defined by the equation   $$f(x,y)=(x^2-y^2,2xy).$$   (a) Show that $f$ is one-to-one on the set of all $(x,y)$ with $x&gt;0$. [Hint: If $f(x,y)=f(a,b)$, then $\|f(x,y)\|=\|f(a,b)\|$.] Mapping ManyToMany with composite Primary key and Annotation: Why do statisticians like "$n-1$" instead of "$n$"? Easy curves in TikZ I find myself needing to draw lots of elegantly curved paths in TikZ. Ideally, I'd just specify a series of points, and TikZ would calculate the extra data itself to draw a nice series of curves passing smoothly through these points, perhaps with an optional "looseness" parameter that I could specify. But the only way I can find to draw nice curves is by explicitly giving control points, or by manually specifying in and out angles.  I can think up a simple algorithm to do this, which would certainly be within TikZ's power to perform: just choose the in and out angles in a simple fashion based on the relative angles between each adjacent pair of line segments.  Is something like this already built-in? Or can someone cook something up that does the job?  Edit: Jake has provided an answer using the plot [smooth] functionality. This is almost perfect! But it can't do what I need, because it doesn't let me specify tangent angles manually where needed, which is especially important at the beginning and end of the curve. I would have thought this would be a natural and straightforward addition to the existing plot [smooth] algorithm: for every coordinate, an optional angle should be able to be specified as an argument, which if supplied is treated as the tangent angle for the curve at that point. And while we're at it, it wouldn't hurt to also allow the tension to be modified along the path.  A minimal extension to the algorithm would just accept two optional parameters, for the curve tangent at the beginning and end. How to set up beamer to grey out past items but hide future items? I know about beamer's  feature \setbeamercovered{invisible} and \setbeamercovered{transparent} to hide or grey out upcoming item's . But is there some way to have past items greyed (transparent) and future items hidden (invisible), so that only the current item is black/highlighted? And the best would be if on or 1 slide after the last all the item's could be visible highlighted. Something like this:   [grey] past 1 [grey] past 2 current [hidden] future 1 [hidden] future 2 [hidden] future 3   Is there any easy way of doing this?