I have been learning java recently and is using eclipse as IDE. But , I am not able to make a .exe file with that . all I get is jar file which I know can be executed on the window machine. My question is that can I use Visual studio and create .exe program or for that matter any software for Microsoft Windows using Java ?
If I have a Java source file (*.java) or a class file (*.class), how can I convert it to a .exe file?  I also need an installer for my program.
Why can't we see light travelling from point A to B? Let's say we have a cloud of dust which is a lightyear across and someone shoots a beam of light from point A to B , why it is not possible for an observer far far away to see the light while it travels through the cloud at the speed of light? How to be granted root rights while browsing files/directories with Nautilus I'd like to know if there a way to be granted root rights while using Nautilus 2.30 ?   For example I'd like to move some old folders of long unused users from the home directory - remains of previous distros (Debian).   Of course I can open a terminal, but I want to know if it is possible to do that with a mouse in Nautilus. $ versus jQuery PHP Echo Line Breaks "Belongs on" comments 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 How to calculate the distribution of the minimum of multiple exponential variables? $X_1$, $X_2$, $X_3$ are independent random variables, each with an exponential distribution, but with means of $2.0, 5.0, 10.0$ respectively. Let $Y$= the smallest or minimum value of these three random variables. Derive and identify the distribution of $Y$. (The distribution function may be useful).    How do I solve this question? Do I plug in each mean to the exponential distribution? I would appreciate it if someone could explain this to me, thanks. Why does du and quota results not match? Subject vs. Object marking for whoever? I know similar questions have been asked before, but I'm having trouble reconciling the following sentence, received in an email:     Can we ask whomever is your contact there to email us a job so we can check backward compatibility?   I could understand using objective case if the sentence were "can we ask (him) to email...", but I get tripped up considering "whomever" to be the subject of "(he) is your contact there". How to install pygments on windows 7 When I run LaTeX file its gives error as  "! Package minted Error: You must have pygmentize installed to use this package", you must have pygments installed. when I go through minted documentation minted itself required pygments, so Tex friend How do I install this pygment, I don't know the procedure to install. Please guid me or any link pygement.exe file Make OS X Terminal commands I type BOLD I'd like to make commands I've typed (input) into terminal stand out from all the output.   For example:  imac:~ buster$ chmod -R g-w myfolder  imac:~ buster$ cd myfolder imac:myfolder buster$ ls -l total 0 drwxr-xr-x   9 root  admin   306 Apr 20  2010 bin drwxr-xr-x   7 root  admin   238 Apr 20  2010 include drwxr-xr-x  73 root  admin  2482 May 18 17:16 lib drwxr-xr-x   6 root  admin   204 Apr 20  2010 man imac:myfolder buster$ echo Go Giants! Go Giants!  bold jumps to mind but I'd accept a color or even highlighting the whole line...   I'm sure there's a way to do this but it's not obvious to me...  thanks! Block network access of a process? Is it possible to block the (outgoing) network access of a single process? How to extract BibTeX entries (as DOI, abstract, etc.) I want to automate personal reading notes generation, I'd like to be able to automatically extract a field's content from a BibTeX entry (other than year or author).  E.g. given a BibTeX entry  @article{einstein05,    author={Albert Einstein},    title={Zur Elektrodynamik bewegter K{\"o}rper},    abstract={Interesting article, which claims that time is relative.},    journal={Annalen der Physik und Chemie},    volume={17},    year={1905},    pages={891--921} }   I want to have a macro in my TeX file (e.g. \citeabstract) that will return the string Interesting article, which claims that time is relative. (the same for DOI, etc.)  I've found the \printfield command that might help me, but I'm not smart enough to know how to use it.  I don't know if it has any influence on the answer, but I'm using BibLaTeX+BibTeX. If a lightning-radio-group has two options, and a user selects one, I want to programmatically have the UI changed back to the other one.  Link to a . (code copied below)  import { LightningElement, track } from 'lwc';  export default class RadioGroupRequired extends LightningElement {     @track value = 'option1';      get options() {         return [             { label: 'Option one', value: 'option1' },             { label: 'Option two', value: 'option2' },         ];     }      handleOnChange(evt) {         this.value = 'option1';         // how to make the UI select Option one?     } }   And the html:  &lt;template&gt;     &lt;lightning-radio-group name="radioGroup"                            label="Radio Group"                            options={options}                            value={value}                            required                            type="button"                            onchange={handleOnChange}&gt;&lt;/lightning-radio-group&gt;                             value={value} &lt;/template&gt;     I think that there is no rerendering because the bind variable, {value} doesn't actually change. However, the selected state of the lightning-radio-group does, and I want to be able to change it back. How can that be achieved?   I did not notice any methods or attributes that would seem to do the trick in the . Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.) When and how did "pretty" enter English as an intensifying adverb? Today I saw an idiomatic road sign: "Pretty Muddy". I found this lack of strict English on a road sign unusual (on par with my "Dead Slow" official speed limit sign in Leeds, pic below), but as it turned out it's a charity race and the signs were merely directions.  This got me thinking however, how do words like "pretty" and "dead" as   end up in English? Taking "pretty" as an example, when did this first appear?     I wonder how fast you have to be going before they stop you. How would this hold up in court? How to combine an SSD boot drive with a HDD data drive? I have an SSD with 64 GB and a 6 GB/s 1 Terabyte hard drive. I want to use the SSD only as a boot device and maybe for swap storage. On the 1 Terabyte hard drive, I want to have the main file system. As in Windows, I liked to have at least 2-3 Partitions example for music games programs and so on. Which partition layout do you recommend to me? I want to be able to have LinkedList.contains() return true for a custom comparator.  Suppose that I have 1 LinkedList and 2 objects  LinkedList&lt;MyObject&gt; myList = new LinkedList&lt;MyObject&gt;();  MyObject a = new MyObject("HELLO"); MyObject b = new MyObject("HELLO");   Technicaly, both objects are identical in terms of comparison (MyObject implements Comparable)   ( a == b ) == true  however, when I do the following, myList does not return true for myList.contains(b)  myList.add(a) myList.contains(b) // == false   I think its because contains will check object reference and see that a and b are 2 distinct objects. Is there any way I can make it so I don't have to extend LinkedList to compare those objects? How does C compute sin() and other math functions? How do I redirect command output to a file?