Macro to shorten a struct member reference in C
is it possible to redefine { and } in C?
what is the relationship between the dynamical casimir effect and virtual particles? Some creatures have Magic Resistance:     The [...] has advantage on saving throws against spells and other magical effects.   We know from  that this works against Turn Undead, but does it work against Stunning Strike, Death Strike, Fey Presence, etc? How do we decide? Remove spaces from std::string in C++ What is the preferred way to remove spaces from a string in C++? I could loop through all the characters and build a new string, but is there a better way? I gave a scammer my telephone number, address, and name. How bad is it? Updated question because old one was embarassing: I saw a spam email from "PayPal" and I didn't realize that it was a phishing attempt. I gave the phishing site my name, telephone number, and email. What could he do with them?  Old question: Well, I saw a spam email, but I didn't realize that it's really a scam at first. I gave him my telephone number, address, and name. What would he do with them? Could he ruin ALL of my reputation with it? Did you know that :  Map&lt;Object,Object&gt; m1 = new HashMap&lt;Object, Object&gt;(); Map&lt;Object,Object&gt; m2 = new HashMap&lt;Object, Object&gt;(); System.out.println("m1.equals(m2) = "+m1.equals(m2)); System.out.println("m1.keySet().equals(m2.keySet()) = "             +m1.keySet().equals(m2.keySet())); System.out.println("m1.entrySet().equals(m2.entrySet()) = "             +m1.entrySet().equals(m2.entrySet())); System.out.println("m1.values().equals(m2.values()) = "             +m1.values().equals(m2.values()));   would output :  m1.equals(m2) = true m1.keySet().equals(m2.keySet()) = true m1.entrySet().equals(m2.entrySet()) = true m1.values().equals(m2.values()) = false   This is caused by the fact that AbstractCollection (which HashMap$Values inherits from) does not overrides #equals().  Do you have an idea why this is so ? For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$ For a polynomial $p(z)=a_0+\ldots+a_nz^n$, prove there exist $R&gt;0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$.  I get $|p(z)|\geq |a_n|R^n$, which makes that factor of $1/2$ useless. So I must have done something wrong.  $$|\frac{\partial^n{f}}{\partial z^n}(P)|\leq \frac{\sup_{z\in \overline{D}(P,r)} |f(z)|k!}{r^k}$$. This is the Cauchy estimate.   $$|\frac{\partial^n{p}}{\partial z^n}(z)| = |a_n n!|$$ $$|\frac{\partial^n{p}}{\partial z^n}(0)|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$|a_n n!|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$R^n |a_n|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|$$  There exist a $z_0$, such that $|z_0|&gt;0$ and $$\sup_{\zeta\in \overline{D}(0,|z_0|)} |p(\zeta)| = |p(z_0)|$$.  This is saying there exist a disk centered at the origin, such that $|p|$ is maximum on some point on the boundary.   Let $R=|z_0|$, we have $R^n |a_n|\leq |p(z_0)|$  Edit: I know what I got wrong. $z_0$ is just one point on the circle $|z| = R$. One can chose another point $z_1$ on the circle, such that $p(z_1)&lt;p(z_0)$. UIDatePicker show only month and day I'd like to have a UIDatePicker where the user can pick a month and a day but not the year. I'm aware that leap years have an extra day, so for simplicity, let's throw that day away.  Is there any way to remove the year column or have 2 reels with month/day that act like the UIDatePicker (grey out days that don't exist in the selected month)? Stop all active ajax requests in jQuery Proof on showing if F(x,y,z)=0 then product of partial derivatives (evaluated at an assigned coordinate) is -1 The task is as follows:     Given: $$F(x,y,z) = 0$$    Goal:  Show    $\frac{\partial  z}{\partial y}|_x \frac{\partial  y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$   Here is my work so far:  (1) Differentiate with respect to y, I get:  $0 + F_2 + F_3  \frac{\partial z}{\partial y} = 0$  So $ F_3  \frac{\partial z}{\partial y} = - F_2$  (2) Differentiate with respect to x, I get:  $F_1 + F_2 \frac{\partial y}{\partial x}  + 0 = 0$  So $ F_2  \frac{\partial y}{\partial x} = - F_1$  (3) Differentiate with respect to z, I get:  $F_1  \frac{\partial x}{\partial z}    + 0  + F_3 = 0$  So $ F_1  \frac{\partial x}{\partial z} = - F_3$  (4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z}  = -1$, so when evaluated with x, z, y respectively, conclusion is still true  My questions are:  (1) Is my proof correct?  (2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.    I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing.  Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too.  Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y.  But is my thought ok?  Would someone please help me on this question? Thank you very much ^_^ Problem with package lineno I'm using package lineno for proofreading of my documents.  So I usually have something like the following MWE:  \documentclass{article}  \usepackage{blindtext}  \usepackage{ifthen}           % to use if then else \newboolean{korrektur}        % for proofreading: true/false  \setboolean{korrektur}{true} % choose one, comment the other %\setboolean{korrektur}{false}  % What we need for proofreading \ifthenelse{\boolean{korrektur}}% proofreading wanted? {% YES:   \usepackage{showframe}          % shows type area   \usepackage{lineno}             % numbers lines   \pagewiselinenumbers            % lines per page }% {% NO: % error if lineno was used before:   \typeout{***** Please delete .aux files if you get errors! *****} % What to add here so that I get no errors?  %&lt;================= }  \begin{document} \Blindtext \end{document}   Run it two times. Everything should be well, blindtext is numbered.  Now please move the % sign from line 8 (false) to line 7 (true).    Because we had had the run before there is an .aux-file, including lines with the macro \@LN.  If you compile now you will get the following error message (and some more):   ***** Please delete .aux files if you get errors! ***** (C:\Users\...\LaTeX\_TeX.SX\lineno\test-lineno.aux ! Undefined control sequence. l.2 \@LN         {0}{0}   Question:  Is it possible to add in line 20 (see %&lt;===========) something to get rid of the error messages? Something that resets the meaning of macro \@LN? Getting Error while starting Google Map v2 I have built a peak detector circuit using op-amps/comparators to amplify the signal from an electret microphone, hold the peak signal and create a sharp logic edge when a peak is detected to be sent to a microcontroller.  I have built the circuit as follows setting up OA1 as a non inverting amplifier with a gain of 1000, but the voltage out from the OA1 remains fixed at about 4.5V no matter how much noise I make into the microphone. The comparator part of the circuit is outputting approximately 2V low and 5V high depending on how I set the potentiometer. All the components appear to be working.         &ndash; Schematic created using Google Chrome: Version 32.0.1700.107 m  This happens when displaying the profile and clicking on the number of visited days/consecutive.  . Are "pushforward" and "pullback by zero section" of diffrentials isomorphic in elliptic curves? Let S be a scheme, $p:E\rightarrow S$ be an elliptic curve with zero section $e:S\rightarrow E$. My question is: Are $p_*(\Omega_{E/S}^1)$ and $e^*(\Omega_{E/S}^1)$ isomorphic? I think it's right because I saw both definitions in different notes and they seem to represent the same thing. Thank you! What would Voldemort have done had he won? This might sound like speculation, but can it be reliably predicted from canon what Voldemort would have done after he won, were he successful in defeating Harry and the Order?  Would he just rule over the wizard world and keep up the masquerade, or would he try to take over the whole Earth by starting a (dark-)wizard vs. muggle war? What is duplicate content and how can I avoid being penalized for it on my site? I have a point $(X,Y)$ on an elliptical curve $E(a,b)$ where $a=-3$ and $B$ is a large number that is in hexadecimal from -51BD. To compress this point oficially in a program, we know that every $X$ on the curve has two $Y$'s, one even and one odd. Therefore, we only need to store whether the corresponding $Y$ point we are storing is even or odd. So to store the point $(X,Y)$ now, we shift $X$ ridding of its most significant bit and add $1$ if the $Y$ point is odd. Therefore, one can think of the compressed point being $2 \cdot X+B$ ($B$ is the bit we added to tell if $Y$ was odd or even) since the multiplication by $2$ is implied because of the shift. I understand this completely. It's recovering the original point that has me confused.  Since a point on the curve is given by the equation $y^2=x^3+ax+b$, we can find square roots officially in $\mathrm{GF}(p)$. The following from my spec is what confuses me.  If $p$ is prime and congruent to $3$ mod $4$, one of its square roots is $z^{(p+1)/4} \bmod p$ and the other is $p-z^{(p+1)/4} \bmod p$. How would I solve this? What is $z$?  For example, I know how to extract the $x$ point so let's say I simplify the right side of the equation to be $7$ and $p$ is $11$. Would I be trying to solve for the square root that's equal to $7$ mod $11$? Proof of dilogarithm reflection formula $\zeta(2)-\log(x)\log(1-x)=\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)$ How to install League of Legends in Ubuntu EEVEE and Cycles render lighting differences I get different lightning render with the same scene camera setup.  Cycles render (both experimental and supported feature set) gets weaker lightning and needs more energy.  How can I get the same lightning and reflection setup with EEVEE and Cycles?