Structures with class fields
Can you have a class in a struct?
How do I move clock to the right end of GNOME 3 Panel from the middle? Need Help Fixing Grub after Installing Ubuntu 13.04 x64 Alongside Windows 8 on a Lenovo IdeaPad Y400 with UEFI I have a Lenovo IdeaPad Y400 that came pre-installed with Windows 8. I would like to run Ubuntu alongside it in a dual-boot configuration. I've spent the past week fiddling around, re-installing both Windows and Linux in various configurations, and I've come awfully close to success, but seem to have hit a snag with Grub.  For my most recent attempt, I took the following steps:   Disabled Secure Boot Re-imaged Windows 8 from a hidden recovery partition, effectively restoring the machine to factory state Activated an option called Legacy Support in the BIOS (or is it UEFI now?) Booted Ubuntu 13.04 x64 from a USB stick, chose "install alongside Windows" and accepted all default options After a restart, found that I could boot Ubuntu, but could not boot Windows Once into Ubuntu, I ran Boot Repair with recommended settings and accepted all defaults. When Boot Repair finished, I was presented with the following message:      The boot files of [The OS now in use - Ubuntu 13.04] are far from the   start of the disk. Your BIOS may not detect them. You may want to   retry after creating a /boot partition (EXT4, >200MB, start of the   disk). This can be performed via tools such as gParted. Then select   this partition via the [Separate /boot partition:] option of [Boot   Repair]. ()    I followed the instructions at the url in the message, creating a 1GB EXT 4 partition near the start of the drive  when I tried to run Boot Repair again, the "Separate /boot partition" option on the "Grub Location" tab was disabled.  The new partition is /dev/sda11 and has the label GRUB. On restart, I get the following error message:      error: file '/boot/grub/i386-pc/normal.mod' not found.    grub rescue>    However, if I manually choose the boot partition by hitting F12, I can boot into Windows no problems. Selecting Ubuntu from the boot device menu gives me a Grub menu with the following options:      Ubuntu      Advanced options for Ubuntu       Windows UEFI recovery bootmgfw.efi      Windows Boot UEFI recovery       Windows UEFI recovery LrsBootmgr.efi      Windows Boot UEFI recovery bkpbootx64.efi       Windows Recovery Environment (loader) (on /dev/sda3)       Windows 8 (loader) (on /dev/sda5) System setup    Selecting Ubuntu from this menu allows me to get to my Ubuntu desktop. I haven't tried the other options yet for fear of losing the progress that I've made.   So it looks like I have successfully dual-booted the machine, but that Grub is broken for some reason or another.   My current partition scheme can be seen at:   Thanks in advance - Jonathan Aufbau principle in modern quantum theory Finite group for which $|\{x:x^m=e\}|\leq m$ for all $m$ is cyclic. Let $G$ be a finite group. For each positive integer $m$, if $x^{m}=e$ has at most $m$ solutions in $G$, $G$ is cyclic.   What I have thought is that $n=\sum_{d\mid n}\phi(d)$ can be used to solve this and showing that $|G|$ order element in $G$ exists is enough. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Save an entire webpage, with all images and css, into just one folder and one file, with wget? Throughout Blade Runner we see Gaff (Edward James Olmos) making origami figures at various points in the movie. He makes a chicken, matchstick men and at the end of the film there is a unicorn waiting outside Deckard's apartment.    There is a   of  out there regarding the unicorn in particular, and how it indicates that Deckard is a replicant. I think an answer regarding the use of origami in general will be interesting new ground to cover.  Why origami? What does the origami represent in Blade Runner? I installed Windows 7 fresh and installed SP1. Now, when I try to check manually for Windows Updates it just hangs on the Checking for updates screen.    I tried running the tools in , but this did not fix the issue either:  No matter what I do it just hangs on the "Checking for updates..." screen and goes no further. I'm looking for a movie that was set in the US, and all the characters (including the robot) spoke English.  I can't recall the exact plot, but mostly it was about an innocent game robot that shoots small objects for fun and becomes dangerous, because its AI made a wrongful choice about 'winning the game'. The robot starts to arm itself with more dangerous objects (not sure but knives and stuff) and starts to attack people.  It was overall supposed to be fun and adventure movie for kids. Color movie, and probably made in 80~90's.   There is a recurring scene when the robot detects an aggressor, it scans and loudly announces "enemy one, enemy one". The robot was somehow looking similar to Wall-E, and had a saucer-like head part. Give an example about limits: $\lim\limits_{x\to0}f(x^{2})$ exists but $\lim\limits_{x\to0}f(x)$ does not. What do ref, val and out mean on method parameters? I'm looking for a clear, concise and accurate answer.    Ideally as the actual answer, although links to good explanations welcome.  This also applies to VB.Net, but the keywords are different - ByRef and ByVal. Here is what it looks like in Blender Render (Not Cycles).   This is how I set it up.   Basically, I'm trying to export it, but it's not including the &quot;Freestyle&quot; lines. When I open up the .dae file in Finder, it looks like this:   What is wrong here? I want it to be exported (COLLIDA .dae) with the black outline. I'm seeming to think that maybe the &quot;Freestyle&quot; is only for the Blender Render. Am I right? How would I go about getting the black outline included in my exported model? Is it an export setting I have to tick? Thanks for any helpful answers. sample code for Dynamically populating dropdown in JSP The sum of powers of two and two's complement – is there a deeper meaning behind this? Probably everyone has once come across the following "theorem" with corresponding "proof": $$\sum_{n=0}^\infty 2^n = -1$$ Proof: $\sum_{n=0}^\infty q^n = 1/(1-q)$. Insert $q=2$ to get the result.  Of course the "proof" neglects the condition on $q$ for this formula, and the sum really diverges. However I now noticed an interesting fact:  If you use two's complement to represent negative numbers on computers, $-1$ is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left.  Now imagine that formally you sign-extend the number $-1$ to infinitely many bits. What you get is an infinite-to-the-left string of $1$s. Which, using the normal base-2 formula $n = \sum_k b_k 2^k$ (where $b_k$ is the bit k positions from the right, i.e. $b_0$ is the rightmost bit), that infinite string of $1$s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation.  Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential. Change elements on different positions within a list Store update, insert, or delete statement affected an unexpected number of rows (0) EntityFramework When is the sum of divisors a perfect square? For $n=3$, $\sigma(n)=4$, a perfect square. Calculating further was not yielding positive results.  I was wondering is there a way to find all such an $n$, like some algorithm?  We know that if $n=p_1^{k_1}p_2^{k_2}\dots p_r^{k_r}$ then $\sigma(n)=\frac {p_1^{k_1+1}-1}{p_1-1} \frac{p_2^{k_2+1}-1}{p_2-1}\dots \frac {p_r^{k_r+1}-1}{p_r-1}$. So we have to find $n$ such that $\sigma(n)=\frac {p_1^{k_1+1}-1}{p_1-1} \frac{p_2^{k_2+1}-1}{p_2-1}\dots \frac {p_r^{k_r+1}-1}{p_r-1}=t^2$ for some integer $t$. If each $\frac{p_i^{k_i+1}-1}{p_i-1}$ is a square we will get some such numbers (not all though), but let us consider $\frac{p_i^{k_i+1}-1}{p_i-1}=m^2$.   Working a example, If I take $p_i=17 $ , (or any prime such that $p_i-1$ is a perfect square), then if I can find a $k_i$ such that ${p_i^{k_i+1}-1}=16.m^2$ I will get an $n$, so now when can this happen, is something I have to try out.   But out of curiosity I am asking has this been worked out before. Problem looks broad to me, Is there a solution to this problem? Finding $\lim_{n\to\infty}\int_0^{\frac{1}{2^n}}(1+\cos(x))^n\,dx$ During our studies on Riemann Integrals we were given the following question in my Calculus II course: $$ \forall n\in\mathbb{N}: \text{let } f_n(x)=(1+\cos(x))^n, f_n:[0,1]\to \mathbb{R}$$ $$\text{ Find}: \lim_{n\to\infty}\int_0^{\frac{1}{2^n}}f_n(x)\, dx $$ I have tried finding the limit using the squeeze theorem. I noticed that the function is defined on $[0,1]$, where $\cos(x)$ is positive and decreasing. I used the following boundaries: $$ 1\leq f_n\leq 2^n \implies \frac{1}{2^n} = \int_0^{\frac{1}{2^n}}1 \leq \int_0^{\frac{1}{2^n}}f_n(x) \leq \int_0^{\frac{1}{2^n}}2^n = 1  $$ but $ \displaystyle{\lim_{n\to\infty}}\frac{1}{2^n} = 0 $ which means I'm unable to apply the squeeze theroem. I've tried using tighter lower boundaries such as $1+\cos(1)$ But it didn't seem to work. I did some trial and error by calculaing the integrals manually for different values of $n$ and it does seem like $\displaystyle{\lim_{n\to\infty}}\int_0^{\frac{1}{2^n}}f_n(x) dx = 1 $ This means that somehow I should find a lower bundary which tends to 1 when $n\to\infty$. Any ideas on how to go forward from here are welcome. Man travelling alone across a planet, thinks about a fearsome creature that he hears behind him About fifteen years ago I read a short story about a man on a planet (possibly Mars or the Moon), travelling alone between bases. It's written in the first person and as he's walking across the planet, he's thinking to himself about this fearsome creature that preys on people. It sneaks up on them, makes some noise and then attacks them. Just as he turns the final corner of the canyon he was walking through and sees the door to the other base, he hears the creature behind him making its noise and the story ends.  It was in a short story collection but I don't remember the name of the story or the book it was in. Any help would be much appreciated. Minecraft LAN "java.net.ConnectException: Connection refused: no further information:" I am trying to play a map on LAN. But every time I try to log on, I get:     java.net.ConnectException: Connection refused: no further information:   This used to work, but for some reason has stopped all of a sudden. I looked it up online, and everything I found said the firewall just wasn't allowing Java through. I followed tutorials to allow Java through the firewall, but even then it won't work. I know for a fact that I'm putting the IP in correctly, and everything seems to be correct, but yet I still get the same error message.  If it helps any, I'm using MC version 1.8.9 (the map requires it), and I don't have any mods.