I have to Read footer of a file and store it into a arraylist in java I have a GMF file. Its last 9 lines are to be considered as footer. The fast line of footer Starts with "Footer" and last line Ends with "End of file".  So far I have written this   while ((sCurrentLine = br.readLine()) != null ) {                     if(headcount&lt;11)                     {                         HeaderArray.add(sCurrentLine);                         headcount++;                         System.out.println("Header["+headcount+"]");                         System.out.println("sCurrentLine"+sCurrentLine);                     }                     else if( headcount&gt;10 &amp;&amp;(sCurrentLine.startsWith("Rerate"))) {                                   RecordsArray.add(sCurrentLine);                                 headcount++;                                 System.out.println("Records");                                 System.out.println("Record"+sCurrentLine);                      }                     else if(sCurrentLine.startsWith("Footer"))                     {                                while(footerFlag!="End_of_file:")                             {                                  FooterArray.add(sCurrentLine);                                 footerFlag=br.readLine();                                 sCurrentLine=footerFlag;                                 System.out.println("Footer");                                 System.out.println("Footer"+sCurrentLine);                               }                     }              }   But this code only goes into an infinite loop. I need all 9 lines of footers in FooterArray. Please help.
How do I compare strings in Java? I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?
Compiz Problems (Plugin Missing, Performance Issues, Development) I've recently set up the compiz cube again, and I was very happy to find that that flickering window glitch is gone! However, the cube reflection and deformation plugins are also gone.  I am running a fresh 12.10 install, and I have every compiz plugin package installed (including compiz-plugins-extra, which supposedly contains everything that compiz-fusion-plugins-extra package used to have), but those two plugins are still missing. I haven't been able to find anyone with the same problem.   Any ideas? (thanks!) So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.  Now, consider the closed interval $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subset (a, b)$.  Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n &gt; \frac{2}{b-a}$.  Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b - \frac{1}{ \frac{2}{b-a}} = b - \frac{b-a}{2}$, and our interval will be degenerate.  Note that $\forall (a,b) \in \mathcal{O}$,   $\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] = (a,b)$.  So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.  But, I don't like this. It looks like I'm still just taking a union of open sets.  Please help me get this to look like the union of closed sets.  (For unbounded intervals, I know what to do - it should follow from this relatively easily). Quickly create a large file on a Linux system How can I quickly create a large file on a Linux () system?   will do the job, but reading from /dev/zero and writing to the drive can take a long time when you need a file several hundreds of GBs in size for testing... If you need to do that repeatedly, the time really adds up.  I don't care about the contents of the file, I just want it to be created quickly. How can this be done?  Using a sparse file won't work for this. I need the file to be allocated disk space. The standard files/tools that report memory seem to have different formats on different Linux distributions. For example, on Arch and Ubuntu.   Arch  $ free               total        used        free      shared  buff/cache   available Mem:        8169312     3870392     2648348       97884     1650572     4110336 Swap:      16777212      389588    16387624   $ head /proc/meminfo  MemTotal:        8169312 kB MemFree:         2625668 kB MemAvailable:    4088520 kB Buffers:          239688 kB Cached:          1224520 kB SwapCached:        17452 kB Active:          4074548 kB Inactive:        1035716 kB Active(anon):    3247948 kB Inactive(anon):   497684 kB  Ubuntu  $ free              total       used       free     shared    buffers     cached Mem:      80642828   69076080   11566748    3063796     150688   58358264 -/+ buffers/cache:   10567128   70075700 Swap:     20971516    5828472   15143044   $ head /proc/meminfo  MemTotal:       80642828 kB MemFree:        11565936 kB Buffers:          150688 kB Cached:         58358264 kB SwapCached:      2173912 kB Active:         27305364 kB Inactive:       40004480 kB Active(anon):    7584320 kB Inactive(anon):  4280400 kB Active(file):   19721044 kB    So, how can I portably (across Linux distros only) and reliably get the amount of memory—excluding swap—that is available for my software to use at a particular time? Presumably that's what's shown as "available" and "MemAvailable" in the output of free and cat /proc/meminfo in Arch but how do I get the same in Ubuntu or another distribution? Counting integer partitions of n into exactly k distinct parts size at most M How can I find the number of partitions of $n$ into exactly $k$ distinct parts, where each part is at most $M$?  The number of partitions $p_k(\leq M,n)$ of $n$ into at most $k$ parts, each of size at most $M$, is given by the generating function: $$  \binom{M+k}{k}_{x} = \prod_{j=1}^{k}\frac{1-x^{M+k-j+1}}{1-x^j}= \sum_{n=0}^{kM} p_{k}(\leq M,n) x^n $$  For the number of the partitions $p_k(\mathcal{D},n)$ of $n$ into at most $k$ parts there is the recurrence relationship: $$   p_{k}(\mathcal{D},n) = p_{k}(\mathcal{D},n-k) + p_{k-1}(\mathcal{D},n) $$  But what, if I want to count only the partitions with distinct parts and restricted number of parts and restricted part size?  Update: Now I know the generating function for the number of distinct restricted partitions $p_k(\leq M, \mathcal{D},n)$ of $n$ into exactly $k$ distinct parts, all at most $M$ is $$    \prod_{j=1}^{M} (1+xq^{j}) = \sum_{k,n=0}^{\infty}p_k(\leq M, \mathcal{D},n)x^{k}q^{n} $$ and there is also a recurrence relation $$    p_k(\leq M, \mathcal{D},n) =         p_{k-1}(\leq M-1, \mathcal{D},n-k) +         p_k(\leq M-1, \mathcal{D},n-k) $$ How can I prove this? Could you recommend a book, where I could read about this? Using "that" and "this" interchangeably How do I get a slanted \sum symbol? I want to use a slanted \sum symbol denoting a different meaning from summation. How do I get a slanted \sum symbol? I know how to , but I'd like to add a bunch of different extensions and avoid the tedious job of adding every single one. Is there a command line tool or registry key I can use to accomplish this? QGIS query syntax differs between attribute table and feature filter Using QGIS 2.16.2, I can successfully select records from a shapefile attribute table using the following expression:  left("start_time", 10) = '2015-08-01'   Please note that start_time is a string field.  However, when I copy and paste the same expression into the shapefile's Layer Properties>General>feature filter>Query Builder I get the following error:    I would expect that the same expression would work everywhere in QGIS.  What am I doing wrong? Does light reflect if incident at exactly the critical angle? A lot of textbooks and exam boards claim that light incident at exactly the critical angle is transmitted along the media boundary (i.e. at right-angles to the normal), but this seems to violate the principle of reversibility in classical physics.  How would a photon or ray travelling in the reverse direction "know" when to enter the higher refracting medium?  It can't know, so I conclude that such light is simply reflected?  Is this correct? Kill process after it's been allowed to run for some time I want to limit the time a grep process command is allowed to run or be alive.  For example. I want to perform the following:  grep -qsRw -m1 "parameter" /var   But before running the grep command I want to limit how long the grep process is to live, say no longer than 30 seconds.  How do I do this?  And if it can, how do I return or reset to have no time limit afterwards. $\delta$ differential notation Various textbooks that I am currently consulting (including Spacecraft Dynamics and Control An Introduction - Anton H.J. De Ruiter | Christopher J. Damaren | James R. Forbes Section 1.4, page 32) use $\delta$, not $d$ or $\partial$ to express an infinitesimal quantity. In the context of the reference text above, the symbol is used as part of a general definition for the derivative of a vector. Specifically, given   $$\mathbf{\vec{r}}=\mathcal{\vec{F^{T}_{1}}}\mathbf{\vec{r_{1}}}$$   where   $$\mathcal{\vec{F^{T}_{1}}}$$   is a vectrix   $$[\mathbf{\vec{x_1}} \phantom{s} \mathbf{\vec{y_1}} \phantom{s} \mathbf{\vec{z_1}}].$$  The time derivative of the vector is defined as   $$\dot{\mathbf{\vec{r}}}\triangleq \lim_{\delta t\to 0}\frac{\delta\mathbf{\vec{r}}}{\delta t}.$$  In this context, what is the difference between $\frac{\delta\mathbf{\vec{r}}}{\delta t}$ and $\frac{d \mathbf{\vec{r}}}{d t}$ ?  Note about thermodynamic use of $\delta$: My understanding is that $\delta$ is used in thermodynamic equations to express path dependence of a scalar quantity such as heat or work. What does it mean in the context of an abstract physical vector? Is it possible to create individual power settings for each hard drive? I have a number of different drives that I use for different purposes but I only use two of them frequently. I would like for the others to enter sleep mode when not in use, but never my main two.  Is there any way to edit power saving settings for individual drives? Book recommendation on plane Euclidean geometry I consider myself relatively good at math, though I don't know it at a high level (yet). One of my problems is that I'm not very comfortable with geometry, unlike algebra, or to restate, I'm much more comfortable with algebra than geometry. I think that's mainly because my geometry education was sparse through the years, lacking in consistency etc. So I'd like to revise (and learn more) all at once, catching the basic axioms, understanding why such is such, etc. Essentially, a moderately rigorous textbook in plain Euclidean geometry (nothing fancy). Please don't say "The Elements" - I have browsed it at the bookstore, it is quite good, but not really what I'm looking for right now. Entire function problem: translation What's the most efficient way to get experience? Now that experience does something besides render regions of a server uninhabitable due to  orb lag, what's the most efficient and safe way to get those excellent enchants?  A mob farmer may rank highly, but what about without tons of setup? What are these three references from episode 9? How to restore Windows 7 MBR without a CD Question about the order of the product of two elements in an abelian group My reference for learning abstract algebra says that in an abelian group $G$ the product of two elements (say $a$ with order $n$ and $b$ with order $m$) has order $mn$ if $gcd(n,m) = 1$.   What I don't understand is where the $gcd(m,n)=1$ comes into use in proving this fact? For example, $ a^{mn} = (a^n)^m = (e)^m = e$ (identity element). A similar argument for $b^{mn}$ together with the fact that $G$ is abelian is enough to show that $(ab)^{mn} = e$. From here, how do I conclude for an integer $r &lt; mn$ that $(ab)^r \neq e$  Also, are there any other conclusion that can be drawn if $gcd(n,m) \neq 1$ ?  Thanks for any feedback in advance ! About free quarks and confinement