NullPointerException (Java)
What is a NullPointerException, and how do I fix it?
tar unpack : Unexpected EOF in archive I have downloaded eclipse-platform-3.7-linux-gtk-x86_64.tar.gz.  Using Ubuntu 10.10  64-bit on Intel 64-bit processor.  While trying to unpack the contents (either double-click or right-click &amp; open-with-archive-manager) I get this error:  gzip: stdin: unexpected end of file&lt;br/&gt; tar: Unexpected EOF in archive&lt;br/&gt;  tar: Error is not recoverable: exiting now   Is there any remedy to repair it? or is the downloaded file corrupt &amp; needs to be re-downloaded? How does the ternary operator work? is the group of rational numbers the fundamental group of some space? pgrouting unexpected output A closed ball in a metric space is a closed set Integral of $\sqrt{1-x^2}$ using integration by parts How to reset postgres' primary key sequence when it falls out of sync? After installing Telegram Desktop on Ubuntu 16.04 by these commands:  sudo add-apt-repository ppa:atareao/telegram sudo apt-get update sudo apt-get install telegram   I get this error:  $ telegram telegram: Panic - Unable to find your tty (pts/2) in /var/run/utmp Why does a wireless network give a more accurate location than a wired network? I build a simple HTML5 website that takes my location (in latitude-longitude), and simply shows my location in Google maps.  I opened the HTML document with my laptop Asus K53E-B1 and I found that when I connected to a wireless network, it showed almost my precise location (about 100 meter of inaccuracy),  while when I plug in the network cable, and test again, it's a bit Inaccurate. (showed a street 12KM away from me).  Can you explain me why is a wireless network gives more accurate location?  Tested with Google Chrome. Let $E$ be a finite dimensional vector space over field $\mathbb R$ with $E^*$ as the dual space of $E$.($\dim E=n$)   $\Omega^P(E):=\{ \alpha\colon \overbrace{E\times\cdots\times E}^{p- times}\rightarrow \mathbb R\  \ , \alpha \text{ is alternating multilinear map}\}$ $\Omega^P(E^*):=\{ u\colon \overbrace{E^*\times\cdots\times E^*}^{p- times}\rightarrow \mathbb R\  \ , u \text{ is alternating multilinear map}\}$  I wish to prove that there exist an $\textbf{unique}$ bilinear form $B\colon \Omega^p(E^*)\times\Omega^p(E)\rightarrow\mathbb R$   such that for all $ f_i\in E^* \ , u_j\in E^{**}\simeq E \qquad  i,j=1,\cdots,p$, it has the following rule:  $B(u_1\wedge\cdots\wedge u_p,f_1\wedge\cdots\wedge f_p)=\det[f_i(u_j)]$.  $\wedge$ is wedge product between alternating maps i.e. $\alpha\wedge\beta(v_1,\cdots,v_{p+p})=\dfrac{1}{p! p!}\displaystyle\sum_{\sigma\in S_{p+p}}sign(\sigma)\alpha(v_{\sigma(1)},\cdots,v_{\sigma(p)})\beta(v_{\sigma(p+1)},\cdots,v_{\sigma(p+p)})$  How can I do this? Do AWS Reserved Instances need to be assigned to the instance? I have an RDS instance running of type db.t2.small.  I bought a reserved instance for 1 year of the same type, I paid the partial upfront rate.  Do I need to assign this subscription in some way to my existing instance to say 'This is the instance that is using the subscription' or does AWS just charge for the instance usage depending which instances you have running and whether an available subscription was active for it at the time which determines the running rate?  The reason that I ask is that I can't see any mention of the reduced rate on my AWS Billing area and it shows that I've used 17 hours of my reservation, but there isn't a row on my bill so far this month that indicates the new partial upfront rate that I should be charged for my instance. Can "whatever" be split into two words? I tend to write, "say whatever they want", but I'm always tempted to write "say what ever they want". Is it acceptable to split the word in this context? What is $48\div2(9+3)$? What's the limit on the no. of partitions I can have? I would like to know how many Primary and Extended Partitions I can create on a x86_64 PC with Linux running on it?     Update  : If there is a limit to the number of partitions, why is that the limit? How to search by key=>value in a multidimensional array in PHP Windows file system is case insensitive. How, given a file/folder name (e.g. "somefile"), I get the actual name of that file/folder (e.g. it should return "SomeFile" if Explorer displays it so)?  Some ways I know, all of which seem quite backwards:   Given the full path, search for each folder on the path (via FindFirstFile). This gives proper cased results of each folder. At the last step, search for the file itself. Get filename from handle (as in ). This requires opening a file, creating file mapping, getting it's name, parsing device names etc. Pretty convoluted. And it does not work for folders or zero-size files.   Am I missing some obvious WinAPI call? The simplest ones, like GetActualPathName() or GetFullPathName() return the name using casing that was passed in (e.g. returns "program files" if that was passed in, even if it should be "Program Files").  I'm looking for a native solution (not .NET one). I recently got a Windows 10 laptop. Opposite to my Macbook, I must swipe right-to-left to go to the right virtual desktop. Is there a way to reverse this, so I swipe left-to-right to go to the right virtual desktop? it can not access popup.js file after creating a chrome extension I want to get random decimal digits out of a 6-faced die.  Let's say the die is fair, with an equal probability for each outcome. (But feel free to share any consequences of that not being true, if you feel like it.)  I will represent the outcome of a die roll as a base-6 number:   In base-6 there are only 6 possible digits: $\{0,1,2,3,4,5\}$. The die has faces numbered 1 to 6. Face $f$ corresponds to digit $f-1$.   Thus, I can represent as many die rolls as I like with a base-6 number:   If the die is rolled once:   Let's say the die stopped on face $5$. The final base-6 number will be $4$.  If the die is rolled twice:   Let's say the die stopped on face $5$. And then on face $3$. The final base-6 number will be $42$.  If I roll the dice 10 times:   Let's say the die stopped sequentially on faces $5$, $3$, $4$, $6$, $2$, $3$, $3$, $3$, $2$, and $4$. The final base-6 number will be $4235122213$.    Since I want decimal digits, I need to convert to base-10.   Base-6: $4235122213$ Base-10: $44753985$   For any given number of die rolls, how many different combinations exist? If there are $r$ die rolls, the total of possible combinations is $6^r$.   $r$ rolls — $6^r$ different combinations. $0$ rolls — only $1$ combination. $1$ roll — $6$ different combinations. $2$ rolls — $36$ different combinations. $3$ rolls — $216$ different combinations. $4$ rolls — $1296$ different combinations. $5$ rolls — $7776$ different combinations. $6$ rolls — $46656$ different combinations. $7$ rolls — $279936$ different combinations. $8$ rolls — $1679616$ different combinations. $9$ rolls — $10077696$ different combinations.   $n_r$ ($\in \mathbb{N_0}$) is the number obtained after $r$ die rolls through the process described previously: "If the die is rolled once, twice, ..."  It is impossible for $n_r$ to reach the number of possible different combinations ($6^r$). It will be always short by least $1$:  $$ 0 \leqslant n_r &lt; 6^r $$   $r$ rolls — $n_r$ never surpasses $6^r-1$. $0$ rolls — $n_r$ never surpasses $0$. $1$ roll — $n_r$ never surpasses $5$. $2$ rolls — $n_r$ never surpasses $35$. $3$ rolls — $n_r$ never surpasses $215$. $4$ rolls — $n_r$ never surpasses $1295$. $5$ rolls — $n_r$ never surpasses $7775$. $6$ rolls — $n_r$ never surpasses $46655$. $7$ rolls — $n_r$ never surpasses $279935$. $8$ rolls — $n_r$ never surpasses $1679615$. $9$ rolls — $n_r$ never surpasses $10077695$.   I can never trust the leftmost nor the rightmost digit (fringe digits), since the possible values are potentially restricted.   For $0$ rolls, the outcome is always $0$. For $1$ roll, the fringe digit never surpasses $5$. For $2$ rolls, the leftmost digit never surpasses $3$, and the rightmost, $5$. For $3$ rolls, the leftmost digit never surpasses $2$, and the rightmost, $5$. For $4$ rolls, the leftmost digit never surpasses $1$, and the rightmost, $5$. For $5$ rolls, the leftmost digit never surpasses $7$, and the rightmost, $5$. For $6$ rolls, the leftmost digit never surpasses $4$, and the rightmost, $5$. For $7$ rolls, the leftmost digit never surpasses $2$, and the rightmost, $5$. For $8$ rolls, the leftmost digit never surpasses $1$, and the rightmost, $5$. For $9$ rolls, the leftmost digit never surpasses $1$, and the rightmost, $5$.   The other digits seem perfectly random to me though. Am I wrong? sort but keep header line at the top I am getting output from a program that first produces one line that is a bunch of column headers, and then a bunch of lines of data. I want to cut various columns of this output and view it sorted according to various columns. Without the headers, the cutting and sorting is easily accomplished via the -k option to sort along with cut or awk to view a subset of the columns. However, this method of sorting mixes the column headers in with the rest of the lines of output. Is there an easy way to keep the headers at the top?