What is wrong with the code:  String maintext = (String) main_text.getText().toString();  if(maintext =="10") {                ToneGenerator toneG = new ToneGenerator(AudioManager.STREAM_ALARM, 100);     toneG.startTone(ToneGenerator.TONE_CDMA_ALERT_CALL_GUARD, 200); }   The code is working without if statement but with the use of if and .getText().toString() is not working at all.
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?
If integral is zero and function is continuous and non negative, then what about the function? Does anyone know?  And a bigger question is what happens when you encounter this maximum? Is this the same number with other Windows OSs such as Vista, XP etc.? Even though I got banned suspended, I'm able to mark a question as a favorite. Is this functionally correct? How can I transform the elliptic curve $E/\mathbb{C}$ of the form $$y^2=4(x-e_1)(x-e_2)(x-e_3)$$ with $e_1&gt;e_2&gt;e_3\in\mathbb{R}$ roots of $E$ into a Weierstrass-Form $y^2=x^3+ax+b$? Regain of reputation from downvoted and deleted answers - update FAQ A formula for the roots of a solvable polynomial Let $F$ be a field and $p(x)\in F[x]$ a separable polynomial, denote $K$ as the splitting field of $p$ and assume that $K/F$ is Galois with a solvable Galois group.  I don't understand if this imply of any formula (in radicals) for the roots of $p$ (however, I do understand how a formula would imply that $p$ is solvable by roots).  Is there some kind of a way to obtain the roots of a solvable polynomial ? Is named indexer property possible? Setting DIV width and height in JavaScript Seating of $n$ people with tickets into $n+k$ chairs with 1st person taking a random seat "$n$ music lovers have reserved seats in a theater containing a total of $n+k$ seats ($k$ seats are unassigned). The first person who enters the theater, however, lost his seat assignment and chooses a seat at random. Subsequently, people enter the theater one at a time and sit in their assigned seat unless it is already occupied. If it is, they choose a seat at random from the remaining empty seats. What is the probability that person $n$, the last person to enter the theater, finds their seat already occupied?"  I could do this problem for small specific values of $n$ and $k$ but as they grow the expressions seem to get really messy really quick with no discernible pattern. How would one solve this problem? From everything I've read about quantum mechanics and quantum entanglement phenomena, it's not obvious to me why quantum entanglement is considered to be an active link. That is, it's stated every time that measurement of one particle affects the other.  In my head, there is a less magic explanation: the entangling measurement affects both particles in a way which makes their states identical, though unknown. In this case measuring one particle will reveal information about state of the other, but without a magical instant modification of remote entangled particle.  Obviously, I'm not the only one who had this idea. What are the problems associated with this view, and why is the magic view preferred? My system was configured to dual boot Ubuntu 12.04.1 LTS and Windows 7. Today I got my hands on the MSDN release of Windows 8 and I installed it over my Windows 7. Now the computer just boots to Windows 8 directly without loading the GRUB screen.  So I followed the steps as suggested in: . Running this command:      ls -l /dev/disk/by-label/   gives the following output:      total 0     lrwxrwxrwx 1 root root 10 Sep 11 07:51 Entertainment -&gt; ../../sda2     lrwxrwxrwx 1 root root 10 Sep 11 02:45 PENDRIVE -&gt; ../../sdb1   Also fdisk -l command gives this as the output:      Disk /dev/sda: 320.1 GB, 320072933376 bytes     255 heads, 63 sectors/track, 38913 cylinders, total 625142448 sectors     Units = sectors of 1 * 512 = 512 bytes     Sector size (logical/physical): 512 bytes / 512 bytes     I/O size (minimum/optimal): 512 bytes / 512 bytes     Disk identifier: 0x1246aa23         Device Boot      Start         End      Blocks   Id  System     /dev/sda1   *        2048   319582199   159790076    7  HPFS/NTFS/exFAT     /dev/sda2       319582208   602906623   141662208    7  HPFS/NTFS/exFAT     /dev/sda3       602908672   625135615    11113472   83  Linux      Disk /dev/sdb: 1939 MB, 1939865600 bytes     64 heads, 63 sectors/track, 939 cylinders, total 3788800 sectors     Units = sectors of 1 * 512 = 512 bytes     Sector size (logical/physical): 512 bytes / 512 bytes     I/O size (minimum/optimal): 512 bytes / 512 bytes     Disk identifier: 0xc3072e18         Device Boot      Start         End      Blocks   Id  System     /dev/sdb1   *        2248     3788799     1893276    c  W95 FAT32 (LBA)   So I assume that I have to run this: sudo grub-install /dev/sda3 to get GRUB up and running. But I am getting this error:      /usr/sbin/grub-probe: error: cannot find a device for /boot/grub (is /dev mounted?).   Can anyone please guide me in the right direction? The current Ubuntu installation is far too customized to my needs to lose it to a boot manager issue! Any help is much appreciated! I keep reading about  ('PPAs') and people answer questions with a link to a PPA. What's the best way to use these?     There are multiple valid answers for this question spanning over several versions of Ubuntu. For your convenience, an index of each is below. If $U$ is unitary operator then spectrum $\sigma(U)$ is contained inside the unit circle In a Hilbert space, let $U$ be a continuous operator which it unitary. Prove $\sigma (U)\subseteq \Bbb{S}^1$.   It is important for me to know how I am doing, and I didn't come by a clear explanation so it would be appreciated to have your correcting and guiding.  Attempt: $U^{*}U=UU^{*}=I$ and therefore $U^{-1}=U$. I have $$ \| U x \|^2 = \langle U x, U x \rangle = \langle U^* U x, x \rangle = \langle x, x \rangle = \|x\|^2 $$ and the same goes for $U^*$ and therefore $\| U \| = \| U^{-1} \| = 1$.   Looking at $U-\lambda$ I can take $$ -\lambda U(U^{-1}-{1\over \lambda}I) =U-\lambda I. $$ Clearly, for $\lambda \ne 0$, $-\lambda U$ is defined and is a linear operator. $(U^{-1}-{1\over \lambda}I)$ is invertible if $|\lambda|\ge 1$ and therefore $\sigma (A^{-1})=\sigma (A)\subseteq \{|z|\le1\}$. But from the RHS, $U-\lambda I$ is invertible if $|\lambda|\le 1$ and therefore $\sigma(A)\subseteq \Bbb{S}^1$. I have noticed my claims tend to have holes in them many times which I cannot spot. If it is substantially true, how can I make sure it is completely formal? Form inside a table Combining  with allowframebreaks in beamer Why shouldn't we allow NULLs? I remember reading this one article about database design and I also remember it said you should have field properties of NOT NULL.  I don't remember why this was the case though.  All I can seem to think of is that, as an application developer, you wouldn't have to test for NULL and a possible nonexistent data value (for instance, an empty string for strings).  But what do you do in the case of dates, datetime, and time (SQL Server 2008)?  You'd have to use some historic or bottomed-out date.  Any ideas on this? $|f(x)-f(y)|\geq k|x-y|$.Then $f$ is bijective and its inverse is continuous. My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:  $$|f(x)-f(y)|\geq k|x-y|$$  Then $f$ is bijective and  its inverse is continuous.  Well, there's a Theorem , Invariance of domain, that says   "If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".  But I'm not knowing how to proceed...need a clue...Thanks for attention!!! ArcGIS ModelBuilder iterator reports only last input shapefile? Using Arc 10.3.1, if I manually enter the names of multiple shapefiles (all from the same folder) into the Check Geometry tool, the (single) output table lists the errors that exist in all of the shapefiles.    That is, only one output table is created, even if multiple input shapefiles have errors.  For example, if two shapefiles have errors, both will be listed in the single output table. Here's a screen print of the input:    However, if I iterate those same shapefiles (using the Feature Classes option), the output table only lists the last shapefile that went through the iteration.  Here's a screenshot of the model:     I suspect that the iterator is running Check Geometry not once, but three times, with a single input shapefile going through each iteration, overwriting the previous output.    How do I get the iterator to list all of the shapefiles with errors, either in separate output tables or aggregated in a single output table? How to prove $(-3) \times (-4) = 12$? How to logically understand the multiplication of two integers ?  Eg:  3 x 4 = 12 (is understandable)  -3 x4 = -12 (is also somewhat understandable)  But , 3 x -4 = -12 (is NOT understandable)  -3 x -4 = 12 (is also NOT understandable)  Now , at this we must assume that commutative property may not be true .  What is the logical explanation for something being multiplied negative times ? What's the best way to store class variables in PHP? I currently have my PHP class variables set up like this:  class someThing {      private $cat;     private $dog;     private $mouse;     private $hamster;     private $zebra;     private $lion;      //getters, setters and other methods }   But I've also seen people using a single array to store all the variables:  class someThing {      private $data = array();      //getters, setters and other methods }   Which do you use, and why? What are the advantages and disadvantages of each?