I have a method rotateImage that returns an int[][] array, I am having trouble understaing how to display the actual values of the array instead of just the hashcode:  [I@548c4f57 [I@1218025c [I@816f27d   I tried doing this:   Arrays.stream(rotateImage(matrix)).forEach(System.out::println);   but that only returns the hashcode, not the actual numbers themselves. I also tried using to Arrays.toString() method in several ways but always get an error.
In Java, arrays don't override toString(), so if you try to print one directly, you get the className + '@' + the hex of the  of the array, as defined by Object.toString():  int[] intArray = new int[] {1, 2, 3, 4, 5}; System.out.println(intArray);     // prints something like '[I@3343c8b3'   But usually, we'd actually want something more like [1, 2, 3, 4, 5]. What's the simplest way of doing that? Here are some example inputs and outputs:  // Array of primitives: int[] intArray = new int[] {1, 2, 3, 4, 5}; //output: [1, 2, 3, 4, 5]  // Array of object references: String[] strArray = new String[] {"John", "Mary", "Bob"}; //output: [John, Mary, Bob]
I was working on a problem that resulted in the calculation: $20 \equiv 10x \pmod{11}$.  I got the answer $x \equiv 2 \pmod{11}$ with the thought process: Since $10$ is a factor of $20$ I can rewrite $20 \pmod{11}$ as $10x \pmod{11}$ with $x=2$. But isn't this basically the same as using division, which we aren't supposed to do in modular spaces?  If I solve by multiplying $20$ by the multiplicative inverse of $10 \pmod{11}$ which is $10$, I get $200 \pmod{11}$ which simplifies to $2 \pmod{11}$ anyway. Did these numbers just happen to be the same or is my original method valid? Sass Interpolation of Mixin, Function, and Variable names Shifting -5 V to +5 V signal to 0 to +5 V for ADC I want to convert the  output voltage of an ECG (-5 to +5) to 0 - +5 volt for ADC. Should i use Op amp ? What is my problem in this picture? I simulated it in Proteus ... but when input voltage is higher than 4 V or lower than -4 V Vout gives an unexpected value.   Below is another simulink that have same problem ...    My idea to make this level of voltage is (Vin +10)/2 that give the range 0-5 V. Is there a "smallest" divergent sum? What permissions should my website files/folders have on a Linux webserver? Why does Earth behave like Natural Bar Magnet? What is the reason for the Earth to behave like a bar magnet and have poles (North and South poles)? Resizing an iframe based on content I am working on an iGoogle-like application. Content from other applications (on other domains) is shown using iframes.   How do I resize the iframes to fit the height of the iframes' content?  I've tried to decipher the javascript Google uses but it's obfuscated, and searching the web has been fruitless so far.  Update: Please note that content is loaded from other domains, so the  applies. Two question on harmonic function Stack overflow links not working I'm exploring the effects of removing the intercept in a logistic regression model.  Assume a model:  $$logit(Y = 1) = \beta_1 x + \beta_2z + 0$$  with $x$ and $z$ being categorical variables with 2 levels each and no intercept.  I understood that having no intercept with categorical predictors produce coefficients that compare the $P(Y = 1)$ in each level of the two predictor against a null case where $P(Y=1) = 0.5$ or $logit(Y=1) = 0$.  I noticed a phenomenon that can understand. Using glm() function in R if you change the order of the variable in the right hand part of the formula, the coefficients change too. But even more oddly, the coefficient of the first variable is always the same.  Here's an R demo:  y &lt;- as.factor(sample(rep(1:2), 30, T)) x &lt;- as.factor(sample(rep(1:2), 30, T)) z &lt;- as.factor(sample(rep(1:2), 30, T))  coef(glm(y ~ x + z - 1, binomial) #        x1         x2         z2  #-0.1764783  0.3260739 -0.1335192  coef(glm(y ~ z + x - 1, binomial)) #        z1         z2         x2  #-0.1764783 -0.3099976  0.5025523    As you can see the first predictors have the same coefficient while the other are different in the two models.  Here is what I expected and instead behave differently than what I though:   Since every level of the two predictors is compared to the same null case, I expected to have the same coefficients in the two models, independently from the order in which I use them. I expected to see the coefficients of every level of every predictor, instead the coefficient for the 1 level of the second predictor is not shown. I therefore assume that only the first variable is compared against the null case, while the second is compared against a reference level; but what is this level? Is it $P(Y = 1 | X = 1 \cap Z = 1)$? Reproducing one of the models WITH the intercept we get:  coef(glm(y ~ x + z - 1, binomial) #        x1         x2         z2  #-0.1764783  0.3260739 -0.1335192  coef(glm(y ~ x + z, binomial)) #(Intercept)         x2          z2  #-0.1764783   0.5025523  -0.1335192    As expected x1 become the intercept, and x2 is likely relative to x1. z1 is missing also in this case and z2 is the same as in the model without intercept.  Thus should I assume that the comparison against the null case $P(Y = 1) = 0.5$ is made only for the first variable in a formula, while the other are compared against the usual intercept? Is this behavior normal? What about the fact that the first coefficient has the same value whichever the order of the predictors in the formula? What if I want to compare all level of each predictor against the null case and have a coefficient for all levels? Or it's theoretically impossible for some reason I don't get? Specific system of differential equations I have the following system of equations:  \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray}  I want to prove that if a solutions starts (at time $t = 0$) in the region   $$ G = \{(x, y) \in \mathbb{R}^2 : x &gt; 0, y &gt; 0 \text{ and } 1 &lt; x^2 + y^2  &lt; 4\} $$  that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 &lt; x &lt; 2$ or the $y$-axis where $0 &lt; y &lt; 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x &gt; 0, y &gt; 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it.  In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} &lt; 0$ in $G$, so $x$ is decreasing.)  I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} &gt; 0$, so after crossing the circle, the solution would go back up to $G$.  Is the circle itself an orbit?  EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} &gt; 0 \Leftrightarrow 0 &lt; r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 &lt; r \leq 1$ satisfies $4r - r^3 &gt; 0$, and thus $\frac{dr}{dt} &gt; 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive.  Can someone tell me if this is correct? This problem has been seen on mid-2012 MacBook Pros on Mavericks and on late-2014 Retina MacBook Pros on Yosemite.  One day while using iMessage on the Mac, all the contacts turned red and messages failed to send, immediately returning a "not delivered" error. It also reported when trying to resend the messages that the contacts were no longer associated with an iMessage account.  Things that didn't fix this behavior:   Deleting the com.apple.ids.madrid Removing the Apple ID information from my phone, turning off iMessage on both of the devices, and placing my iPhone in airplane mode for several minutes before turning it on logging in again.   Could someone help me figure out how to solve this? How do I show for an ideal transformer $M^2=L_1L_2$? I've been stuck on this problem for about an hour.     In an ideal transformer, the same flux passes through all turns of the primary and of the secondary . Show that in this case $M^2=L_1L_2$, where $M$ is the mutual inductance of the coils, and $L_1$, $L_2$ are their individual self-inductances.   My first instinct was the following: If we isolate the response of each coil to the other one, we get the equations $N_1\Phi_1=MI_2$ and $N_2\Phi_2=MI_1$, where $\Phi$ is magnetic flux going through a single turn of the coil and $I$ is current. Likewise, if we isolate the response of each coil to itself, we get the equations $N_1\Phi_1=L_1I_1$ and $N_2\Phi_2=L_2I_2$. I would now naturally want to say "Now just multiply $N_1\Phi_1$ and $N_2\Phi_2$ both ways, and get the desired result!" i.e.,  $$(N_1\Phi_1)(N_2\Phi_2)=(MI_2)(MI_1)=(L_1I_1)(L_2I_2)\implies M^2=L_1L_2$$  But that's incorrect, because the response of the coils to themselves and to the other coil are different quantities (i.e. not necessarily equal).  Realizing this, I write down the more general equations.  $$N_1\Phi_1=MI_2+L_1I_1$$ $$N_2\Phi_2=MI_1+L_2I_2$$  Now even with the hypothesis that $\Phi_1=\Phi_2=\Phi$, I still can't see how the identity $M^2=L_1L_2$ follows. Could you help clear up my confusion?  *Taken from D. Griffiths Introduction to Electrodynamics (4th ed.), Problem 7.58. How long would the conference be for? I want to know if this is grammatically correct " How long would the conference be for?". No connection could be made because the target machine actively refused it (PHP / WAMP) Let $f:\mathbb{R}\to\mathbb{R}$, $f(kx)=f(x)$. Show that if $f$ is continuous at $0$ then $f$ is a Constant function Let $f:\mathbb{R}\to\mathbb{R}$ , there is $k \in \mathbb{R}$, $|k|\ne 1$, so that for every $x\in \mathbb{R} , f(kx)=f(x).$ Show that if $f$ continuous at $0$ then $f$ is a constant function.   I tried to find something to show it from the definition of continuous function but I didn't know how to continue. We have that $f$ continuous at $0$ if for every $\epsilon&gt;0$ there is $\delta&gt;0$ such that $|x-0|&lt;\delta \implies |f(x)-f(0)|&lt;\epsilon$ And if $f$ not continuous on $0$ is $f$ still a constant function? Does a shortcode with a single attribute have to use an array? permission denied BASH I get the following error when trying to ls *.txt | wc -l a directory that contains many files:  -bash: /bin/ls: Argument list too long   Does the threshold of this "Argument list" dependent on distro or computer's spec?  Usually, I'd pipe the result of such big result to some other commands (wc -l for example), so I'm not concerned with limits of the terminal. Is it bad to defrag a ssd? Or just pointless? I've read on some websites that it's bad to defrag an ssd. But I'm reading on this website now and one person was like "You didnt do any harm its just not necessary to defrag it." I feel like that's wrong since every other website I've read says it will damage the ssd.