Java thread implementation vs declaration i'm currently studying Java and i want to know what is the difference between implementing Runnable to a class and declaring a new thread and executing what you need inside.   Thread t = new Thread(){      public void run() {     //code here     } }; t.start();   vs  Thread t = new Thread(My_class_with_runnable_implementation); t.start();
"implements Runnable" vs "extends Thread" in Java From what time I've spent with threads in Java, I've found these two ways to write threads:  With implements Runnable:  public class MyRunnable implements Runnable {     public void run() {         //Code     } } //Started with a "new Thread(new MyRunnable()).start()" call   Or, with extends Thread:  public class MyThread extends Thread {     public MyThread() {         super("MyThread");     }     public void run() {         //Code     } } //Started with a "new MyThread().start()" call   Is there any significant difference in these two blocks of code ?
How can I know if a branch has been already merged into master? I have a git repository with multiple branches.   How can I know which branches are already merged into the master branch? I want to be able to  write the following in LaTeX      We make the further assumption that \begin{equation} \label{eqn:assumption} There is a bounded linear operator     $B:X\to X^*$ such that $B^*=B$ and the operator $JB$ is an extension of $T'(0)$\end{equation}   and get a nice output, in such a way that this whole chunk of text along with the math can be referenced later. Of course I can write text inside the equation environment using \text{}, but when I tried this the spacing was all wrong. Can anyone suggest a solution to this? Why does 64-bit Windows need a separate "Program Files (x86)" folder? Can I get a rebate after using my HSA? This is Exercise 3.1.A. in      Suppose that $\pi: X\rightarrow Y$ is a continuous map of differentiable manifolds. Show that $\pi$ is differentiable if differentialble functions pull back to differentiable functions, i.e., if pullback by $\pi$ gives mpa $\mathcal{O}_Y\rightarrow \pi_*\mathcal{O}_X$.   Let $f: V\subseteq Y\rightarrow \mathbb{R}$ be a differentiable function on an open subset of $Y$. Let $(U,\phi)$ be a chart of $X$, $(V,\psi)$ be a chart of $Y$. Then $f\circ\pi(\phi^{-1})$ is differentiable and $f\circ\psi^{-1}$ is differentiable. We need to show that $\psi\circ\pi\circ\phi^{-1}$ is differentiable. I don't know how to prove this.   Can we say that since $(U,\psi\circ\pi)$ constructs a chart of $X$, by the compatibility, $\psi\circ\pi\circ\phi^{-1}$ is differentiable? This seems true but this does not use the fact that $f$ is differentiable.  Sorry I have no background of Differentiable Manifolds. Any help would be appreciated.  I also saw this post: , but I don't understand how $\gamma$ is smooth in the answer. $\gamma=\beta\circ\rho$, where $\rho$ is defined to be smooth, but $\beta$ is not. Each package name is in the form: package-XubuntuY*_amd64(i386)  So my questions are:   What does the X stand for? What does the Y stand for?   for example binutils_2.22-6ubuntu1.3_amd64  What is the meaning of the numbers 6 and 1.3 How can I get a complete list of fields for a standard object? Root cause for "Entity is not org-accessible" for sharing object? How to mix three different shaders on the same material? Should we always override equals? How to install a LibreOffice Language Tools (Spelling Check, Thesaurus...) What are the steps to install the Spelling Check and other language supports for LibreOffice. For example spanish, french, italian, etc.. It is always told as a fact without explaining the reason. Why do two objects get charged by rubbing? Why one object get negative charge and other get positive ? Why didn't Elizabeth get cursed? Clearing all cookies with JavaScript I want to buy a home, but I'm not interested in using a realtor as a buyer's agent.  I know that in MLS systems, when a realtor lists a house they specify what commission a buyer's agent will be paid.  Since I am the one who is bringing a qualified buyer to the listing agent, I feel that I would should get the commission (even if it is in the form of a discount on the house).    Essentially, the seller would get the same amount of money at closing... the listing agent would get the same commission as they normally would for showing the house to someone with a buyer's agent... but I would pay less for the house.  Is there a tactful way to negotiate this with the listing realtor?  For example, should I make a side contract with the realtor to pay me the commission at closing?  Or perhaps I can specify in my offer that they should reduce their fees to the seller by the commission amount if the offer is accepted, in order to make my offer more attractive?  Are these types of deals ever done, and if so, is there a conventional way to make it happen?  Thanks What is the relationship between symmetry and degeneracy in quantum mechanics? Inverting direction of mouse scroll wheel Is there a way to make a mouse scroll wheel invert its direction? I'd like to scroll upwards and have that action scroll downwards and vice-versa. There is no setting on the Mouse control panel that makes this possible.   Any pointers to a hack or a particular mouse model that has such a setting would be appreciated. I am using Windows 7. How to pass javascript value to controller..? I want to pass the java-script value to controller.I am using repeat on VF page.  VF page      &lt;apex:repeat value="{!listAdj}" var="Adj" id="hrs"&gt;       &lt;td&gt;&lt;div class="requiredInput"&gt;&lt;div class="requiredBlock"&gt;&lt;/div&gt;     &lt;apex:inputField id="decimal" value="{!Adj.Hours_Decimal__c}" onchange="setHidden5(this);" &gt;       &lt;/apex:inputField&gt;     &lt;/div&gt;&lt;/td&gt;             &lt;apex:inputField id="hours" value="{!Adj.Hours__c}" /&gt; &lt;apex:inputField id="decimal" value="{!Adj.Hours_Decimal__c}" onchange="setHidden5(this);"/&gt;              &lt;/apex:repeat&gt;   Javascript code  function setHidden5(element){     //split the id of the inputField by ':' and then use the 5th index in the array to exact position of the inputField in the apex:repeat.   var idArray = element.id.split(':');     var hiddenRep1 =document.getElementById("thePage:form:hrs:hrs:"+idArray[4]+":decimal");          var hiddenRep2 = document.getElementById("thePage:form:hrs:hrs:"+idArray[4]+":hours");       if(element.value.length &gt;=1){        document.getElementById("thePage:form:hrs:hrs:"+idArray[4]+":hours").disabled = true; **hiddenRep2.value=hidden1.value;**                }                             }     }   I want to pass the hiddenRep2.value to controller.How to pass that. Hidden2.value is same as hidden1.value + some addition subtraction.i dont know how to pass that value. Can someone me sort out this.  Thanks Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.  $$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$  Perhaps someone can help me out I don't understand how to move forward from here: $$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$ proof and explanation would greatly be appreciated :) Thanks :)  EDIT sorry meant GE not = fixed :) An unclear sentence in the book "Griffiths, Harris-Principles of algebraic geometry" Let $X$ be a complex manifold and consider a holomorphic vector bundle $\pi:E\to X$ of rank $r$. By the usual correspondence between vector bundles and locally free sheaves, we consider $E$ as a sheaf by abuse of notation. We denote with $\mathscr A_X^{p,q}$ the sheaf of $C^\infty$ $(p,q)$-differential forms on $X$ and we define: $$A^{p,q}(E):=\Gamma(X,\mathscr A^{p,q}\otimes_{\mathscr O_X} E)$$ $$A^k(E):=\bigoplus_{p+q=k}A^{p,q}(E)$$  Now suppose that $\nabla:A^\bullet(E)\to A^\bullet(E)$ is a linear connection on $E$, then the curvature operator is defined as: $$\nabla^2:A^0(E)\to A^2(E)$$ where $\nabla^2$ is the composition of the maps $\nabla: A^0(E)\to A^2(E)$   and $\nabla: A^1(E)\to A^2(E)$. One can show that $\nabla^2$ is $C^\infty(X)$-linear.    I don't understand the following sentence from the Griffiths-Harris book at page 75 (I rephrase it by using my notation):     Since $\nabla^2$ is $C^\infty(X)$-linear, then it is induced by a bundle map $E\to\bigwedge^2T^*X\otimes E$, or in other words, $\nabla^2$ corresponds to a global section $\Theta$ of the bundle $\bigwedge^2T^*X\otimes\operatorname{Hom}(E,E)$.   I absolutely don't understand how I can see the operator $\nabla^2$ as the section $\Theta$.