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  • is the + in += on a Map a prefix operator of =?

    - by Steve
    In the book "Programming in Scala" from Martin Odersky there is a simple example in the first chapter: var capital = Map("US" -> "Washington", "France" -> "Paris") capital += ("Japan" -> "Tokyo") The second line can also be written as capital = capital + ("Japan" -> "Tokyo") I am curious about the += notation. In the class Map, I didn't found a += method. I was able to the same behaviour in an own example like class Foo() { def +(value:String) = { println(value) this } } object Main { def main(args: Array[String]) = { var foo = new Foo() foo = foo + "bar" foo += "bar" } } I am questioning myself, why the += notation is possible. It doesn't work if the method in the class Foo is called test for example. This lead me to the prefix notation. Is the + a prefix notation for the assignment sign (=)? Can somebody explain this behaviour?

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  • Sending message to multiple contacts of mobile by providing search facility in J2ME

    - by learn
    I wan to send the message to multiple contacts in the contactlist for(int j=0;j<vector.size();j++){ listofContacts=new ListofContacts(); listofContacts=(ListofContacts)vector.elementAt(j); list.setFitPolicy(1); list.append(listofContacts.contactname + " "+ listofContacts.contactno,null); System.out.println(listofContacts.contactname + " "+ listofContacts.contactno); } here i have taken all the contacts of contact list in vector and the listofcontacts is the class containing the name and number. To show the list of contacts for selection i am using list control with multiple choice. The code is working fine and message is sent to all the contacts which are selected by the user but as we know there may be 1000 of contacts in phonebook and in these case to select a particular user we have to scroll down the list. Now how to keep the search facility so that we can directly go to the required contact and if it is not possible with the list control which control is to be used so that multiple contacts can be selected and also search facility is available.

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  • Java Collections.Rotate with an array doesn't work

    - by steve_72
    I have the following java code: import java.util.Arrays; import java.util.Collections; public class Test { public static void main(String[] args) { int[] test = {1,2,3,4,5}; Collections.rotate(Arrays.asList(test), -1); for(int i = 0; i < test.length; i++) { System.out.println(test[i]); } } } I want the array to be rotated, but the output I get is 1 2 3 4 5 Why is this? And is there an alternative solution?

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  • question on HyperOperation

    - by davit-datuashvili
    i am trying to solve following recurence program http://en.wikipedia.org/wiki/Hyper_operator here is my code i know it has mistakes but i have done what i could public class hyper{ public static int Hyper(int a,int b,int n){ int t=0; if ( n==0) return b+1; if ((n==1) &&(b==0)) return a; if ((n==2) && (b==0)) return 0; if ((n>=3) && (b==0)) return 1; t=Hyper(a,b-1,n); return Hyper (a,t,n-1); } public static void main(String[]args){ int n=3; int a=5; int b=7; System.out.println(Hyper(a,b,n)); } } please help

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  • SEND VALUE TO SERVLET WITH THE JAVASCRIPT

    - by kawtousse
    hi everyone, In my JavaScript function I do like this in order to redirect parameters to servlet: var ids1=document.getElementById("projet").value; document.location.href("http://localhost:8080/Opc_Web_App/ServletAffectation?ids1="+ids1); and in the servlet I do the following to get Value: String idprojet= request.getParameter("ids1"); System.out.println("le projet selectionné est :" +idprojet); the problem that i didn't have the result of System.out.print in my screen; so in other terms the servlet didn't get the parameter. I can not see the problem until now. Please help. Thank you.

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  • Why doesn't list.get(0).equals(null) work?

    - by Jessy
    The first index is set to null (empty), but it doesn't print the right output, why? //set the first index as null and the rest as "High" String a []= {null,"High","High","High","High","High"}; //add array to arraylist ArrayList<Object> choice = new ArrayList<Object>(Arrays.asList(a)); for(int i=0; i<choice.size(); i++){ if(i==0){ if(choice.get(0).equals(null)) System.out.println("I am empty"); //it doesn't print this output } }

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  • Send value to servlet with javascript

    - by kawtousse
    hi everyone, In my JavaScript function I do like this in order to redirect parameters to servlet: var ids1=document.getElementById("projet").value; document.location.href("http://localhost:8080/Opc_Web_App/ServletAffectation?ids1="+ids1); and in the servlet I do the following to get Value: String idprojet= request.getParameter("ids1"); System.out.println("le projet selectionné est :" +idprojet); the problem that i didn't have the result of System.out.print in my screen; so in other terms the servlet didn't get the parameter. I can not see the problem until now. Please help. Thank you.

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  • i have done code so please help

    - by davit-datuashvili
    public class bitap{ public static void main(String[]args){ String text="tbillisi"; String pattern="tbilxiri"; int k=2; int m=pattern.length(); long pattern_mask[]=new long[Character.MAX_VALUE+1]; String result=""; boolean[]R=new boolean[m+1]; long i,d; for (i=0;i<=k;i++){ R[i]=~1; } for (i=0;i if (0==(R[k]& (1< System.out.println(result); } } http://en.wikipedia.org/wiki/Bitap_algorithm from this site

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  • List files starting with a specific name using java

    - by user3610075
    i want to list files starting with a name like "Report" from a folder. i found this in google to list all files but i don't how to list file starting with a name. Thank you File directory = new File("C:\\Users\\kiki\\Downloads"); File[] files = directory.listFiles(); for (int index = 0; index < files.length; index++) { //Print out the name of files in the directory System.out.println(files[index].toString()); }

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  • Functions without arguments, with unit as argument in scala

    - by scout
    def foo(x:Int, f:Unit=>Int) = println(f()) foo(2, {Unit => 3+4} //case1 def loop:Int = 7 foo(2, loop) //does not compile changing loop to //case 2 def loop():Int = 7 foo(2, loop) // does not compile changing loop to //case 3 def loop(x:Unit): Int = 7 //changing according to Don's Comments foo(2,loop) // compiles and works fine should'nt case 1 and case 2 also work? why are they not working? defining foo as def foo(x:Int, y:()=>Int) then case 2 works but not case 1. Arent they all supposed to work, defining the functions either way. //also i think ()=Int in foo is a bad style, y:=Int does not work, comments??

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  • Trying to use a list iterator to print out entire linked list in Java. Infinite loop for some reaso

    - by Matt
    I created my list: private static List list = new LinkedList(); and my iterator: ListIterator itr = list.listIterator(); and use this code to try to print out the list... Only problem is, it never comes out of the loop. When it reaches the tail, shouldn't it come out of the loop, because there is no next? Or is it going back to the head like a circular linked list? It is printing so quickly and my computer locks up shortly after, so I can't really tell what is going on. while (itr.hasNext()) System.out.println(itr.next());

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  • How do I setup Eclipse to stop on the line an exception occured?

    - by Chris Persichetti
    Hi, How can I setup Eclipse to stop at the point an exception occurred. I have an Eclipse breakpoint setup to break on an exception. In the code example below, the problem I'm having is Eclipse tries to open the Integer source code. Is there any way to just have debugger break at the point shown in my code example? If I move down the stack trace, I will get to this line, it'd be nice if there's a way to do this without the "Source not found" window coming up. This can be done in Visual Studio, so it's driving me crazy not being able to find a way to do this in Eclipse. package com.test; public class QuickTest { public static void main(String[] args) { try { test(); } catch(NumberFormatException e) { System.out.println(e.getMessage()); } } private static void test() { String str = "notAnumber"; Integer.parseInt(str);//<----I want debugger to stop here } }

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  • Will Sytem.currentTimeMillis always return a value >= previous calls?

    - by 1984isnotamanual
    http://java.sun.com/j2se/1.4.2/docs/api/java/lang/System.html#currentTimeMillis() says: Returns the current time in milliseconds. Note that while the unit of time of the return value is a millisecond, the granularity of the value depends on the underlying operating system and may be larger. For example, many operating systems measure time in units of tens of milliseconds. It is not clear to me if I am guaranteed that this code will always print ever increasing (or the same) numbers. while (1) { System.out.println(System.currentTimeMillis() ); }

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  • C++ string how to

    - by typoknig
    This is a very simple question and I feel stupid for asking it, but I am pressed for time and I need to figure it out :) I just need to know how to make a string that contains text and other variables. For instance in Java I can just do this: String someString; for(int i = 0; i>10; i++){ someString = ("this text has printed " + i + " times"); //how do I create this line in C++? System.out.println(someString); i++; }

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  • the PrintWriter servlet Buffer and displayind data from a jsp

    - by nabilaloui
    Hello all, I need really your help please. What I do is to build a table in html tags in my servlet then when trying to send this table to a servlet for the display using: Response.sendRedirect this did not work. I have an error but I don't know the cause. I search to how do it since I use: response.setContentType("text/html"); PrintWriter out = response.getWriter(); out.println("<...");.... thinks a lot for your help

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  • Can a thread call wait() on two locks at once in Java (6)

    - by Dr. Monkey
    I've just been messing around with threads in Java to get my head around them (it seems like the best way to do so) and now understand what's going on with synchronize, wait() and notify(). I'm curious about whether there's a way to wait() on two resources at once. I think the following won't quite do what I'm thinking of: synchronized(token1) { synchronized(token2) { token1.wait(); token2.wait(); //won't run until token1 is returned System.out.println("I got both tokens back"); } } In this (very contrived) case token2 will be held until token1 is returned, then token1 will be held until token2 is returned. The goal is to release both token1 and token2, then resume when both are available (note that moving the token1.wait() outside the inner synchronized loop is not what I'm getting at). A loop checking whether both are available might be more appropriate to achieve this behaviour (would this be getting near the idea of double-check locking?), but would use up extra resources - I'm not after a definitive solution since this is just to satisfy my curiosity.

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  • Why javabeans framework create the IndexedPropertyDescriptor for the NON index method

    - by George Macus
    I'm not familiar with java beans framework, in the below scenario, I got the IndexedPropertyDescriptor for the method getFooWithX, could someone explain why? public class IntrospectorTest { public static void main(String[] args) throws IntrospectionException { BeanInfo info = Introspector.getBeanInfo(SubClass.class); PropertyDescriptor[] descriptors = info.getPropertyDescriptors(); for (int i = 0; i < descriptors.length; i++) { System.out.println(descriptors[i].getClass().getName() + ":" + descriptors[i].getName()); } } } abstract class BaseClass { public abstract Object getFoo(); } abstract class SubClass extends BaseClass { public Object getFooWithX(int x) { return null; } } and the result will be: java.beans.PropertyDescriptor:class java.beans.PropertyDescriptor:foo java.beans.IndexedPropertyDescriptor:fooWithX Why?

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  • How to sort an array by (smallest, largest, second smallest, second, largest) etc?

    - by Binka
    Any ideas? I can sort an array. But not in this pattern? It needs to sort by the pattern I mentioned above. public void wackySort2(int[] nums) { int sign = 0; int temp = 0; int temp2 = 0; for (int i = 0; i < nums.length; i++) { for (int j = 0; j < nums.length - 1; j++) { if (nums[j] > nums[j + 1]) { temp = nums[j]; nums[j] = nums[j + 1]; nums[j + 1] = temp; //sign = 1; System.out.println("Something has been done"); } } } }

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  • What does this java output mean?!

    - by Phil
    public class Arrys { private int[] nums; //Step 3 public Arrys (int arrySize) { nums = new int[arrySize]; } public int [] getNums (){ return nums; } } Test class: public class TestArrys { public static void main(String args[]) { //Step 4 Arrys arry = new Arrys(10); System.out.println("\nStep4 "); for(int index = 0; index < arry.getNums().length; index++) { System.out.print(arry.getNums()); } } } It's incredibly simple, that is why I think I'm doing something fundamentally wrong. All I want is to display the value of the array. This is what I get back. I am totally lost, there is nothing in my book that explains this nor does googling it help. Step4 [I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440[I@1ac88440

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  • please help me to solve problem

    - by davit-datuashvili
    i have operation on heap fixdown operation below is code public class HEap{ public static void fixdown(int a[],int k,int n){ while(2*k<=n){ int j=2*k; if (j<n && a[j]<a[j+1]) j++; if (!(a[k]<a[j])) break; int t=a[k]; a[k]=a[j]; a[j]=k; k=j; } } public static void main(String[]args){ int a[]=new int[]{12,15,20,29,23,22,17,40,26,35,19,51}; fixdown(a,1,a.length); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } } } //and result is 12 29 20 40 23 22 17 3 26 35 19 51 i am interested why is 3 in the list?in the array is not

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  • How to check null element if it is integer array in Java?

    - by masato-san
    I'm quite new to Java and having an issue checking null element in integer array. I'm using Eclipse for editor and the line that checks null element is showing error: Line that complains: if(a[i] != null) { Error msg from Eclipse: The operator != is undefined for the argument type(s) int, null In PHP, this works without any problem but in Java it seems like I have to change the array type from integer to Object to make the line not complain (like below) Object[] a = new Object[3]; So my question is if I still want to declare as integer array and still want to check null, what is the syntax for it? Code: public void test() { int[] a = new int[3]; for(int i=0; i<a.length; i++) { if(a[i] != null) { //this line complains... System.out.println('null!'); } } }

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  • How to validate phone number(US format) in Java?

    - by Maxood
    I just want to know where am i wrong here: import java.io.*; class Tokens{ public static void main(String[] args) { //String[] result = "this is a test".split(""); String[] result = "4543 6546 6556".split(""); boolean flag= true; String num[] = {"0","1","2","3","4","5","6","7","8","9"}; String specialChars[] = {"-","@","#","*"," "}; for (int x=1; x<result.length; x++) { for (int y=0; y<num.length; y++) { if ((result[x].equals(num[y]))) { flag = false; continue; } else { flag = true; } if (flag == true) break; } if (flag == false) break; } System.out.println(flag); } }

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  • Args error in main method for client-server program

    - by socket
    Hi I have a client and server program, all the coding is done and compiles, the client has a GUI and the server is command line. The program uses sockets. But when I run the client to connect to the server it keeps coming with the error message: "Usage: TodoClient []", rather than connecting to the server and starting up. This is where the problem lies: public static void main(String[] args) { TodoClient client; if (args.length > 2 || args.length == 0) { System.err.println("Usage: TodoClient <host> [<port>]"); } else if (args.length == 1) { client = new TodoClient(args[0], DEFAULT_PORT); } else { client = new TodoClient(args[0], Integer.parseInt(args[1])); } } Thank You

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  • is this code correct? [closed]

    - by davit-datuashvili
    hi i have poste this code from this title http://stackoverflow.com/questions/2896363/hi-i-have-question-here-is-pseudo-code-about-sift-up-and-sift-down-on-heaps i have following code of siftup on heap is it correct?i have put here because i have changed at old place my question and it became unreadable so i have posted here public class siftup{ public static void main(String[]args){ int p; int n=12; int a[]=new int[]{15,20,12,29,23,17,22,35,40,26,51,19}; int i=n-1; while (i!=0){ if (i==1) break; p=i/2; if (a[p]<=a[i]){ int t=a[p]; a[p]=a[i]; a[i]=t; } i=p; } for (int j=0;j<n;j++){ System.out.println(a[j]); } } } //result is this 15 20 19 29 23 12 22 35 40 26 51 17 is it correct?

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  • question about Tetration

    - by davit-datuashvili
    i have question how write program which calculates following procedures http://en.wikipedia.org/wiki/Tetration i have exponential program which returns x^n here is code public class Exp{ public static long exp(long x,long n){ long t=0; if (n==0){ t= 1; } else{ if (n %2==0){ t= exp(x,n/2)* exp(x,n/2); } else{ t= x*exp(x,n-1); } } return t; } public static void main(String[]args){ long x=5L; long n=4L; System.out.println(exp(x,n)); } } but how use it in Tetration program?please help

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