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  • C++ Template problem adding two data types

    - by Sara
    I have a template class with an overloaded + operator. This is working fine when I am adding two ints or two doubles. How do I get it to add and int and a double and return the double? template <class T> class TemplateTest { private: T x; public: TemplateTest<T> operator+(const TemplateTest<T>& t1)const { return TemplateTest<T>(x + t1.x); } } in my main function i have void main() { TemplateTest intTt1 = TemplateTest<int>(2); TemplateTest intTt2 = TemplateTest<int>(4); TemplateTest doubleTt1 = TemplateTest<double>(2.1d); TemplateTest doubleTt2 = TemplateTest<double>(2.5d); std::cout << intTt1 + intTt2 << /n; std::cout << doubleTt1 + doubleTt2 << /n; } I want to be able to also do this std::cout << doubleTt1 + intTt2 << /n;

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  • Chache problem running two consecutive HTTP GET requests from an APP1 to an APP2

    - by user502052
    I use Ruby on Rails 3 and I have 2 applications (APP1 and APP2) working on two subdomains: app1.domain.local app2.domain.local and I am tryng to run two consecutive HTTP GET requests from APP1 to APP2 like this: Code in APP1 (request): response1 = Net::HTTP.get( URI.parse("http://app2.domain.local?test=first&id=1") ) response2 = Net::HTTP.get( URI.parse("http://app2.domain.local/test=second&id=1") ) Code in APP2 (response): respond_to do |format| if <model_name>.find(params[:id]).<field_name> == "first" <model_name>.find(params[:id]).update_attribute ( <field_name>, <field_value> ) format.xml { render :xml => <model_name>.find(params[:id]).<field_name> } elsif <model_name>.find(params[:id]).<field_name> == "second" format.xml { render :xml => <model_name>.find(params[:id]).<field_name> } end end After the first request I get the correct XML (response1 is what I expect), but on the second it isn't (response2 isn't what I expect). Doing some tests I found that the second time that <model_name>.find(params[:id]).<field_name> run (for the elsif statements) it returns always a blank value so that the code in the elseif statement is never run. Is it possible that the problem is related on caching <model_name>.find(params[:id]).<field_name>? P.S.: I read about eTag and Conditional GET, but I am not sure that I must use that approach. I would like to keep all simple.

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  • Why is CoRegisterClassObject creating two extra threads?

    - by Stijn Sanders
    I'm trying to fix a problem that only recently happens on a number of machine's on a VPN. They each run a client application I wrote that exposes a COM automation object. For some strange reason I haven't been able to discover yet, one thread in the application takes up all of the available CPU time, slowing other operation on the machine. In observing the application's strange behaviour, I've noticed it's the third thread started, and if I debug on my machine I notice the first call to CoRegisterClassObject created two extra threads. If the second of these two threads is the one that gets into an infinite loop, I'm not at all shure how to fix this. Where could I check next about what's wrong? Could it have started by one of the recent patches rolled out by Microsoft this last 'patch tuesday'? I had a go with ProcessExplorer to extract a stack trace of the thread: ntoskrnl.exe!ExReleaseResourceLite+0x1a3 ntoskrnl.exe!PsGetContextThread+0x329 WLDAP32.dll!Ordinal325+0x1231 WLDAP32.dll!Ordinal325+0x129e WLDAP32.dll!Ordinal325+0x1178 ntdll.dll!LdrInitializeThunk+0x24 ntdll.dll!LdrShutdownThread+0xe9 kernel32.dll!ExitThread+0x3e kernel32.dll!FreeLibraryAndExitThread+0x1e ole32.dll!StringFromGUID2+0x65d kernel32.dll!GetModuleFileNameA+0x1ba

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  • How to combine two separate unrelated Git repositories into one with single history timeline

    - by Antony
    I have two unrelated (not sharing any ancestor check in) Git repositories, one is a super repository which consists a number of smaller projects (Lets call it repository A). Another one is just a makeshift local Git repository for a smaller project (lets call it repository B). Graphically, it would look like this A0-B0-C0-D0-E0-F0-G0-HEAD (repo A) A0-B0-C0-D0-E0-F0-G0-HEAD (remote/master bare repo pulled & pushed from repo A) A1-B1-C1-D1-E1-HEAD (repo B) Ideally, I would really like to merge repo B into repo A with a single history timeline. So it would appear that I originally started project in repo A. Graphically, this would be the ideal end result A0-A1-B1-B0-D1-C0-D0-E0-F0-G0-E1-H(from repo B)-HEAD (new repo A) A0-A1-B1-B0-D1-C0-D0-E0-F0-G0-E1-H(from repo B)-HEAD (remote/master bare repo pulled & pushed from repo A) I have been doing some reading with submodules and subtree (Pro Git is a pretty good book by the way), but both of them seem to cater solution towards maintaining two separate branch with sub module being able to pull changes from upstream and subtree being slightly less headache. Both solution require additional and specialized git commands to handle check ins and sync between master and sub tree/module branch. Both solution also result in multiple time-lines (with --squash you even get 3 timelines with subtree). The closest solution from SO seems to talk about "graft", but is that really it? The goal is to have a single unified repository where I can pull/push check-ins, so that there are no more repo B, just repo A in the end.

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  • Seperating two graphs based on connectivity and coordinates

    - by martin
    I would like to separate existing data of vertices and edges into two or more graphs that are not connected. I would like to give the following as example: Imagine two hexagons on top of each other but are lying in different Z. Hexagon 1 has the following vertices A(0,0,1), B(1,0,2), C(2,1,2), D(1,2,1), E(0,2,1), F(-1,2,1). The connectivity is as following: A-B, B-C, C-D, D-E, E-F, F-A. This part of Graph 1 as all the vertices are connected in this layer. Hexagon2 has the following vertices A1(0,0,6), B1(1,0,7), C1(2,1,7), D1(1,2,8), E1(0,2,7), F1(-1,2,6). The connectivity is as following: A1-B1, B1-C1, C1-D1, D1-E1, E1-F1, F1-A1. This is part of Graph 2 My data is in the following form: list of Vertices and list of Edges that i can form graphs with. I would like to eliminate graph 2 and give only vertices and connectivity of graph 1 to polygon determination part of my algorithm. My real data contains around 1000 connected polygons as graph 1 and around 100 (much larger in area) polygons as graph 2. I would like to eliminate graph 2. I am programming this in python. Thanks in advance.

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  • SQL Binary Microsoft Access - Combining two tables if specific field values are equal

    - by Jordan
    I am new to Microsoft Access and SQL but have a decent programming background and I believe this problem should be relatively simple. I have two tables that I have imported into Access. I will give you a little context. One table is huge and contains generic, global data. The other table is still big but contains specific, regional data. There is only one common field (or column) between the two tables. Let’s call this common field CF. The other fields in both tables are different. I’ll take you through one iteration of what I need to do. I need to take each CF value in the regional, smaller table and find the common CF value in the larger, global table. After finding the match, I need to take the whole “record” or “row” from the global data and copy it over to the corresponding record in the smaller regional table (This should involve creating the new fields). I need to do this for all CF values in the regional, smaller table. I was recommended to use SQL and a binary search, but I am unfamiliar. Let me know if you have any questions. I appreciate the help!

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  • jQuery select by two name roots and perform one of two function depending on which root was selected

    - by RetroCoder
    I'm trying to get this code to work in jQuery and I'm trying to make sure that for each iteration of each root element, its alternate root element for that same iteration doesn't contain anything. Otherwise it sets the .val("") property to an empty string. Looking for a simple solution if possible using search, find, or swap. Each matching number is on the same row level and the same iteration count. I have two input types of input text elements with two different root names like so: 1st Root is "rootA" <input type="text" name="rootA1" /> <input type="text" name="rootA2 /> <input type="text" name="rootA3" /> 2nd Root is "rootB" <input type="text" name="rootB1" /> <input type="text" name="rootB2 /> <input type="text" name="rootB3" /> On blur if any of rootA is called call function fnRootA();. On blur if any of rootB is called call function fnRootB();. Again, I'm trying to make sure that for each iteration like 1 that the alternate root doesn't contain anything, else it sets the .val("") property to an empty string of the root being blurred. My current code works for a single element but wanted to use find or search but not sure how to construct it.. $("input[name='rootA1']").blur(function(e) { fnRootA(1); // this code just removes rootA1's value val("") //if rootB1 has something in it value property // the (1) in parenthesis is the iteration number });

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  • Synchronize write to two collections

    - by glaz666
    I need to put some value to maps if it is not there yet. The key-value (if set) should always be in two collections (that is put should happen in two maps atomically). I have tried to implement this as follows: private final ConcurrentMap<String, Object> map1 = new ConcurrentHashMap<String, Object>(); private final ConcurrentMap<String, Object> map2 = new ConcurrentHashMap<String, Object>(); public Object putIfAbsent(String key) { Object retval = map1.get(key); if (retval == null) { synchronized (map1) { retval = map1.get(key); if (retval == null) { Object value = new Object(); //or get it somewhere synchronized (map2) { map1.put(key, value); map2.put(key, new Object()); } retval = value; } } } return retval; } public void doSomething(String key) { Object obj1 = map1.get(key); Object obj2 = map2.get(key); //do smth } Will that work fine in all cases? Thanks

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  • Difference between two PHP times as years, months and days for PHP version 5.2

    - by Dominor Novus
    Forward: I've scanned through the existing questions/answers on this matter. This is not a duplicitous question; I cannot find a working solution from the accepted answers. The main questions/answers I've reviewed can be found here: How to calculate the difference between two dates using PHP? What I need: A calucalation of the difference between two dates expressed as years, months and days that works with PHP version: 5.2. <?php $current_date = date('d-M-Y'); $future_date = '2012-11-01'; ?> What I've tried: Most answers I find online don't seem to exact in that they don't factor in leap years. This highly rated answer won't work because DateTime-diff() is php 5.3+. This accepted answer results in (i.e. the second block of code aimed at PHP 5.2) results in the following being parsed: Array ( [y] = 25 [m] = 11 [d] = 7 [h] = 3 [i] = 15 [s] = 19 [invert] = 0 [days] = 9473 ) Array ( [y] = 25 [m] = 11 [d] = 7 [h] = 3 [i] = 15 [s] = 19 [invert] = 1 [days] = 9473 ) I can't tell if I've incorrectly applied the code or it's simply a case of me not knowing how to manipulate the array.

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  • How to combine two rows and calculate the time difference between two timestamp values in MySQL?

    - by Nadar
    I have a situation that I'm sure is quite common and it's really bothering me that I can't figure out how to do it or what to search for to find a relevant example/solution. I'm relatively new to MySQL (have been using MSSQL and PostgreSQL earlier) and every approach I can think of is blocked by some feature lacking in MySQL. I have a "log" table that simply lists many different events with their timestamp (stored as datetime type). There's lots of data and columns in the table not relevant to this problem, so lets say we have a simple table like this: CREATE TABLE log ( id INT NOT NULL AUTO_INCREMENT, name VARCHAR(16), ts DATETIME NOT NULL, eventtype VARCHAR(25), PRIMARY KEY (id) ) Let's say that some rows have an eventtype = 'start' and others have an eventtype = 'stop'. What I want to do is to somehow couple each "startrow" with each "stoprow" and find the time difference between the two (and then sum the durations per each name, but that's not where the problem lies). Each "start" event should have a corresponding "stop" event occuring at some stage later then the "start" event, but because of problems/bugs/crashed with the data collector it could be that some are missing. In that case I would like to disregard the event without a "partner". That means that given the data: foo, 2010-06-10 19:45, start foo, 2010-06-10 19:47, start foo, 2010-06-10 20:13, stop ..I would like to just disregard the 19:45 start event and not just get two result rows both using the 20:13 stop event as the stop time. I've tried to join the table with itself in different ways, but the key problems for me seems to be to find a way to correctly identify the corresponding "stop" event to the "start" event for the given "name". The problem is exactly the same as you would have if you had table with employees stamping in and out of work and wanted to find out how much they actually were at work. I'm sure there must be well known solutions to this, but I can't seem to find them...

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  • C# property exactly the same, defined in two places

    - by Sarah Vessels
    I have the following classes: Defect - represents a type of data that can be found in a database FilterQuery - provides a way of querying the database by setting simple Boolean filters Both Defect and FilterQuery implement the same interface: IDefectProperties. This interface specifies particular fields that are in the database. Different classes have methods that return lists of Defect instances. With FilterQuery, you specify some filters for the particular properties implemented as part of IDefectProperties, and then you run the query and get back a list of Defect instances. My problem is that I end up implementing some properties exactly the same in FilterQuery and Defect. The two are inherently different classes, they just share some of the same properties. For example: public DateTime SubmitDateAsDate { get { return DateTime.Parse(SubmitDate); } set { SubmitDate = value.ToString(); } } This is a property required by IDefectProperties that depends on a different property, SubmitDate, which returns a string instead of a DateTime. Now SubmitDate is implemented differently in Defect and FilterQuery, but SubmitDateAsDate is exactly the same. Is there a way that I can define SubmitDateAsDate in only place, but both Defect and FilterQuery provide it as a property? FilterQuery and Defect already inherit from two different classes, and it wouldn't make sense for them to share an ancestor anyway, I think. I am open to suggestions as to my design here as well.

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  • C++ Recursive function that reverses the order of an array's indexes between two bounds

    - by CPT Kirk
    I am trying to write a recursive function that has three arguments; an array and two array indexes. The function should reverse the order of the values between the two indexes. I would like to understand what is happening instead of just being told an answer. Here is my code so far: #include <iostream> using namespace std; char switchAroo(char a[], int b1, int b2); int main() { char a[6] {'A', 'B', 'C', 'D', 'E', '\0'}; cout << a; switchAroo(a, 2, 5); return 0; } char switchAroo(char a [], int b1, int b2) { char temp; if (b1 == b2) cout << "The array after switchAroo is " << a << endl; else { temp = a[b1]; a[b1] = a[b2]; a[b2] = temp; b1++; b2--; return switchAroo(a, b1, b2); } } I am getting the following warning code: warning C4715: 'switchAroo' : not all control paths return a value Any help would be greatly appreciated.

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  • Bind two images together to be dragged

    - by Ryan Beaulieu
    I'm looking for some help with a script to drag two images together at once. I currently have a script that allows me to drag thumbnail images into a collection bin to be saved later. However, some of my thumbnails have an image positioned over the top of them to represent these thumbnail images as "unknown" plants. I was wondering if someone could point me in the right direction as to how I would go about binding these two images together to be dragged. Here is my code: $(document).ready(function() { var limit = 16; var counter = 0; $("#mainBin1, #mainBin2, #mainBin3, #mainBin4, #mainBin5, #bin_One_Hd, #bin_Two_Hd, #bin_Three_Hd, #bin_Four_Hd, #bin_Five_Hd").droppable({ accept: ".selector, .plant_Unknown", drop: function(event, ui) { counter++; if (counter == limit) { $(this).droppable("disable"); } $(this).append($(ui.draggable).clone()); $("#cbOptions").show(); $(".item").draggable({ containment: "parent", grid: [72,72], }); } }); $(".selector").draggable({ helper: "clone", revert: "invalid", revertDuration: 700, opacity: 0.75, }); $(".plant_Unknown").draggable({ helper: "clone", revert: "invalid", revertDuration: 700, opacity: 0.75, }); }); Any help would be greatly appreciated. Thanks. EDIT: Website

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  • initalizing two pointers to same value in "for" loop

    - by MCP
    I'm working with a linked list and am trying to initalize two pointers equal to the "first"/"head" pointer. I'm trying to do this cleanly in a "for" loop. The point of all this being so that I can run two pointers through the linked list, one right behind the other (so that I can modify as needed)... Something like: //listHead = main pointer to the linked list for (blockT *front, *back = listHead; front != NULL; front = front->next) //...// back = back->next; The idea being I can increment front early so that it's one ahead, doing the work, and not incrementing "back" until the bottom of the code block in case I need to backup in order to modify the linked list... Regardless as to the "why" of this, in addition to the above I've tried: for (blockT *front = *back = listHead; /.../ for (blockT *front = listHead, blockT *back = listHead; /.../ I would like to avoid pointer to a pointer. Do I just need to initialize these before the loop? As always, thanks!

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  • two view controllers and reusability with delegate

    - by netcharmer
    Newbie question about design patterns in objC. I'm writing a functionality for my iphone app which I plan to use in other apps too. The functionality is written over two classes - Viewcontroller1 and Viewcontroller2. Viewcontroller1 is the root view of a navigation controller and it can push Viewcontroller2. Rest of the app will use only ViewController1 and will never access Viewcontroller2 directly. However, triggered by user events, Viewcontroller2 has to send a message to the rest of the app. My question is what is the best way of achieving it? Currently, I use two level of delegation to send the message out from Viewcontroller2. First send it to Viewcontroller1 and then let Viewcontroller1 send it to rest of the app or the application delegate. So my code looks like - //Viewcontroller1.h @protocol bellDelegate -(int)bellRang:(int)size; @end @interface Viewcontroller1 : UITableViewController <dummydelegate> { id <bellDelegate> delegate; @end //Viewcontroller1.m @implementation Viewcontroller1 -(void)viewDidLoad { //some stuff here Viewcontroller2 *vc2 = [[Viewcontroller2 alloc] init]; vc2.delegate = self; [self.navigationController pushViewController:vc2 animated:YES]; } -(int)dummyBell:(int)size { return([self.delegate bellRang:size]); } //Viewcontroller2.h @protocol dummyDelegate -(int)dummyBell:(int)size; @end @interface Viewcontroller2 : UITableViewController { id <dummyDelegate> delegate; @end //Viewcontroller2.m @implementation Viewcontroller2 -(int)eventFoo:(int)size { rval = [self.delegate dummyBell:size]; } @end

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  • Syncing objects between two devices with different system times

    - by Mike Weller
    Hi there. I'm syncing objects between two devices. Objects have a lastModified property. If both devices have modified an object, then during the next sync the version of the object with the most recent lastModified is chosen on both devices. So we don't do fine-grained merging, only 'most recent version' merging. The problem is this. When one device receives a list of changed objects it can't reliably compare the lastModified of received objects to its own because the system times on the two devices may be different. I considered having each device send its current date/time during the sync. Then each calculates the difference between the remote time and the local time to compare the dates properly. But if there is lag between sending a date and the remote device receiving it, this causes incorrect comparisons with objects that were modified at the same time (or very close together in time). i.e. both devices think the remote object is newer and they end up with different objects. I hope I have explained this clearly enough. There must be a common solution to this kind of problem but my brain isn't coming up with anything. Any suggestions? Thanks in advance...

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  • Why is the operation address incremented by two?

    - by Gavin Jones
    I am looking at a Javascript emulator of a NES to try and understand how it works. On this line: addr = this.load(opaddr+2); The opcode is incremented by two. However, the documentation (see appendix E) I'm reading says: Zero page addressing uses a single operand which serves as a pointer to an address in zero page ($0000-$00FF) where the data to be operated on can be found. By using zero page addressing, only one byte is needed for the operand, so the instruction is shorter and, therefore, faster to execute than with addressing modes which take two operands. An example of a zero page instruction is AND $12. So if the operand's argument is only one byte, shouldn't it appear directly after it, and be + 1 instead of + 2? Why +2? This is how I think it works, which may be incorrect. Suppose our memory looks like: ------------------------- | 0 | 1 | 2 | 3 | 4 | 5 | <- index ------------------------- | a | b | c | d | e | f | <- memory ------------------------- ^ \ PC and our PC is 0, pointing to a. For this cycle, we say that the opcode: var pc= 0; //for example's sake var opcode= memory[pc]; //a So shouldn't the first operand be the next slot, i.e. b? var first_operand = memory[pc + 1]; //b

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  • Union of two or more (hash)maps

    - by javierfp
    I have two Maps that contain the same type of Objects: Map<String, TaskJSO> a = new HashMap<String, TaskJSO>(); Map<String, TaskJSO> b = new HashMap<String, TaskJSO>(); public class TaskJSO { String id; } The map keys are the "id" properties. a.put(taskJSO.getId(), taskJSO); I want to obtain a list with: all values in "Map b" + all values in "Map a" that are not in "Map b". What is the fastest way of doing this operation? Thanks EDIT: The comparaison is done by id. So, two TaskJSOs are considered as equal if they have the same id (equals method is overrided). My intention is to know which is the fastest way of doing this operation from a performance point of view. For instance, is there any difference if I do the "comparaison" in a map (as suggested by Peter): Map<String, TaskJSO> ab = new HashMap<String, TaskJSO>(a); ab.putAll(b); ab.values() or if instead I use a set (as suggested by Nishant): Set s = new Hashset(); s.addAll(a.values()); s.addAll(b.values());

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  • find the difference between two very large list

    - by user157195
    I have two large list(could be a hundred million items), the source of each list can be either from a database table or a flat file. both lists are of comparable sizes, both unsorted. I need to find the difference between them. so I have 3 scenarios: 1. List1 is a database table(assume each row simply have one item(key) that is a string), List2 is a large file. 2. Both lists are from 2 db tables. 3. both lists are from two files. in case 2, I plan to use: select a.item from MyTable a where a.item not in (select b.item form MyTable b) this clearly is inefficient, is there a better way? Another approach is: I plan to sort each list, and then walk down both of them to find the diff. If the list is from a file, I have to read it into a db table first, then use db sorting to output the list. Is the run time complexity still O(nlogn) in db sorting? either approach is a pain and seems would be very slow when the list involved has hundreds of millions of items. any suggestions?

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  • Android text layout question: two textviews, side-by-side, with different layout alignments and weights

    - by thx1200
    I'm still a bit of an Android noob, forgive me if this is simple and I'm just not seeing it. There are two portions of text in a view that spans the entire width horizontally, but is only as high as one line of text. The left side must always be displayed in full, but should take no more horizontal space than it needs. The right side should be pushed over by the left side and fill up the remainder of the screen width. If the right side text is smaller than this width, the text should be right-aligned horizontally. If the text is greater than the width, it should scroll horizontally. The text on the right side will be updated frequently and should slide up with new text when the app tells it (explaining the TextSwitcher in the layout). I have tried two different layout styles. In both situations, I can get the left side to "push" the layout, the right side to scroll, but I can't figure out how to get the right side to right align. It is always left aligned. Here is a picture showing what is happening... http://img10.imageshack.us/img10/5599/androidlayout.png In addition (but less important), in my layout code I have android:fadingEdge="none" on the TextViews, but it still has a faded edge on the left and right side when it scrolls. Why is that? Here are the two layouts I created, which yield the results shown, but not the results I want. Using a horizontal LinearLayout... <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/LinearLayoutStatusBar" android:orientation="horizontal" android:layout_width="fill_parent" android:layout_height="wrap_content" android:layout_margin="2px" android:background="#555555" > <TextView android:id="@+id/TextViewTimer" android:textSize="18px" android:textColor="#FFFFFF" android:layout_gravity="left" android:layout_weight="0" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_marginLeft="0px" android:layout_marginRight="3px" android:text="Left Side" > </TextView> <TextSwitcher android:id="@+id/TextSwitcherDetails" android:inAnimation="@anim/push_up_in" android:outAnimation="@anim/push_up_out" android:layout_weight="1" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_gravity="right" android:layout_marginLeft="3px" android:layout_marginRight="0px" > <TextView android:id="@+id/TextViewDetails1" android:textSize="18px" android:layout_width="match_parent" android:layout_height="match_parent" android:layout_gravity="right" android:singleLine="true" android:ellipsize="marquee" android:marqueeRepeatLimit="marquee_forever" android:scrollHorizontally="true" android:focusable="true" android:focusableInTouchMode="true" android:fadingEdge="none" android:text="Right Side 1" > </TextView> <TextView android:id="@+id/TextViewDetails2" android:textSize="18px" android:layout_width="match_parent" android:layout_height="match_parent" android:layout_gravity="right" android:singleLine="true" android:ellipsize="marquee" android:marqueeRepeatLimit="marquee_forever" android:scrollHorizontally="true" android:focusable="true" android:focusableInTouchMode="true" android:fadingEdge="none" android:text="Right Side 2 - This is a really long text this is long and fun and fun and long" > </TextView> </TextSwitcher> </LinearLayout> And the RelativeLayout style... <?xml version="1.0" encoding="utf-8"?> <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/LinearLayoutStatusBar" android:layout_width="fill_parent" android:layout_height="wrap_content" android:layout_margin="2px" android:background="#555555" > <TextView android:id="@+id/TextViewTimer" android:textSize="18px" android:textColor="#FFFFFF" android:layout_gravity="left" android:layout_weight="0" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_marginLeft="0px" android:layout_marginRight="3px" android:layout_alignParentLeft="true" android:text="Left Side" > </TextView> <TextSwitcher android:id="@+id/TextSwitcherDetails" android:inAnimation="@anim/push_up_in" android:outAnimation="@anim/push_up_out" android:layout_weight="1" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_marginLeft="3px" android:layout_marginRight="0px" android:layout_toRightOf="@+id/TextViewTimer" android:layout_alignParentRight="true" android:fadingEdge="none" android:fadingEdgeLength="0px" > <TextView android:id="@+id/TextViewDetails1" android:textSize="18px" android:layout_width="fill_parent" android:layout_height="wrap_content" android:layout_gravity="right" android:singleLine="true" android:ellipsize="marquee" android:marqueeRepeatLimit="marquee_forever" android:scrollHorizontally="true" android:focusable="true" android:focusableInTouchMode="true" android:fadingEdge="none" android:fadingEdgeLength="0px" android:text="Right Side 1" > </TextView> <TextView android:id="@+id/TextViewDetails2" android:textSize="18px" android:layout_width="fill_parent" android:layout_height="wrap_content" android:layout_gravity="right" android:singleLine="true" android:ellipsize="marquee" android:marqueeRepeatLimit="marquee_forever" android:scrollHorizontally="true" android:focusable="true" android:focusableInTouchMode="true" android:fadingEdge="none" android:fadingEdgeLength="0px" android:text="Right Side 2 - This is a really long text this is long and fun and fun and long" > </TextView> </TextSwitcher> </RelativeLayout> So how do I get that text on the right side to right-align. Thanks!

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  • Finding all the shortest paths between two nodes in unweighted directed graphs using BFS algorithm

    - by andra-isan
    Hi All, I am working on a problem that I need to find all the shortest path between two nodes in a given directed unweighted graph. I have used BFS algorithm to do the job, but unfortunately I can only print one shortest path not all of them, for example if they are 4 paths having lenght 3, my algorithm only prints the first one but I would like it to print all the four shortest paths. I was wondering in the following code, how should I change it so that all the shortest paths between two nodes could be printed out? class graphNode{ public: int id; string name; bool status; double weight;}; map<int, map<int,graphNode>* > graph; int Graph::BFS(graphNode &v, graphNode &w){ queue <int> q; map <int, int> map1; // this is to check if the node has been visited or not. std::string str= ""; map<int,int> inQ; // just to check that we do not insert the same iterm twice in the queue map <int, map<int, graphNode>* >::iterator pos; pos = graph.find(v.id); if(pos == graph.end()) { cout << v.id << " does not exists in the graph " <<endl; return 1; } int parents[graph.size()+1]; // this vector keeps track of the parents for the node parents[v.id] = -1; // there is a direct path between these two words, simply print that path as the shortest path if (findDirectEdge(v.id,w.id) == 1 ){ cout << " Shortest Path: " << v.id << " -> " << w.id << endl; return 1; } //if else{ int gn; map <int, map<int, graphNode>* >::iterator pos; q.push(v.id); inQ.insert(make_pair(v.id, v.id)); while (!q.empty()){ gn = q.front(); q.pop(); map<int, int>::iterator it; cout << " Popping: " << gn <<endl; map1.insert(make_pair(gn,gn)); //backtracing to print all the nodes if gn is the same as our target node such as w.id if (gn == w.id){ int current = w.id; cout << current << " - > "; while (current!=v.id){ current = parents[current]; cout << current << " -> "; } cout <<endl; } if ((pos = graph.find(gn)) == graph.end()) { cout << " pos is empty " <<endl; continue; } map<int, graphNode>* pn = pos->second; map<int, graphNode>::iterator p = pn->begin(); while(p != pn->end()) { map<int, int>::iterator it; //map1 keeps track of the visited nodes it = map1.find(p->first); graphNode gn1= p->second; if (it== map1.end()) { map<int, int>::iterator it1; //if the node already exits in the inQ, we do not insert it twice it1 = inQ.find(p->first); if (it1== inQ.end()){ parents[p->first] = gn; cout << " inserting " << p->first << " into the queue " <<endl; q.push(p->first); // add it to the queue } //if } //if p++; } //while } //while } I do appreciate all your great help Thanks, Andra

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  • How to merge two different child nodes in the same XML file

    - by user814698
    I have an XML file and I would like to merge two different CONTACT child nodes. I have checked these websites it shows how to merge two different xml files into a single file. http://www2.informatik.hu-berlin.de/~obecker/XSLT/#merge Merge XML documents In my case this is my first contact in the xml file: <CONTACT> <PDE-Identity>N65539</PDE-Identity> <FirstName>Arun_niit</FirstName> <LastName>Arun_niit</LastName> <Facebook-ID/> <EMAILS> <EMail> <Type>yahoo</Type> <Value>[email protected]</Value> </EMail> </EMAILS> </CONTACT> This is the second contact in the file: <PDE-Identity>N65567</PDE-Identity> <FirstName>Arun_niit</FirstName> <LastName>Ramanathan</LastName> <Facebook-ID/> <EMAILS> <EMail> <Type>gmail</Type> <Value>[email protected]</Value> </EMail> <EMail> <Type>yahoo</Type> <Value>[email protected]</Value> </EMail> </EMAILS> </CONTACT> <CONTACT> I know both of the contacts are belongs to the same person. How can i merge these two contacts in the same xml file. Original XML File: <?xml version="1.0" encoding="UTF-8"?> <CONTACTS> <CONTACT> <PDE-Identity>N65539</PDE-Identity> <FirstName>Arun_niit</FirstName> <LastName>Arun_niit</LastName> <Facebook-ID/> <EMAILS> <EMail> <Type>yahoo</Type> <Value>[email protected]</Value> </EMail> </EMAILS> </CONTACT> <CONTACT> <PDE-Identity>N65567</PDE-Identity> <FirstName>Arun_niit</FirstName> <LastName>Ramanathan</LastName> <Facebook-ID/> <EMAILS> <EMail> <Type>gmail</Type> <Value>[email protected]</Value> </EMail> <EMail> <Type>yahoo</Type> <Value>[email protected]</Value> </EMail> </EMAILS> </CONTACT> <CONTACT> <PDE-Identity>N65567</PDE-Identity> <FirstName>Rangarajkarthik</FirstName> <LastName>karthik Rangaraj</LastName> <Facebook-ID/> <EMAILS> <EMail> <Type>gmail</Type> <Value>[email protected]</Value> </EMail> <EMail> <Type>yahoo</Type> <Value>[email protected]</Value> </EMail> </EMAILS> </CONTACT> <CONTACTS>

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  • Hosting WCF service in IIS 7 (WAS) with net.tcp binding on TWO tcp ports

    - by Yuri
    By default IIS 7 Web site has net.tcp binding with "808:" binding information string. If i add another net.tcp binding with "xxx:" exception occurs: This collection already contains an address with scheme net.tcp. There can be at most one address per scheme in this collection. Parameter name: item How can i solve this problem and listen my service at TWO ports?

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