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• How can I pause a BackgroundWorker? Or something similar...

  - by juan

  I was using a BackgroundWorker to download some web sites by calling WebClient.DownloadString inside a loop. I wanted the option for the user to cancel in the middle of downloading stuff, so I called CancelAsync whenever I found that CancellationPending was on in the middle of the loop. But now I noticed that the function DownloadString kinda freezes sometimes, so I decided to use DownloadStringAsync instead (all this inside the other thread created with BackgroundWorker). An since and don't want to rewrite my whole code by having to exit the loop and the function after calling DownloadStringAsync, I made a while loop right after calling it that does nothing but checks for a variable bool Stop that I turn true either when the DownloadStringCompleted event handler is called or when the user request to cancel the operation. Now, the weird thing is that it works fine on the debug version, but on the release one, the program freezes in the while loop like if it were the main thread.

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• Exalogic 2.0.1 Tea Break Snippets - Modifying the Default Shipped Template

  - by The Old Toxophilist

  Having installed your Exalogic Virtual environment by default you have a single template which can be used to create your vServers. Although this template is suitable for creating simple test or development vServers it is recommended that you look at creating your own custom vServers that match the environment you wish to build and deploy. Therefore this Tea Time Snippet will take you through the simple process of modifying the standard template. Before You Start To edit the template you will need the Oracle ModifyJeos Utility which can be downloaded from the eDelivery Site. Once the ModifyJeos Utility has been downloaded we can install the rpms onto either an existing vServer or one of the Control vServers. rpm -ivh ovm-modify-jeos-1.0.1-10.el5.noarch.rpm rpm -ivh ovm-template-config-1.0.1-5.el5.noarch.rpm Alternatively you can install the modify jeos packages on a none Exalogic OEL installation or a VirtualBox image. If you are doing this, assuming OEL 5u8, you will need the following rpms. rpm -ivh ovm-modify-jeos-1.0.1-10.el5.noarch.rpm rpm -ivh ovm-el5u2-xvm-jeos-1.0.1-5.el5.i386.rpm rpm -ivh ovm-template-config-1.0.1-5.el5.noarch.rpm Base Template If you have installed the modify onto a vServer running on the Exalogic then simply mount the /export/common/images from the ZFS storage and you will be able to find the el_x2-2_base_linux_guest_vm_template_2.0.1.1.0_64.tgz (or similar depending which version you have) template file. Alternatively the latest can be downloaded from the eDelivery Site. Now we have the Template tgz we will need the extract it as follows: tar -zxvf  el_x2-2_base_linux_guest_vm_template_2.0.1.1.0_64.tgz This will create a directory called BASE which will contain the System.img (VServer image) and vm.cfg (VServer Config information). This directory should be renamed to something more meaning full that indicates what we have done to the template and then the Simple name / name in the vm.cfg editted for the same reason. Modifying the Template Resizing Root File System By default the shipped template has a root size of 4 GB which will leave a vServer created from it running at 90% full on the root disk. We can simply resize the template by executing the following: modifyjeos -f System.img -T <New Size MB>) For example to imcrease the default 4 GB to 40 GB we would execute: modifyjeos -f System.img -T 40960) Resizing Swap We can modify the size of the swap space within a template by executing the following: modifyjeos -f System.img -S <New Size MB>) For example to increase the swap from the default 512 MB to 4 GB we would execute: modifyjeos -f System.img -S 4096) Changing RPMs Adding RPMs To add RPMs using modifyjeos, complete the following steps: Add the names of the new RPMs in a list file, such as addrpms.lst. In this file, you should list each new RPM in a separate line. Ensure that all of the new RPMs are in a single directory, such as rpms. Run the following command to add the new RPMs: modifyjeos -f System.img -a <path_to_addrpms.lst> -m <path_to_rpms> -nogpg Where <path_to_addrpms.lst> is the path to the location of the addrpms.lst file, and <path_to_rpms> is the path to the directory that contains the RPMs. The -nogpg option eliminates signature check on the RPMs. Removing RPMs To remove RPM s using modifyjeos, complete the following steps: Add the names of the RPMs (the ones you want to remove) in a list file, such as removerpms.lst. In this file, you should list each RPM in a separate line. The Oracle Exalogic Elastic Cloud Administrator's Guide provides a list of all RPMs that must not be removed from the vServer. Run the following command to remove the RPMs: modifyjeos -f System.img -e <path_to_removerpms.lst> Where <path_to_removerpms.lst> is the path to the location of the removerpms.lst file. Mounting the System.img For all other modifications that are not supported by the modifyjeos command (adding you custom yum repositories, pre configuring NTP, modify default NFSv4 Nobody functionality, etc) we can mount the System.img and access it directly. To facititate quick and easy mounting/unmounting of the System.img I have put together the simple scripts below. MountSystemImg.sh #!/bin/sh # The script assumes it's being run from the directory containing the System.img # Export for later i.e. during unmount export LOOP=`losetup -f` export SYSTEMIMG=/mnt/elsystem # Make Temp Mount Directory mkdir -p $SYSTEMIMG # Create Loop for the System Image losetup $LOOP System.img kpartx -a $LOOP mount /dev/mapper/`basename $LOOP`p2 $SYSTEMIMG #Change Dir into mounted Image cd $SYSTEMIMG UnmountSystemImg.sh #!/bin/sh # The script assumes it's being run from the directory containing the System.img # Assume the $LOOP & $SYSTEMIMG exist from a previous run on the MountSystemImg.sh umount $SYSTEMIMG kpartx -d $LOOP losetup -d $LOOP Packaging the Template Once you have finished modifying the template it can be simply repackaged and then imported into EMOC as described in "Exalogic 2.0.1 Tea Break Snippets - Importing Public Server Template". To do this we will simply cd to the directory above that containing the modified files and execute the following: tar -zcvf <New Template Directory> <New Template Name>.tgz The resulting.tgz file can be copied to the images directory on the ZFS and uploadd using the IB network. This entry was originally posted on the The Old Toxophilist Site.

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• Implement loops for python 3

  - by Alex

  Implement this loop: total up the product of the numbers from 1 to x. Implement this loop: total up the product of the numbers from a to b. Implement this loop: total up the sum of the numbers from a to b. Implement this loop: total up the sum of the numbers from 1 to x. Implement this loop: count the number of characters in a string s. i'm very lost on implementing loops these are just some examples that i am having trouble with-- if someone could help me understand how to do them that would be awesome

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• MS SQL Server Job with precise timing

  - by TcKs

  Hi, I have a DB with game data (map, players, etc...) and I have a game core mechanics writen in T-SQL stored procedure. I need process game loop (via the stored procedure) every "X" seconds. I tried used the SQL Job, but when I set the interval to seconds, the SQL server stops responding. If I set the interval greater than one minute, all was ok. I need game loop precise in time, e.g. the game loop will run only once and will be executed every "X" precisely (tolerance should be less than one second). Can I do it with MS SQL Server capabilities? Or should I create a windows service which will repeatly execute game loop procedure? Or should I go another way? Thanks! EDIT: The game loop stored procedure takes less than the interval.

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• breaking the look in jython

  - by kdev

  Hi everyone , I have a written a loop for and if condition and i want to break to loop completely when if gets satisfied. while count: i=0 for line in read_record: #print str.strip(count[1:28]) #print str.strip(read_record[i]) if string.find(str.strip(read_record[i]),str.strip(count[1:28]))>0: code=str.strip(read_record[i+19])+str.strip(read_record[i+20]) print code[25:] break i=i+1 so here if the if string.find condition gets satified i want to go to while loop flow. Please tell me what will be the best place to break and how should i modify the program so that once the if condition is satified iam out of the loop and start again with while loop .

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• Setting WPF RichTextBox width and height according to the size of a monospace font

  - by oxeb

  I am trying to fit a WPF RichTextBox to exactly accommodate a grid of characters in a particular monospace font. I am currently using FormattedText to determine the width and height of my RichTextBox, but the measurements it is providing me with are too small--specifically two characters in width too small. Is there a better way to perform this task? This does not seem to be an appropriate way to determine the size of my control. RichTextBox rtb; rtb = new RichTextBox(); FontFamily fontFamily = new FontFamily("Consolas"); double fontSize = 16; char standardizationCharacter = 'X'; String standardizationLine = ""; for(long loop = 0; loop < columns; loop ++) { standardizationLine += standardizationCharacter; } standardizationLine += Environment.NewLine; String standardizationString = ""; for(long loop = 0; loop < rows; loop ++) { standardizationString += standardizationLine; } Typeface typeface = new Typeface(fontFamily, FontStyles.Normal, FontWeights.Normal, FontStretches.Normal); FormattedText formattedText = new FormattedText(standardizationString, CultureInfo.CurrentCulture, FlowDirection.LeftToRight, typeface, fontSize, Brushes.Black); rtb.Width = formattedText.Width; rtb.Height = formattedText.Height;

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• Why does this cast to Base class in virtual function give a segmentation fault?

  - by dehmann

  I want to print out a derived class using the operator<<. When I print the derived class, I want to first print its base and then its own content. But I ran into some trouble (see segfault below): class Base { public: friend std::ostream& operator<<(std::ostream&, const Base&); virtual void Print(std::ostream& out) const { out << "BASE!"; } }; std::ostream& operator<<(std::ostream& out, const Base& b) { b.Print(out); return out; } class Derived : public Base { public: virtual void Print(std::ostream& out) const { out << "My base: "; //((const Base*)this)->Print(out); // infinite, calls this fct recursively //((Base*)this)->Print(out); // segfault (from infinite loop?) ((Base)*this).Print(out); // OK out << " ... and myself."; } }; int main(int argc, char** argv){ Derived d; std::cout << d; return 0; } Why can't I cast in one of these ways? ((const Base*)this)->Print(out); // infinite, calls this fct recursively ((Base*)this)->Print(out); // segfault (from infinite loop?)

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• Variable Assignment and loops (Java)

  - by Raven Dreamer

  Greetings Stack Overflowers, A while back, I was working on a program that hashed values into a hashtable (I don't remember the specifics, and the specifics themselves are irrelevant to the question at hand). Anyway, I had the following code as part of a "recordInput" method. tempElement = new hashElement(someInt); while(in.hasNext() == true) { int firstVal = in.nextInt(); if (firstVal == -911) { break; } tempElement.setKeyValue(firstVal, 0); for(int i = 1; i<numKeyValues;i++) { tempElement.setKeyValue(in.nextInt(), i); } elementArray[placeValue] = tempElement; placeValue++; } // close while loop } // close method This part of the code was giving me a very nasty bug -- no matter how I finagled it, no matter what input I gave the program, it would always produce an array full of only a single value -- the last one. The problem, as I later determined it, was that because I had not created the tempElement variable within the loop, and because values were not being assigned to elementArray[] until after the loop had ended -- every term was defined rather as "tempElement" -- when the loop terminated, every slot in the array was filled with the last value tempElement had taken. I was able to fix this bug by moving the declaration of tempElement within the while loop. My question to you, Stackoverflow, is whether there is another (read: better) way to avoid this bug while keeping the variable declaration of tempElement outside the while loop. (suggestions for better title and tags also appreciated)

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• Why doesn't gcc remove this check of a non-volatile variable?

  - by Thomas

  This question is mostly academic. I ask out of curiosity, not because this poses an actual problem for me. Consider the following incorrect C program. #include <signal.h> #include <stdio.h> static int running = 1; void handler(int u) { running = 0; } int main() { signal(SIGTERM, handler); while (running) ; printf("Bye!\n"); return 0; } This program is incorrect because the handler interrupts the program flow, so running can be modified at any time and should therefore be declared volatile. But let's say the programmer forgot that. gcc 4.3.3, with the -O3 flag, compiles the loop body (after one initial check of the running flag) down to the infinite loop .L7: jmp .L7 which was to be expected. Now we put something trivial inside the while loop, like: while (running) putchar('.'); And suddenly, gcc does not optimize the loop condition anymore! The loop body's assembly now looks like this (again at -O3): .L7: movq stdout(%rip), %rsi movl $46, %edi call _IO_putc movl running(%rip), %eax testl %eax, %eax jne .L7 We see that running is re-loaded from memory each time through the loop; it is not even cached in a register. Apparently gcc now thinks that the value of running could have changed. So why does gcc suddenly decide that it needs to re-check the value of running in this case?

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• how useful is Turing completeness? are neural nets turing complete?

  - by Albert

  While reading some papers about the Turing completeness of recurrent neural nets (for example: Turing computability with neural nets, Hava T. Siegelmann and Eduardo D. Sontag, 1991), I got the feeling that the proof which was given there was not really that practical. For example the referenced paper needs a neural network which neuron activity must be of infinity exactness (to reliable represent any rational number). Other proofs need a neural network of infinite size. Clearly, that is not really that practical. But I started to wonder now if it does make sense at all to ask for Turing completeness. By the strict definition, no computer system nowadays is Turing complete because none of them will be able to simulate the infinite tape. Interestingly, programming language specification leaves it most often open if they are turing complete or not. It all boils down to the question if they will always be able to allocate more memory and if the function call stack size is infinite. Most specification don't really specify this. Of course all available implementations are limited here, so all practical implementations of programming languages are not Turing complete. So, what you can say is that all computer systems are just equally powerful as finite state machines and not more. And that brings me to the question: How useful is the term Turing complete at all? And back to neural nets: For any practical implementation of a neural net (including our own brain), they will not be able to represent an infinite number of states, i.e. by the strict definition of Turing completeness, they are not Turing complete. So does the question if neural nets are Turing complete make sense at all? The question if they are as powerful as finite state machines was answered already much earlier (1954 by Minsky, the answer of course: yes) and also seems easier to answer. I.e., at least in theory, that was already the proof that they are as powerful as any computer.

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• PHP, MySQL prepared statements - can you use results of execute more than once by calling data_seek(

  - by Carvell Fenton

  Hello, I have a case where I want to use the results of a prepared statement more than once in a nested loop. The outer loop processes the results of another query, and the inner loop is the results of the prepared statement query. So the code would be something like this (just "pseudoish" to demonstrate the concept): // not showing the outer query, it is just a basic SELECT, not prepared statement // we'll call it $outer_query $obj_array = array(); // going to save objects in this $ids = array(18,19,20); // just example id numbers $query = "SELECT field1, field2 FROM table1 WHERE id=?"; $stmt = $db->prepare($query); foreach ($ids as $id) { $stmt->bind_param("i", $id); $stmt->execute(); $stmt->bind_result($var1, $var2); $stmt->store_result(); // I think I need this for data_seek while ($q1 = $outer_query->fetch_object()) { while ($stmt->fetch()) { if ($q1->field1 == $var1) { // looking for a match $obj = new stdClass(); $obj->var1 = $var1; $obj->var2 = $var2; $obj_array[] = $obj; $stmt->data_seek(0); // reset for outer loop break; // found match, so leave inner } } } } The problem I seem to be experiencing is that the values are not getting bound in the variables as I would expect after the first time I use fetch in the inner loop. Specifically, in one example I ran with 3 ids for the foreach, the first id was processed correctly, the second was processed incorrectly (matches were not found in the inner loop even though they existed), and then the third was processed correctly. Is there something wrong with the prepared statment function calls in the sequence I am doing above, or is this an invalid way to use the results of the prepared statement? Thanks.

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• visual studio 2008 (VB)

  - by Ousman

  hello guys, Am writing a programming with visual basic 2008 and i want the programme to be able to read and loop through a text file line by line and showing the event of the loop on a textbox or label until a button is press and the loop will stop on any number that happend to be at the loop event and when a button is press again the loop will continue from where it starts. this is my codes below and having problem with it and any help will be really great. tanks ==========================my codes======================= Imports System.IO '========================================================================================== Public Class Form1 '====================================================================================== 'Dim nFileNum As Integer = FreeFile() ' Get a free file number Dim strFileName As String = "C:\scb.txt" Dim objFilename As FileStream = New FileStream(strFileName, FileMode.Open, FileAccess.Read, FileShare.Read) Dim objFileRead As StreamReader = New StreamReader(objFilename) 'Dim lLineCount As Long 'Dim sNextLine As String '====================================================================================== '======================================================================================== Private Sub btStart_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btStart.Click Try If objFileRead.ReadLine = Nothing Then MsgBox("No Accounts Available to show!", MsgBoxStyle.Information, MsgBoxStyle.DefaultButton2 = MsgBoxStyle.OkOnly) Return Else Do While (objFileRead.Peek() > -1) Loop lblAccounts.Text = objFileRead.ReadLine() 'objFileRead.Close() 'objFilename.Close() End If Catch ex As Exception MessageBox.Show(ex.Message) Finally 'objFileRead.Close() 'objFilename.Close() End Try End Sub Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load End Sub End Class

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• Do applescript repeat loops reflect changes instantly?

  - by user1159454

  I'm making a script to open multiple folders in a directory, and it's not working as planned. I've tried outlining it and walking through the steps one by one pretending I'm the computer executing it but when I run it the outcome is very different. It uses repeat and repeat with a lot. The repeat repeats for as long as there is ANYTHING in a certain array (I mean List) and the repeat with is INSIDE of the first repeat, which repeats it's own loop with everything in the array at that time. Now, one of the actions of the repeat with loop is to change the array. Which I think would change the loop would it not? Example (foldList is A, B, C) repeat until {} repeat with folder_name in foldList do something set foldList to 1, 2, 3 end repeat end repeat What I would THINK that does is iterate through the first loop as "A", but before hitting the end it would change foldList to 1, 2, 3. So instead of going through the next loop as "B" I'd think it would go as "1" instead. But if it did that then I don't think my manual walkthrough would be off by so much. So I'm under the assumption that in Applescript when you're in a repeat with, regardless of changing the List you WILL end that loop on the last item of the first List (before the list was replaced.) Is this right?

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• A F# tail-recursive question

  - by ksharp

  Recently, I'm learning F#. I try to solve problem in different ways. Like this: (* [0;1;2;3;4;5;6;7;8] -> [(0,1,2);(3,4,5);(6,7,8)] *) //head-recursive let rec toTriplet_v1 list= match list with | a::b::c::t -> (a,b,c)::(toTriplet_v1 t) | _ -> [] //tail-recursive let toTriplet_v2 list= let rec loop lst acc= match lst with | a::b::c::t -> loop t ((a,b,c)::acc) | _ -> acc loop list [] //tail-recursive(???) let toTriplet_v3 list= let rec loop lst accfun= match lst with | a::b::c::t -> loop t (fun ls -> accfun ((a,b,c)::ls)) | _ -> accfun [] loop list (fun x -> x) let funs = [toTriplet_v1; toTriplet_v2; toTriplet_v3]; funs |> List.map (fun x -> x [0..8]) |> List.iteri (fun i x -> printfn "V%d : %A" (i+1) x) I thought the results of V2 and V3 should be the same. But, I get the result below: V1 : [(0, 1, 2); (3, 4, 5); (6, 7, 8)] V2 : [(6, 7, 8); (3, 4, 5); (0, 1, 2)] V3 : [(0, 1, 2); (3, 4, 5); (6, 7, 8)] Why the results of V2 and V3 are different?

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• Replacing repetitively occuring loops with eval in Javascript - good or bad?

  - by Herc

  Hello stackoverflow! I have a certain loop occurring several times in various functions in my code. To illustrate with an example, it's pretty much along the lines of the following: for (var i=0;i<= 5; i++) { function1(function2(arr[i],i),$('div'+i)); $('span'+i).value = function3(arr[i]); } Where i is the loop counter of course. For the sake of reducing my code size and avoid repeating the loop declaration, I thought I should replace it with the following: function loop(s) { for (var i=0;i<= 5; i++) { eval(s); } } [...] loop("function1(function2(arr[i],i),$('div'+i));$('span'+i).value = function3(arr[i]);"); Or should I? I've heard a lot about eval() slowing code execution and I'd like it to work as fast as a proper loop even in the Nintendo DSi browser, but I'd also like to cut down on code. What would you suggest? Thank you in advance!

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• How to identify end of InputStream in java

  - by Vardhaman

  I am trying to read bytes from server using Socket program, ie I am using InputStream to read the bytes. If I pass the length size I am able to read the bytes, but I am not sure what may be the length. So I am not able initialize the byte array. Also I tried while (in.read() !=-1), I observered it loop works fine when the data is sent , but the next line after the loop is not executable , I feel its still looking for the data in the stream but there is no ata. If I close the Server connection , then my client will execute the next line followed to the loop. I am not sure where I am going wrong? this.in = socket.getInputStream(); int dataInt = this.in.read(); while(dataInt != -1){ System.out.print(","+i+"--"+dataInt); i++; dataInt = this.in.read(); } System.out.print("End Of loop"); I get the output as:- ,1--0,2--62,3--96,4--131,5--142,6--1,7--133,8--2,9--16,10--48,11--56,12--1,13--0,14--14,15--128,16--0,17--0,18--0,19--48,20--0,21--0,22--0,23--0,24--0,25--1,26--0,27--0,28--38,29--114,30--23,31--20,32--70,33--3,34--20,35--1,36--133,37--48,38--51,39--49,40--52,41--49,42--55,43--49,44--52,45--52,46--54,47--55,48--50,49--51,50--52,51--48,52--53,53--56,54--51,55--48,56--48,57--57,58--57,59--57,60--57,61--57,62--57,63--57,64--56 But no output for :- End Of loop Please guide how shall I close the loop? Looking forward for you response. Thanking you all in advance.

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• Empty value when iterating a dictionary with .iteritems() method

  - by ptpatil

  I am having some weird trouble with dictionaries, I am trying to iterate pairs from a dictionary to pass to another function. The loop for the iterator though for some reason always returns empty values. Here is the code: def LinktoCentral(self, linkmethod): if linkmethod == 'sim': linkworker = Linker.SimilarityLinker() matchlist = [] for k,v in self.ToBeMatchedTable.iteritems(): matchlist.append(k, linkworker.GetBestMatch(v, self.CentralDataTable.items())) Now if I insert a print line above the for loop: matchlist = [] print self.ToBeMatchedTable.items() for k,v in self.ToBeMatchedTable.iteritems(): matchlist.append(k, linkworker.GetBestMatch(v, self.CentralDataTable.items())) I get the data that is supposed to be in the dictionary printed out. The values of the dictionary are list objects. An example tuple I get from the dictionary when printing just above the for loop: >>> (1, ['AARP/United Health Care', '8002277789', 'PO Box 740819', 'Atlanta', 'GA', '30374-0819', 'Paper', '3676']) However, the for loop gives empty lists to the linkworker.GetBestMatch method. If I put a print line just below the for loop, here is what I get: Code: matchlist = [] for k,v in self.ToBeMatchedTable.iteritems(): print self.ToBeMatchedTable.items() matchlist.append(k, linkworker.GetBestMatch(v, self.CentralDataTable.items())) ## Place holder for line to send match list to display window return matchlist Result of first iteration: >>> (0, ['', '', '', '', '', '', '', '']) I literally have no idea whats going on, there is nothing else going on while this loop is executed. Any stupid mistakes I made?

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• Interpolation using a sprite's previous frame and current frame

  - by user22241

  Overview I'm currently using a method which has been pointed out to me is extrapolation rather than interolation. As a result, I'm also now looking into the possibility of using another method which is based on a sprite's position at it's last (rendered) frame and it's current one. Assuming an interpolation value of 0.5 this is, (visually), how I understand it should affect my sprite's position.... This is how I'm obtaining an inerpolation value: public void onDrawFrame(GL10 gl) { // Set/re-set loop back to 0 to start counting again loops=0; while(System.currentTimeMillis() > nextGameTick && loops < maxFrameskip) { SceneManager.getInstance().getCurrentScene().updateLogic(); nextGameTick += skipTicks; timeCorrection += (1000d / ticksPerSecond) % 1; nextGameTick += timeCorrection; timeCorrection %= 1; loops++; tics++; } interpolation = (float)(System.currentTimeMillis() + skipTicks - nextGameTick) / (float)skipTicks; render(interpolation); } I am then applying it like so (in my rendering call): render(float interpolation) { spriteScreenX = (spriteScreenX - spritePreviousX) * interpolation + spritePreviousX; spritePreviousX = spriteScreenX; // update and store this for next time } Results This unfortunately does nothing to smooth the movement of my sprite. It's pretty much the same as without the interpolation code. I can't get my head around how this is supposed to work and I honestly can't find any decent resources which explain this in any detail. My understanding of extrapolation is that when we arrive at the rendering call, we calculate the time between the last update call and the render call, and then adjust the sprite's position to reflect this time (moving the sprite forward) - And yet, this (Interpolation) is moving the sprite back, so how can this produce smooth results? Any advise on this would be very much appreciated. Edit I've implemented the code from OriginalDaemon's answer like so: @Override public void onDrawFrame(GL10 gl) { newTime = System.currentTimeMillis()*0.001; frameTime = newTime - currentTime; if ( frameTime > (dt*25)) frameTime = (dt*25); currentTime = newTime; accumulator += frameTime; while ( accumulator >= dt ) { SceneManager.getInstance().getCurrentScene().updateLogic(); previousState = currentState; t += dt; accumulator -= dt; } interpolation = (float) (accumulator / dt); render(); } Interpolation values are now being produced between 0 and 1 as expected (similar to how they were in my original loop) - however, the results are the same as my original loop (my original loop allowed frames to skip if they took too long to draw which I think this loop is also doing). I appear to have made a mistake in my previous logging, it is logging as I would expect it to (interpolated position does appear to be inbetween the previous and current positions) - however, the sprites are most definitely choppy when the render() skipping happens.

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• More Fun With Math

  - by PointsToShare

  More Fun with Math   The runaway student – three different ways of solving one problem Here is a problem I read in a Russian site: A student is running away. He is moving at 1 mph. Pursuing him are a lion, a tiger and his math teacher. The lion is 40 miles behind and moving at 6 mph. The tiger is 28 miles behind and moving at 4 mph. His math teacher is 30 miles behind and moving at 5 mph. Who will catch him first? Analysis Obviously we have a set of three problems. They are all basically the same, but the details are different. The problems are of the same class. Here is a little excursion into computer science. One of the things we strive to do is to create solutions for classes of problems rather than individual problems. In your daily routine, you call it re-usability. Not all classes of problems have such solutions. If a class has a general (re-usable) solution, it is called computable. Otherwise it is unsolvable. Within unsolvable classes, we may still solve individual (some but not all) problems, albeit with different approaches to each. Luckily the vast majority of our daily problems are computable, and the 3 problems of our runaway student belong to a computable class. So, let’s solve for the catch-up time by the math teacher, after all she is the most frightening. She might even make the poor runaway solve this very problem – perish the thought! Method 1 – numerical analysis. At 30 miles and 5 mph, it’ll take her 6 hours to come to where the student was to begin with. But by then the student has advanced by 6 miles. 6 miles require 6/5 hours, but by then the student advanced by another 6/5 of a mile as well. And so on and so forth. So what are we to do? One way is to write code and iterate it until we have solved it. But this is an infinite process so we’ll end up with an infinite loop. So what to do? We’ll use the principles of numerical analysis. Any calculator – your computer included – has a limited number of digits. A double floating point number is good for about 14 digits. Nothing can be computed at a greater accuracy than that. This means that we will not iterate ad infinidum, but rather to the point where 2 consecutive iterations yield the same result. When we do financial computations, we don’t even have to go that far. We stop at the 10th of a penny.  It behooves us here to stop at a 10th of a second (100 milliseconds) and this will how we will avoid an infinite loop. Interestingly this alludes to the Zeno paradoxes of motion – in particular “Achilles and the Tortoise”. Zeno says exactly the same. To catch the tortoise, Achilles must always first come to where the tortoise was, but the tortoise keeps moving – hence Achilles will never catch the tortoise and our math teacher (or lion, or tiger) will never catch the student, or the policeman the thief. Here is my resolution to the paradox. The distance and time in each step are smaller and smaller, so the student will be caught. The only thing that is infinite is the iterative solution. The race is a convergent geometric process so the steps are diminishing, but each step in the solution takes the same amount of effort and time so with an infinite number of steps, we’ll spend an eternity solving it.  This BTW is an original thought that I have never seen before. But I digress. Let’s simply write the code to solve the problem. To make sure that it runs everywhere, I’ll do it in JavaScript. function LongCatchUpTime(D, PV, FV) // D is Distance; PV is Pursuers Velocity; FV is Fugitive’ Velocity {     var t = 0;     var T = 0;     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     t = d / pv;     while (t > 0.000001) //a 10th of a second is 1/36,000 of an hour, I used 1/100,000     {         T = T + t;         d = t * fv;         t = d / pv;     }     return T;     } By and large, the higher the Pursuer’s velocity relative to the fugitive, the faster the calculation. Solving this with the 10th of a second limit yields: 7.499999232000001 Method 2 – Geometric Series. Each step in the iteration above is smaller than the next. As you saw, we stopped iterating when the last step was small enough, small enough not to really matter.  When we have a sequence of numbers in which the ratio of each number to its predecessor is fixed we call the sequence geometric. When we are looking at the sum of sequence, we call the sequence of sums series.  Now let’s look at our student and teacher. The teacher runs 5 times faster than the student, so with each iteration the distance between them shrinks to a fifth of what it was before. This is a fixed ratio so we deal with a geometric series.  We normally designate this ratio as q and when q is less than 1 (0 < q < 1) the sum of  + … +  is  – 1) / (q – 1). When q is less than 1, it is easier to use ) / (1 - q). Now, the steps are 6 hours then 6/5 hours then 6/5*5 and so on, so q = 1/5. And the whole series is multiplied by 6. Also because q is less than 1 , 1/  diminishes to 0. So the sum is just  / (1 - q). or 1/ (1 – 1/5) = 1 / (4/5) = 5/4. This times 6 yields 7.5 hours. We can now continue with some algebra and take it back to a simpler formula. This is arduous and I am not going to do it here. Instead let’s do some simpler algebra. Method 3 – Simple Algebra. If the time to capture the fugitive is T and the fugitive travels at 1 mph, then by the time the pursuer catches him he travelled additional T miles. Time is distance divided by speed, so…. (D + T)/V = T  thus D + T = VT  and D = VT – T = (V – 1)T  and T = D/(V – 1) This “strangely” coincides with the solution we just got from the geometric sequence. This is simpler ad faster. Here is the corresponding code. function ShortCatchUpTime(D, PV, FV) {     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     return d / (pv - fv); } The code above, for both the iterative solution and the algebraic solution are actually for a larger class of problems.  In our original problem the student’s velocity (speed) is 1 mph. In the code it may be anything as long as it is less than the pursuer’s velocity. As long as PV > FV, the pursuer will catch up. Here is the really general formula: T = D / (PV – FV) Finally, let’s run the program for each of the pursuers.  It could not be worse. I know he’d rather be eaten alive than suffering through yet another math lesson. See the code run? Select  “Catch Up Time” in www.mgsltns.com/games.htm The host is running on Unix, so the link is case sensitive. That’s All Folks

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• Create USB installer from the command line?

  - by j-g-faustus

  I'm trying to create a bootable USB image to install Ubuntu on a new computer. I have done this before following the "create USB drive" instructions for Ubuntu desktop, but I don't have an Ubuntu desktop available. How can I do the same using only the command line? Things I've tried: Create bootable USB on Mac OS X following the ubuntu.com "create USB drive" instructions for Mac: Doesn't boot. usb-creator: According to apt-cache search usb-creator and Wikipedia usb-creator only exists as a graphical tool. "Create manually" instructions at help.ubuntu.com: None of the files and directories described (e.g. casper, filesystem.manifest, menu.lst) exist in the ISO image, and I don't know what has replaced them. unetbootin scripting: Requires X server (graphics support) to run, even when fully scripted. (The command sudo unetbootin lang=en method=diskimage isofile=~/ubuntu-10.10-server-amd64.iso installtype=USB targetdrive=/dev/sdg1 autoinstall=yes gives an error message unetbootin: cannot connect to X server.) Update Also tried GRUB fiddling: Merging information from pendrivelinux.com a related question on the Linux Stackexchange and a grub configuration example I was able to get halfway there - it booted from USB, displayed the grub menu and started the installation, but installation did not complete. For reference, this is the closest I got: sudo su # mount USB pen mount /dev/sd[X]1 /media/usb # install GRUB grub-install --force --no-floppy --root-directory=/media/usb /dev/sd[X] # copy ISO image to USB cp ~/ubuntu-10.10-server-amd64.iso /media/usb # mount ISO image, copy existing grub.cfg mount ~/ubuntu-10.10-server-amd64.iso /media/iso/ -o loop cp /media/iso/boot/grub/grub.cfg /media/usb/boot/grub/ I then edited /media/usb/boot/grub.cfg to add an .iso loopback, example grub entry: menuentry "Install Ubuntu Server" { set gfxpayload=keep loopback loop /ubuntu-10.10-server-amd64.iso linux (loop)/install/vmlinuz file=(loop)/preseed/ubuntu-server.seed iso-scan/filename=/ubuntu-10.10-server-amd64.iso quiet -- initrd (loop)/install/initrd.gz } When booting from USB, this would give me the Grub boot menu and start the installer, but the installer gave up after a couple of screens complaining that it couldn't find the CD-ROM drive. (Naturally, as the box I'm installing on doesn't have an optical drive.) I resolved this particular issue by giving up and doing the "create USB drive" routine using the Ubuntu Live desktop CD (on a computer that does have an optical drive), then the USB install works. But I expect that there is some way to do this from the command line of an Ubuntu system without X server and without an optical drive, so the question still stands. Does anyone know how?

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• Can parser combination be made efficient?

  - by Jon Harrop

  Around 6 years ago, I benchmarked my own parser combinators in OCaml and found that they were ~5× slower than the parser generators on offer at the time. I recently revisited this subject and benchmarked Haskell's Parsec vs a simple hand-rolled precedence climbing parser written in F# and was surprised to find the F# to be 25× faster than the Haskell. Here's the Haskell code I used to read a large mathematical expression from file, parse and evaluate it: import Control.Applicative import Text.Parsec hiding ((<|>)) expr = chainl1 term ((+) <$ char '+' <|> (-) <$ char '-') term = chainl1 fact ((*) <$ char '*' <|> div <$ char '/') fact = read <$> many1 digit <|> char '(' *> expr <* char ')' eval :: String -> Int eval = either (error . show) id . parse expr "" . filter (/= ' ') main :: IO () main = do file <- readFile "expr" putStr $ show $ eval file putStr "\n" and here's my self-contained precedence climbing parser in F#: let rec (|Expr|) (P(f, xs)) = Expr(loop (' ', f, xs)) and shift oop f op (P(g, xs)) = let h, xs = loop (op, g, xs) loop (oop, f h, xs) and loop = function | ' ' as oop, f, ('+' | '-' as op)::P(g, xs) | (' ' | '+' | '-' as oop), f, ('*' | '/' as op)::P(g, xs) | oop, f, ('^' as op)::P(g, xs) -> let h, xs = loop (op, g, xs) let op = match op with | '+' -> (+) | '-' -> (-) | '*' -> (*) | '/' -> (/) | '^' -> pown loop (oop, op f h, xs) | _, f, xs -> f, xs and (|P|) = function | '-'::P(f, xs) -> let f, xs = loop ('~', f, xs) P(-f, xs) | '('::Expr(f, ')'::xs) -> P(f, xs) | c::xs when '0' <= c && c <= '9' -> P(int(string c), xs) My impression is that even state-of-the-art parser combinators waste a lot of time back tracking. Is that correct? If so, is it possible to write parser combinators that generate state machines to obtain competitive performance or is it necessary to use code generation?

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• Big O Complexity of a method

  - by timeNomad

  I have this method: public static int what(String str, char start, char end) { int count=0; for(int i=0;i<str.length(); i++) { if(str.charAt(i) == start) { for(int j=i+1;j<str.length(); j++) { if(str.charAt(j) == end) count++; } } } return count; } What I need to find is: 1) What is it doing? Answer: counting the total number of end occurrences after EACH (or is it? Not specified in the assignment, point 3 depends on this) start. 2) What is its complexity? Answer: the first loops iterates over the string completely, so it's at least O(n), the second loop executes only if start char is found and even then partially (index at which start was found + 1). Although, big O is all about worst case no? So in the worst case, start is the 1st char & the inner iteration iterates over the string n-1 times, the -1 is a constant so it's n. But, the inner loop won't be executed every outer iteration pass, statistically, but since big O is about worst case, is it correct to say the complexity of it is O(n^2)? Ignoring any constants and the fact that in 99.99% of times the inner loop won't execute every outer loop pass. 3) Rewrite it so that complexity is lower. What I'm not sure of is whether start occurs at most once or more, if once at most, then method can be rewritten using one loop (having a flag indicating whether start has been encountered and from there on incrementing count at each end occurrence), yielding a complexity of O(n). In case though, that start can appear multiple times, which most likely it is, because assignment is of a Java course and I don't think they would make such ambiguity. Solving, in this case, is not possible using one loop... WAIT! Yes it is..! Just have a variable, say, inc to be incremented each time start is encountered & used to increment count each time end is encountered after the 1st start was found: inc = 0, count = 0 if (current char == start) inc++ if (inc > 0 && current char == end) count += inc This would also yield a complexity of O(n)? Because there is only 1 loop. Yes I realize I wrote a lot hehe, but what I also realized is that I understand a lot better by forming my thoughts into words...

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• OpenSSL Handshake Failure (14094410) - Erroneous Client Certificate Check from Mobile Phone

  - by Clayton Sims

  I'm running a proxy server through Apache with modssl, which we're using to proxy POSTs from mobile devices to another internal server. This works successfully for most clients, but requests from a specific phone model (Nokia 2690) are showing a bizarre handshake failure. It looks as though OpenSSL is either requesting (or attempting to read an unsolicited) client certificate from the phone (which is especially bizarre because j2me's kssl implementation doesn't support client certs). I've disabled client certificates with the SSLVerifyClient none directive in both the virtual host conf and the modssl conf. The trace from error.log on debug level is (details redacted): [client 41.220.207.10] Connection to child 0 established (server www.myserver.org:443) [info] Seeding PRNG with 656 bytes of entropy [debug] ssl_engine_kernel.c(1866): OpenSSL: Handshake: start [debug] ssl_engine_kernel.c(1874): OpenSSL: Loop: before/accept initialization [debug] ssl_engine_io.c(1882): OpenSSL: read 11/11 bytes from BIO#7fe3fbaf17a0 [mem: 7fe3fbaf90d0] (BIO dump follows) [debug] ssl_engine_io.c(1815): +-------------------------------------------------------------------------+ [debug] ssl_engine_io.c(1860): +-------------------------------------------------------------------------+ [debug] ssl_engine_io.c(1882): OpenSSL: read 49/49 bytes from BIO#7fe3fbaf17a0 [mem: 7fe3fbaf90db] (BIO dump follows) [debug] ssl_engine_io.c(1815): +-------------------------------------------------------------------------+ [debug] ssl_engine_io.c(1860): +-------------------------------------------------------------------------+ [debug] ssl_engine_kernel.c(1874): OpenSSL: Loop: SSLv3 read client hello A [debug] ssl_engine_kernel.c(1874): OpenSSL: Loop: SSLv3 write server hello A [debug] ssl_engine_kernel.c(1874): OpenSSL: Loop: SSLv3 write certificate A [debug] ssl_engine_kernel.c(1874): OpenSSL: Loop: SSLv3 write server done A [debug] ssl_engine_kernel.c(1874): OpenSSL: Loop: SSLv3 flush data [debug] ssl_engine_io.c(1882): OpenSSL: read 5/5 bytes from BIO#7fe3fbaf17a0 [mem: 7fe3fbaf90d0] (BIO dump follows) [debug] ssl_engine_io.c(1815): +-------------------------------------------------------------------------+ [debug] ssl_engine_io.c(1860): +-------------------------------------------------------------------------+ [debug] ssl_engine_io.c(1882): OpenSSL: read 2/2 bytes from BIO#7fe3fbaf17a0 [mem: 7fe3fbaf90d5] (BIO dump follows) [debug] ssl_engine_io.c(1815): +-------------------------------------------------------------------------+ [debug] ssl_engine_io.c(1860): +-------------------------------------------------------------------------+ [debug] ssl_engine_kernel.c(1879): OpenSSL: Read: SSLv3 read client certificate A [debug] ssl_engine_kernel.c(1898): OpenSSL: Exit: failed in SSLv3 read client certificate A [client 41.220.207.10] SSL library error 1 in handshake (server www.myserver.org:443) [info] SSL Library Error: 336151568 error:14094410:SSL routines:SSL3_READ_BYTES:sslv3 alert handshake failure [client 41.220.207.10] Connection closed to child 0 with abortive shutdown (server www.myserver.org:443) I've tried enabling all ciphers and all protocols temporarily with modssl, neither of which seemed to be the issue. The phone should be using RSA_RC4_128_MD5 and SSLv3, all of which are available. Am I missing something more fundamental about what's failing here? It seemed like the certificate request might have been part of a renegotiation failure. I tried enabling SSLInsecureRenegotiation On on the virtual host, in case it was an issue of the phone's SSL not supporting the new protocol, but to no avail. Currently running: Apache/2.2.16 (Ubuntu) mod_ssl/2.2.16 OpenSSL/0.9.8o Apache proxy_html/3.0.1

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• Debian Squeeze vzquota

  - by benjamin

  Hello, Apparently, I got Debian Squeeze (Debian 6) to work on a VPS using debootstrap and chroot as described here. Subsequent installation of the harden, exim4, mysql-server packages failed partially. Relevant information: insserv: warning: script 'S10vzquota' missing LSB tags and overrides insserv: warning: script is corrupt or invalid: /etc/init.d/../rc6.d/S00vzreboot insserv: warning: script 'vzquota' missing LSB tags and overrides insserv: There is a loop between service vzquota and stop-bootlogd if started insserv: loop involving service stop-bootlogd at depth 2 insserv: loop involving service vzquota at depth 1 insserv: loop involving service rsyslog at depth 1 insserv: Starting vzquota depends on stop-bootlogd and therefore on system facility `$all' which can not be true! insserv: Starting vzquota depends on stop-bootlogd and therefore on system facility `$all' which can not be true! insserv: There is a loop between service vzquota and stop-bootlogd if started insserv: Starting vzquota depends on stop-bootlogd and therefore on system facility `$all' which can not be true! insserv: Starting vzquota depends on stop-bootlogd and therefore on system facility `$all' which can not be true! insserv: exiting now without changing boot order! update-rc.d: error: insserv rejected the script header dpkg: error processing exim4-base (--configure): subprocess installed post-installation script returned error exit status 1 Any suggestions? Keywords: vzquota debian squeeze installation vps, virtual private server.

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• Varnish gets in a restart loop and causes the system to lock up; how can I fix?

  - by chrism2671

  Here is an extract from the syslog. Mar 2 14:01:10 ip-10-226-34-17 varnishd[20204]: Child (20205) not responding to ping, killing it. Mar 2 14:01:16 ip-10-226-34-17 varnishd[20204]: Child (20205) not responding to ping, killing it. Mar 2 14:01:16 ip-10-226-34-17 varnishd[20204]: Child (20205) died signal=3 Mar 2 14:01:21 ip-10-226-34-17 varnishd[20204]: Child cleanup complete Mar 2 14:01:21 ip-10-226-34-17 varnishd[20204]: child (13224) Started Mar 2 14:01:21 ip-10-226-34-17 varnishd[20204]: Child (13224) said Closed fds: 4 5 6 10 11 13 14 Mar 2 14:01:21 ip-10-226-34-17 varnishd[20204]: Child (13224) said Child starts Mar 2 14:01:21 ip-10-226-34-17 varnishd[20204]: Child (13224) said managed to mmap 536870912 bytes of 536870912 Mar 2 14:01:21 ip-10-226-34-17 varnishd[20204]: Child (13224) said Ready Mar 2 14:01:35 ip-10-226-34-17 varnishd[20204]: Child (13224) not responding to ping, killing it. Mar 2 14:02:10 ip-10-226-34-17 last message repeated 7 times Mar 2 14:03:15 ip-10-226-34-17 last message repeated 13 times Mar 2 14:03:20 ip-10-226-34-17 varnishd[20204]: Child (13224) not responding to ping, killing it. Mar 2 14:05:53 ip-10-226-34-17 varnishd[20204]: Child (13224) not responding to ping, killing it. Mar 2 14:05:53 ip-10-226-34-17 varnishd[20204]: Child (13224) not responding to ping, killing it. Mar 2 14:05:53 ip-10-226-34-17 varnishd[20204]: Child (13224) died signal=3 Mar 2 14:05:53 ip-10-226-34-17 varnishd[20204]: Child cleanup complete Mar 2 14:05:53 ip-10-226-34-17 varnishd[20204]: child (13288) Started I'm not expecting a solution here but any help just to decode what each line is doing would be very instructive. Many thanks!

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