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  • HP devient numéro 1 mondial du marché des serveurs, à la fois en chiffre d'affaires et en nombre d'unités vendues

    HP devient numéro 1 mondial du marché des serveurs A la fois en chiffre d'affaires et en nombre d'unités vendues Selon un nouveau rapport publié par IDC, HP est devenu numéro 1 mondial du marché des serveurs en 2010. Et ce à la fois en termes de chiffre d'affaires qu'en nombre d'unités vendues. Selon IDC, HP totalise 39% de part de marché en chiffre d'affaires sur le segment des serveurs x86 en 2010, avec une croissance de 34% par rapport à 2009 (de 29 % suppérieure à celle du marcché). « Cela fait désormais plus de 14 années consécutives (59 trimestres), depuis qu'IDC étudie la part de marché des serveurs x86, que les serveurs HP ProLiant occupent la première place de ...

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  • Mandriva Linux 2010.2 est disponible, et apporte de nombreux correctifs et améliorations autour du noyau Linux 2.6.33.7

    Mandriva Linux 2010.2 est disponible, elle apporte de nombreux correctifs et améliorations autour du noyau Linux 2.6.33.7 Mise à jour du 24.12.2010 par Katleen La distribution Linux Mandriva est désormais disponible dans une nouvelle version, la 2010.2. Il s'agit d'une mise à jour majeure et incrémentielle de Mandriva Linux 2010 Spring. Elle apporte plusieurs amélioration de sécurité ainsi que des correctifs de bogues. Cette version est la seconde dite "d'entretien". Elle est construite autour du noyau Linux 2.6.33.7 et intègre de nombreux environnements et outils : - GNOME 2.30.0 - KDE 4.4.3 - XFCE 4.6.1 - OpenOffice 3.2 - Thunderbird 3.0 -...

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  • IBM, numéro 1 des dépôts de brevets pour la 18eme année consécutive avec un record de 5.896 brevets, Microsoft dans le top 3

    IBM, numéro 1 des dépôts de brevets pour la 18eme année consécutive Avec un record de 5 896 brevets, Samsung et Microsoft dans le top 3 IBM est une fois de plus le numéro 1 dans le classement annuel des entreprises ayant déposé le plus de brevets au cours de l'année 2010. Les chercheurs de la firme ont réalisé un record avec un dépôt de 5 896 brevets au cours de l'année 2010, soit une augmentation de 10 % par rapport à l'année précédente. Ce chiffre fait de « Big Blue » la première ayant déposé plus de 5000 brevets au cours d'une année. C'est la 18eme année consécutive qu'IBM occupe cette place de numéro 1. Ces brevets touchent des domaines très variés. La fi...

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  • Playing with Windows Phone Developer Tools CTP

    tweetmeme_source = 'alpascual'; Installation tips. If Visual Studio 2010 Professional or higher is already installed on your development computer, an add-in for Visual Studio 2010 Professional is automatically installed as well. The installation took an hour on a Windows 7 with 4 GB of RAM and rebooted the computer once. Something tells me the installer still needs some work.   Lets Write some code Everything installed, lets check if Visual Studio 2008 still works with Silverlight...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Change AccountName/LoginName for a SharePoint User (SPUser)

    - by Rohit Gupta
    Consider the following: We have an account named MYDOMAIN\eholz. This accounts Active Directory Login Name changes to MYDOMAIN\eburrell Now this user was a active user in a Sharepoint 2010 team Site, and had a userProfile using the Account name MYDOMAIN\eholz. Since the AD LoginName changed to eburrell hence we need to update the Sharepoint User (SPUser object) as well update the userprofile to reflect the new account name. To update the Sharepoint User LoginName we can run the following stsadm command on the Server: STSADM –o migrateuser –oldlogin MYDOMAIN\eholz –newlogin MYDOMAIN\eburrell –ignoresidhistory However to update the Sharepoint 2010 UserProfile, i first tried running a Incremental/Full Synchronization using the User Profile Synchronization service… this did not work. To enable me to update the AccountName field (which is a read only field) of the UserProfile, I had to first delete the User Profile for MYDOMAIN\eholz and then run a FULL Synchronization using the User Profile Synchronization service which synchronizes the Sharepoint User Profiles with the AD profiles.

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  • And just like that, there went February

    - by Enrique Lima
    The best intentions were having me at perhaps posting more than twice a month, and here we are and … well it has been busy. There has been a lot of SharePoint 2010 already in this first number of weeks of the year, a good amount of TFS 2010, but even better working quite actively on two topics that I enjoy quite a bit … Windows Azure and Application Lifecycle Management.  And there is more to come around those two. Through the influencers program with Geekswithblogs and Discountasp.net I got access to a hosted TFS solution that I am currently testing and will be posting my findings and documenting a good amount of information on that process. Another great resource has been fpweb.net, and there will be more details on this too.  Pretty exciting stuff! That is what is going on and what will be coming on here shortly.

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  • Buy ReSharper 6 - Get Version 7 Free!?

    - by TATWORTH
    A tip that has just been passed to me by my good friends at Jet Brains.JetBrains ReSharper is approaching its new major release later this summer. We're delighted to announce a limited 2-in-1 offer: all new and upgrade ReSharper 6 licenses purchased on or after June 1, 2012, are entitled to a free upgrade for the upcoming ReSharper 7. Below is a list of features and improvements that will be included in ReSharper 7: Visual Studio 2012 Release Candidate support. Visual Studio 2012 RTM support will be provided as soon as it is available.Continued support for Visual Studio 2005, 2008 and 2010.Support for Windows 8 and for developing the new trend of Metro style applications.New code inspections and quick-fixes for different languages, including C# and VB.NET.Multiple JavaScript support improvements.Enhanced XAML development support pack.More ReSharper functionality for SharePoint, ASP.NET 4.5, ASP.NET MVC 4, and Silverlight 5.Unit testing improvements, including support for MSTest 11, NUnit 2.6, Jasmine and PhantomJS.Compatibility with dark schemes in Visual Studio 2010 and 12, and overall support for custom themes.More improvements in quick-fixes, code annotations, code hierarchy views, and refactorings. Enjoy ReSharper 7 free, when you upgrade to ReSharper 6 or buy new licenses now.

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  • Trimestre financier record pour Adobe qui franchit la barre du milliard de dollars de chiffre d'affaires trimestriel

    Trimestre financier record pour Adobe qui franchit la barre du milliard De dollars de chiffre d'affaires trimestriel Adobe vient d'annoncer un bilan financier resplendissant en ce quatrième et dernier trimestre financier de l'année 2010. L'entreprise se réjouit surtout de rejoindre le club des entreprises IT enregistrant plus d'un milliard de dollars de chiffre d'affaire (CA) trimestriel. Soit une croissance de 33% par rapport au même trimestre de l'année passée où l'entreprise n'avait enregistré "que" 757.3 millions de dollars de revenus. Pour l'ensemble de l'année 2010, l'entreprise a réalisé 3.8 milliards de dollars de CA (+29%). Des perf...

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  • Hacker Croll ??

    - by Haruto Kitano, CISSP-ISSJP
    ??????????????????????????????????? ???4?16??HP???HP Security DAY 2010????????????????ID????????????????????????2????????????????????????????????????????PCIDSS?????12??????????????????????????????????????????????????????????????????? ???????···

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  • ??????????·??????????? - DBSC???!??????·??????

    - by Yusuke.Yamamoto
    ????? ??:2010/11/09 ??:???? ?????????????????·??????(??????)????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? ???????????·????????????????????·????????????????????????????????????????????????????????????????????????????????????????????????????????????????RBAC ????????? ????????????????? http://thinkit.co.jp/story/2010/11/09/1852

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  • ???????????? -Windows?-

    - by atsuko.nishihata
    Windows7/Windows Server 2008 R2???Oracle Database????????????Tips???????????????? 2010?4?28?(?) 11:00~12:00 Windows 7 / Windows Server 2008 R2 ?????? Oracle ! Oracle Direct ??????????????????????????????????????????????/???????????? ?Wondows7/Windows Server 2008 R2?? Oracle Database???????·???????? ?????? ????????! ????????Oracle Database 11g Release 2??Windows Server 2008 R2???Windows 7????? ??????????????????????? ?Oracle Database 11g Release 2??...2010?5?18????Microsoft Windows Server 2008 R2???Microsoft Windows 7?????????????·??????????????????????????????????????????????????? (??????????) ???OTN???????????????????????? ???????????????????????

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  • Selecting unique records in XSLT/XPath

    - by Daniel I-S
    I have to select only unique records from an XML document, in the context of an <xsl:for-each> loop. I am limited by Visual Studio to using XSL 1.0. <availList> <item> <schDate>2010-06-24</schDate> <schFrmTime>10:00:00</schFrmTime> <schToTime>13:00:00</schToTime> <variousOtherElements></variousOtherElements> </item> <item> <schDate>2010-06-24</schDate> <schFrmTime>10:00:00</schFrmTime> <schToTime>13:00:00</schToTime> <variousOtherElements></variousOtherElements> </item> <item> <schDate>2010-06-25</schDate> <schFrmTime>10:00:00</schFrmTime> <schToTime>12:00:00</schToTime> <variousOtherElements></variousOtherElements> </item> <item> <schDate>2010-06-26</schDate> <schFrmTime>13:00:00</schFrmTime> <schToTime>14:00:00</schToTime> <variousOtherElements></variousOtherElements> </item> <item> <schDate>2010-06-26</schDate> <schFrmTime>10:00:00</schFrmTime> <schToTime>12:00:00</schToTime> <variousOtherElements></variousOtherElements> </item> </availList> The uniqueness must be based on the value of the three child elements: schDate, schFrmTime and schToTime. If two item elements have the same values for all three child elements, they are duplicates. In the above XML, items one and two are duplicates. The rest are unique. As indicated above, each item contains other elements that we do not wish to include in the comparison. 'Uniqueness' should be a factor of those three elements, and those alone. I have attempted to accomplish this through the following: availList/item[not(schDate = preceding:: schDate and schFrmTime = preceding:: schFrmTime and schToTime = preceding:: schToTime)] The idea behind this is to select records where there is no preceding element with the same schDate, schFrmTime and schToTime. However, its output is missing the last item. This is because my XPath is actually excluding items where all of the child element values are matched within the entire preceding document. No single item matches all of the last item's child elements - but because each element's value is individually present in another item, the last item gets excluded. I could get the correct result by comparing all child values as a concatenated string to the same concatenated values for each preceding item. Does anybody know of a way I could do this?

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  • .NET Framework generates strange DCOM error

    - by Anders Oestergaard Jensen
    Hello, I am creating a simple application that enables merging of key-value pairs fields in a Word and/or Excel document. Until this day, the application has worked out just fine. I am using the latest version of .NET Framework 4.0 (since it provides a nice wrapper API for Interop). My sample merging method looks like this: public byte[] ProcessWordDocument(string path, List<KeyValuePair<string, string>> kvs) { logger.InfoFormat("ProcessWordDocument: path = {0}", path); var localWordapp = new Word.Application(); localWordapp.Visible = false; Word.Document doc = null; try { doc = localWordapp.Documents.Open(path, ReadOnly: false); logger.Debug("Executing Find->Replace..."); foreach (Word.Range r in doc.StoryRanges) { foreach (KeyValuePair<string, string> kv in kvs) { r.Find.Execute(Replace: Word.WdReplace.wdReplaceAll, FindText: kv.Key, ReplaceWith: kv.Value, Wrap: Word.WdFindWrap.wdFindContinue); } } logger.Debug("Done! Saving document and cleaning up"); doc.Save(); doc.Close(); System.Runtime.InteropServices.Marshal.ReleaseComObject(doc); localWordapp.Quit(); System.Runtime.InteropServices.Marshal.ReleaseComObject(localWordapp); logger.Debug("Done."); return System.IO.File.ReadAllBytes(path); } catch (Exception ex) { // Logging... // doc.Close(); if (doc != null) { doc.Close(); System.Runtime.InteropServices.Marshal.ReleaseComObject(doc); } localWordapp.Quit(); System.Runtime.InteropServices.Marshal.ReleaseComObject(localWordapp); throw; } } The above C# snippet has worked all fine (compiled and deployed unto a Windows Server 2008 x64) with latest updates installed. But now, suddenly, I get the following strange error: System.Runtime.InteropServices.COMException (0x80080005): Retrieving the COM class factory for component with CLSID {000209FF-0000-0000-C000-000000000046} failed due to the following error: 80080005 Server execution failed (Exception from HRESULT: 0x80080005 (CO_E_SERVER_EXEC_FAILURE)). at System.RuntimeTypeHandle.CreateInstance(RuntimeType type, Boolean publicOnly, Boolean noCheck, Boolean& canBeCached, RuntimeMethodHandleInternal& ctor, Boolean& bNeedSecurityCheck) at System.RuntimeType.CreateInstanceSlow(Boolean publicOnly, Boolean skipCheckThis, Boolean fillCache) at System.RuntimeType.CreateInstanceDefaultCtor(Boolean publicOnly, Boolean skipVisibilityChecks, Boolean skipCheckThis, Boolean fillCache) at System.Activator.CreateInstance(Type type, Boolean nonPublic) at Meeho.Integration.OfficeHelper.ProcessWordDocument(String path, List`1 kvs) in C:\meeho\src\webservices\Meeho.Integration\OfficeHelper.cs:line 30 at Meeho.IntegrationService.ConvertDocument(Byte[] template, String ext, String[] fields, String[] values) in C:\meeho\src\webservices\MeehoService\IntegrationService.asmx.cs:line 49 -- I googled the COM error, but it returns nothing of particular value. I even gave the right permissions for the COM dll's using mmc -32, where I allocated the Word and Excel documents respectively and set the execution rights to the Administrator. I could not, however, locate the dll's by the exact COM CLSID given above. Very frustrating. Please, please, please help me as the application is currently pulled out of production. Anders EDIT: output from the Windows event log: Faulting application name: WINWORD.EXE, version: 12.0.6514.5000, time stamp: 0x4a89d533 Faulting module name: unknown, version: 0.0.0.0, time stamp: 0x00000000 Exception code: 0xc0000005 Fault offset: 0x00000000 Faulting process id: 0x720 Faulting application start time: 0x01cac571c4f82a7b Faulting application path: C:\Program Files (x86)\Microsoft Office\Office12\WINWORD.EXE Faulting module path: unknown Report Id: 041dd5f9-3165-11df-b96a-0025643cefe6 - 1000 2 100 0x80000000000000 2963 Application meeho3 - WINWORD.EXE 12.0.6514.5000 4a89d533 unknown 0.0.0.0 00000000 c0000005 00000000 720 01cac571c4f82a7b C:\Program Files (x86)\Microsoft Office\Office12\WINWORD.EXE unknown 041dd5f9-3165-11df-b96a-0025643cefe6

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  • Getting Error System.Runtime.InteropServices.COMException

    - by Savan Parmar
    Hey All, I am using Vb.Net to Create Labels in Microsoft. For that i am using below mantioend Code. Public Sub CreateLabel(ByVal StrFilter As String, ByVal Path As String) WordApp = CreateObject("Word.Application") ''Add a new document. WordDoc = WordApp.Documents.Add() Dim oConn As SqlConnection = New SqlConnection(connSTR) oConn.Open() Dim oCmd As SqlCommand Dim oDR As SqlDataReader oCmd = New SqlCommand(StrFilter, oConn) oDR = oCmd.ExecuteReader Dim intI As Integer Dim FilePath As String = "" With WordDoc.MailMerge With .Fields Do While oDR.Read For intI = 0 To oDR.FieldCount - 1 .Add(WordApp.Selection.Range, oDR.Item(intI)) Next Loop End With Dim objAutoText As Word.AutoTextEntry = WordApp.NormalTemplate.AutoTextEntries.Add("MyLabelLayout", WordDoc.Content) WordDoc.Content.Delete() .MainDocumentType = Word.WdMailMergeMainDocType.wdMailingLabels FilePath = CreateSource(StrFilter) .OpenDataSource(FilePath) Dim NewLabel As Word.CustomLabel = WordApp.MailingLabel.CustomLabels.Add("MyLabel", False) WordApp.MailingLabel.CreateNewDocument(Name:="MyLabel", Address:="", AutoText:="MyLabelLayout") objAutoText.Delete() .Destination = Word.WdMailMergeDestination.wdSendToNewDocument WordApp.Visible = True .Execute() End With oConn.Close() WordDoc.Close() End Sub Private Function CreateSource(ByVal StrFilter As String) As String Dim CnnUser As SqlConnection = New SqlConnection(connSTR) Dim sw As StreamWriter = File.CreateText("C:\Mail.Txt") Dim Path As String = "C:\Mail.Txt" Dim StrHeader As String = "" Try Dim SelectCMD As SqlCommand = New SqlCommand(StrFilter, CnnUser) Dim oDR As SqlDataReader Dim IntI As Integer SelectCMD.CommandType = CommandType.Text CnnUser.Open() oDR = SelectCMD.ExecuteReader For IntI = 0 To oDR.FieldCount - 1 StrHeader &= oDR.GetName(IntI) & " ," Next StrHeader = Mid(StrHeader, 1, Len(StrHeader) - 2) sw.WriteLine(StrHeader) sw.Flush() sw.Close() StrHeader = "" Do While oDR.Read For IntJ As Integer = 0 To oDR.FieldCount - 1 StrHeader &= oDR.GetString(IntJ) & " ," Next Loop StrHeader = Mid(StrHeader, 1, Len(StrHeader) - 2) sw = File.AppendText(Path) sw.WriteLine(StrHeader) CnnUser.Close() sw.Flush() sw.Close() Catch ex As Exception MessageBox.Show(ex.Message, "TempID", MessageBoxButtons.OK, MessageBoxIcon.Error) End Try Return Path End Function Now when i am running the programm i am getting this error.I tried hard but not able to locate what could be the problem the error is:- System.Runtime.InteropServices.COMException --Horizontal and vertical pitch must be greater than or equal to the label width and height, respectively. Even though i tried to set the Horizontal and vertical pitch programatically but it gives same err. Plz if any one can help

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  • Delphi Phrase Count / Keyword Density

    - by Brad
    Does anyone know how to or have some code on counting the number of unique phrases in a document? (Single word, two word phrases, three word phrases). Thanks Example of what I'm looking for: What I mean is I have a text document, and i need to see what the most popular word phrases are. Example text I took the car to the car wash. I : 1 took : 1 the : 2 car: 2 to : 1 wash : 1 I took : 1 took the : 1 the car : 2 car to : 1 to the : 1 car wash : 1 I took the : 1 took the car : 1 the car to : 1 car to the : 1 to the car : 1 the car wash : 1 I took the car to : 1 took the car to the : 1 the car to the car : 1 car to the car wash : 1 I need the phrase, and the count that it shows up. Any help would be appreciated. The closet thing I found to this was a PHP script from http://tools.seobook.com/general/keyword-density/source.php I used to have some code for this, but I cannot find it.

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  • Problem calling linux C code from FIQ handler

    - by fastmonkeywheels
    I'm working on an armv6 core and have an FIQ hander that works great when I do all of my work in it. However I need to branch to some additional code that's too large for the FIQ memory area. The FIQ handler gets copied from fiq_start to fiq_end to 0xFFFF001C when registered static void test_fiq_handler(void) { asm volatile("\ .global fiq_start\n\ fiq_start:"); // clear gpio irq asm("ldr r10, GPIO_BASE_ISR"); asm("ldr r9, [r10]"); asm("orr r9, #0x04"); asm("str r9, [r10]"); // clear force register asm("ldr r10, AVIC_BASE_INTFRCH"); asm("ldr r9, [r10]"); asm("mov r9, #0"); asm("str r9, [r10]"); // prepare branch register asm(" ldr r11, fiq_handler"); // save all registers, build sp and branch to C asm(" adr r9, regpool"); asm(" stmia r9, {r0 - r8, r14}"); asm(" adr sp, fiq_sp"); asm(" ldr sp, [sp]"); asm(" add lr, pc,#4"); asm(" mov pc, r11"); #if 0 asm("ldr r10, IOMUX_ADDR12"); asm("ldr r9, [r10]"); asm("orr r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); asm("bic r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); #endif asm(" adr r9, regpool"); asm(" ldmia r9, {r0 - r8, r14}"); // return asm("subs pc, r14, #4"); asm("IOMUX_ADDR12: .word 0xFC2A4000"); asm("AVIC_BASE_INTCNTL: .word 0xFC400000"); asm("AVIC_BASE_INTENNUM: .word 0xFC400008"); asm("AVIC_BASE_INTDISNUM: .word 0xFC40000C"); asm("AVIC_BASE_FIVECSR: .word 0xFC400044"); asm("AVIC_BASE_INTFRCH: .word 0xFC400050"); asm("GPIO_BASE_ISR: .word 0xFC2CC018"); asm(".globl fiq_handler"); asm("fiq_sp: .long fiq_stack+120"); asm("fiq_handler: .long 0"); asm("regpool: .space 40"); asm(".pool"); asm(".align 5"); asm("fiq_stack: .space 124"); asm(".global fiq_end"); asm("fiq_end:"); } fiq_hander gets set to the following function: static void fiq_flip_pins(void) { asm("ldr r10, IOMUX_ADDR12_k"); asm("ldr r9, [r10]"); asm("orr r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); asm("bic r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); asm("IOMUX_ADDR12_k: .word 0xFC2A4000"); } EXPORT_SYMBOL(fiq_flip_pins); I know that since the FIQ handler operates outside of any normal kernel API's and that it is a rather high priority interrupt I must ensure that whatever I call is already swapped into memory. I do this by having the fiq_flip_pins function defined in the monolithic kernel and not as a module which gets vmalloc. If I don't branch to the fiq_flip_pins function, and instead do the work in the test_fiq_handler function everything works as expected. It's the branching that's causing me problems at the moment. Right after branching I get a kernel panic about a paging request. I don't understand why I'm getting the paging request. fiq_flip_pins is in the kernel at: c00307ec t fiq_flip_pins Unable to handle kernel paging request at virtual address 736e6f63 pgd = c3dd0000 [736e6f63] *pgd=00000000 Internal error: Oops: 5 [#1] PREEMPT Modules linked in: hello_1 CPU: 0 Not tainted (2.6.31-207-g7286c01-svn4 #122) PC is at strnlen+0x10/0x28 LR is at string+0x38/0xcc pc : [<c016b004>] lr : [<c016c754>] psr: a00001d3 sp : c3817ea0 ip : 736e6f63 fp : 00000400 r10: c03cab5c r9 : c0339ae0 r8 : 736e6f63 r7 : c03caf5c r6 : c03cab6b r5 : ffffffff r4 : 00000000 r3 : 00000004 r2 : 00000000 r1 : ffffffff r0 : 736e6f63 Flags: NzCv IRQs off FIQs off Mode SVC_32 ISA ARM Segment user Control: 00c5387d Table: 83dd0008 DAC: 00000015 Process sh (pid: 1663, stack limit = 0xc3816268) Stack: (0xc3817ea0 to 0xc3818000) Since there are no API calls in my code I have to assume that something is going wrong in the C call and back. Any help solving this is appreciated.

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  • Java: If vs. Switch

    - by _ande_turner_
    I have a piece of code with a) which I replaced with b) purely for legibility ... a) if ( WORD[ INDEX ] == 'A' ) branch = BRANCH.A; /* B through to Y */ if ( WORD[ INDEX ] == 'Z' ) branch = BRANCH.Z; b) switch ( WORD[ INDEX ] ) { case 'A' : branch = BRANCH.A; break; /* B through to Y */ case 'Z' : branch = BRANCH.Z; break; } ... will the switch version cascade through all the permutations or jump to a case ? EDIT: Some of the answers below regard alternative approaches to the approach above. I have included the following to provide context for its use. The reason I asked, the Question above, was because the speed of adding words empirically improved. This isn't production code by any means, and was hacked together quickly as a PoC. The following seems to be a confirmation of failure for a thought experiment. I may need a much bigger corpus of words than the one I am currently using though. The failure arises from the fact I did not account for the null references still requiring memory. ( doh ! ) public class Dictionary { private static Dictionary ROOT; private boolean terminus; private Dictionary A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z; private static Dictionary instantiate( final Dictionary DICTIONARY ) { return ( DICTIONARY == null ) ? new Dictionary() : DICTIONARY; } private Dictionary() { this.terminus = false; this.A = this.B = this.C = this.D = this.E = this.F = this.G = this.H = this.I = this.J = this.K = this.L = this.M = this.N = this.O = this.P = this.Q = this.R = this.S = this.T = this.U = this.V = this.W = this.X = this.Y = this.Z = null; } public static void add( final String...STRINGS ) { Dictionary.ROOT = Dictionary.instantiate( Dictionary.ROOT ); for ( final String STRING : STRINGS ) Dictionary.add( STRING.toUpperCase().toCharArray(), Dictionary.ROOT , 0, STRING.length() - 1 ); } private static void add( final char[] WORD, final Dictionary BRANCH, final int INDEX, final int INDEX_LIMIT ) { Dictionary branch = null; switch ( WORD[ INDEX ] ) { case 'A' : branch = BRANCH.A = Dictionary.instantiate( BRANCH.A ); break; case 'B' : branch = BRANCH.B = Dictionary.instantiate( BRANCH.B ); break; case 'C' : branch = BRANCH.C = Dictionary.instantiate( BRANCH.C ); break; case 'D' : branch = BRANCH.D = Dictionary.instantiate( BRANCH.D ); break; case 'E' : branch = BRANCH.E = Dictionary.instantiate( BRANCH.E ); break; case 'F' : branch = BRANCH.F = Dictionary.instantiate( BRANCH.F ); break; case 'G' : branch = BRANCH.G = Dictionary.instantiate( BRANCH.G ); break; case 'H' : branch = BRANCH.H = Dictionary.instantiate( BRANCH.H ); break; case 'I' : branch = BRANCH.I = Dictionary.instantiate( BRANCH.I ); break; case 'J' : branch = BRANCH.J = Dictionary.instantiate( BRANCH.J ); break; case 'K' : branch = BRANCH.K = Dictionary.instantiate( BRANCH.K ); break; case 'L' : branch = BRANCH.L = Dictionary.instantiate( BRANCH.L ); break; case 'M' : branch = BRANCH.M = Dictionary.instantiate( BRANCH.M ); break; case 'N' : branch = BRANCH.N = Dictionary.instantiate( BRANCH.N ); break; case 'O' : branch = BRANCH.O = Dictionary.instantiate( BRANCH.O ); break; case 'P' : branch = BRANCH.P = Dictionary.instantiate( BRANCH.P ); break; case 'Q' : branch = BRANCH.Q = Dictionary.instantiate( BRANCH.Q ); break; case 'R' : branch = BRANCH.R = Dictionary.instantiate( BRANCH.R ); break; case 'S' : branch = BRANCH.S = Dictionary.instantiate( BRANCH.S ); break; case 'T' : branch = BRANCH.T = Dictionary.instantiate( BRANCH.T ); break; case 'U' : branch = BRANCH.U = Dictionary.instantiate( BRANCH.U ); break; case 'V' : branch = BRANCH.V = Dictionary.instantiate( BRANCH.V ); break; case 'W' : branch = BRANCH.W = Dictionary.instantiate( BRANCH.W ); break; case 'X' : branch = BRANCH.X = Dictionary.instantiate( BRANCH.X ); break; case 'Y' : branch = BRANCH.Y = Dictionary.instantiate( BRANCH.Y ); break; case 'Z' : branch = BRANCH.Z = Dictionary.instantiate( BRANCH.Z ); break; } if ( INDEX == INDEX_LIMIT ) branch.terminus = true; else Dictionary.add( WORD, branch, INDEX + 1, INDEX_LIMIT ); } public static boolean is( final String STRING ) { Dictionary.ROOT = Dictionary.instantiate( Dictionary.ROOT ); return Dictionary.is( STRING.toUpperCase().toCharArray(), Dictionary.ROOT, 0, STRING.length() - 1 ); } private static boolean is( final char[] WORD, final Dictionary BRANCH, final int INDEX, final int INDEX_LIMIT ) { Dictionary branch = null; switch ( WORD[ INDEX ] ) { case 'A' : branch = BRANCH.A; break; case 'B' : branch = BRANCH.B; break; case 'C' : branch = BRANCH.C; break; case 'D' : branch = BRANCH.D; break; case 'E' : branch = BRANCH.E; break; case 'F' : branch = BRANCH.F; break; case 'G' : branch = BRANCH.G; break; case 'H' : branch = BRANCH.H; break; case 'I' : branch = BRANCH.I; break; case 'J' : branch = BRANCH.J; break; case 'K' : branch = BRANCH.K; break; case 'L' : branch = BRANCH.L; break; case 'M' : branch = BRANCH.M; break; case 'N' : branch = BRANCH.N; break; case 'O' : branch = BRANCH.O; break; case 'P' : branch = BRANCH.P; break; case 'Q' : branch = BRANCH.Q; break; case 'R' : branch = BRANCH.R; break; case 'S' : branch = BRANCH.S; break; case 'T' : branch = BRANCH.T; break; case 'U' : branch = BRANCH.U; break; case 'V' : branch = BRANCH.V; break; case 'W' : branch = BRANCH.W; break; case 'X' : branch = BRANCH.X; break; case 'Y' : branch = BRANCH.Y; break; case 'Z' : branch = BRANCH.Z; break; } if ( branch == null ) return false; if ( INDEX == INDEX_LIMIT ) return branch.terminus; else return Dictionary.is( WORD, branch, INDEX + 1, INDEX_LIMIT ); } }

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  • Link List Problem,

    - by david
    OK i have a problem with a Link List program i'm trying to Do, the link List is working fine. Here is my code #include <iostream> using namespace std; struct record { string word; struct record * link; }; typedef struct record node; node * insert_Node( node * head, node * previous, string key ); node * search( node *head, string key, int *found); void displayList(node *head); node * delete_node( node *head, node * previous, string key); int main() { node * previous, * head = NULL; int found = 0; string node1Data,newNodeData, nextData,lastData; //set up first node cout <<"Depature"<<endl; cin >>node1Data; previous = search( head, node1Data, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, node1Data); cout <<"Depature inserted"<<endl; //insert node between first node and head cout <<"Destination"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Destinationinserted"<<endl; //insert node between second node and head cout <<"Cost"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Cost inserted"<<endl; cout <<"Number of Seats Required"<<endl; //place node between new node and first node cin >>nextData; previous = search( head, nextData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, nextData); cout <<"Number of Seats Required inserted"<<endl; //insert node between first node and head cout <<"Name"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Name inserted"<<endl; //insert node between node and head cout <<"Address "<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Address inserted"<<endl; //place node as very last node cin >>lastData; previous = search( head, lastData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, lastData); cout <<"C"<<endl; displayList(head); char Ans = 'y'; //Delete nodes do { cout <<"Enter Keyword to be delete"<<endl; cin >>nextData; previous = search( head, nextData, &found); if (found == 1) head = delete_node( head, previous,nextData); displayList(head); cout <<"Do you want to Delete more y /n "<<endl; cin >> Ans; } while( Ans =='y'); int choice, i=0, counter=0; int fclass[10]; int coach[10]; printf("Welcome to the booking program"); printf("\n-----------------"); do{ printf("\n Please pick one of the following option:"); printf("\n 1) Reserve a first class seat on Flight 101."); printf("\n 2) Reserve a coach seat on Flight 101."); printf("\n 3) Quit "); printf("\n ---------------------------------------------------------------------"); printf("\nYour choice?"); scanf("%d",&choice); switch(choice) { case 1: i++; if (i <10){ printf("Here is your seat: %d " , fclass[i]); } else if (i = 10) { printf("Sorry there is no more seats on First Class. Please wait for the next flight"); } break; case 2: if (i <10){ printf("Here is your Seat Coach: %d " , coach[i]); } else if ( i = 10) { printf("Sorry their is no more Seats on Coach. Please wait for the next flight"); } break; case 3: printf("Thank you and goodbye\n"); //exit(0); } } while (choice != 3); } /******************************************************* search function to return previous position of node ******************************************************/ node * search( node *head, string key, int *found) { node * previous, * current; current = head; previous = current; *found = 0;//not found //if (current->word < key) move through links until the next link //matches or current_word > key while( current !=NULL) { //compare exactly if (key ==current->word ) { *found = 1; break; } //if key is less than word else if ( key < current->word ) break; else { //previous stays one link behind current previous = current; current = previous -> link; } } return previous; } /******************************************************** display function as used with createList ******************************************************/ void displayList(node *head) { node * current; //current now contains the address held of the 1st node similar //to head current = head; cout << "\n\n"; if( current ==NULL) cout << "Empty List\n\n"; else { /*Keep going displaying the contents of the list and set current to the address of the next node. When set to null, there are no more nodes */ while(current !=NULL) { cout << current->word<<endl; current = current ->link; } } } /************************************************************ insert node used to position node (i) empty list head = NULL (ii) to position node before the first node key < head->word (iii) every other position including the end of the list This is done using the following steps (a) Pass in all the details to create the node either details or a whole record (b) Pass the details over to fill the node (C) Use the if statement to add the node to the list **********************************************************/ node * insert_Node( node * head, node * previous, string key ) { node * new_node, * temp; new_node = new node; //create the node new_node ->word = key; new_node -> link = NULL; if (head == NULL || key < head->word ) //empty list { //give address of head to temp temp = head; //head now points to the new_node head = new_node; //new_node now points to what head was pointing at new_node -> link = temp; } else { //pass address held in link to temp temp = previous-> link; //put address of new node to link of previous previous -> link = new_node; //pass address of temp to link of new node new_node -> link = temp; } return head; } node * delete_node( node *head, node * previous, string key) { /* this function will delete a node but will not return its contents */ node * temp; if(key == head->word) //delete node at head of list { temp = head; //point head at the next node head = head -> link; } else { //holds the address of the node after the one // to be deleted temp = previous-> link; /*assign the previous to the address of the present node to be deleted which holds the address of the next node */ previous-> link = previous-> link-> link; } delete temp; return head; }//end delete The problem i have is when i Enter in the Number 2 in the Node(Seats) i like to get a Counter Taken 2 off of 50, some thing like what i have here enter code here int choice, i=0, counter=0; int fclass[10]; int coach[10]; printf("Welcome to the booking program"); printf("\n-----------------"); do{ printf("\n Please pick one of the following option:"); printf("\n 1) Reserve a first class seat on Flight 101."); printf("\n 2) Reserve a coach seat on Flight 101."); printf("\n 3) Quit "); printf("\n ---------------------------------------------------------------------"); printf("\nYour choice?"); scanf("%d",&choice); switch(choice) { case 1: i++; if (i <10){ printf("Here is your seat: %d " , fclass[i]); } else if (i = 10) { printf("Sorry there is no more seats on First Class. Please wait for the next flight"); } break; case 2: if (i <10){ printf("Here is your Seat Coach: %d " , coach[i]); } else if ( i = 10) { printf("Sorry their is no more Seats on Coach. Please wait for the next flight"); } break; case 3: printf("Thank you and goodbye\n"); //exit(0); } } while (choice != 3); How can i get what the User enters into number of Seats into this function

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