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  • git merge should ignore one directory

    - by dorelal
    I have tons of data in directory called reports. While doing git merge with another branch I am getting lots of conflicts for files under reports directory. I would like git merge to ignore files under reports. In another words I would like all the data from reports from master and not from lab branch. Is that possible? This is what I am doing right now. git checkout master git merge lab

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  • Subversion merging, tree merge

    - by krystan honour
    I need to merge changes from a branch back into trunk but want to continue work on the existing branch. I was going to use a re-integrate merge but realised this is not suitable as I will need to recreate my branch etc which for a variety of reasons is not desirable. What I really want to do is merge the current revisions in the branch down to head and then keep people working on their current working copies. So my question is , can tree merge be used to solve this or do I have to reintegrate and recreate.

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  • Git merge command

    - by Bialecki
    I'm reading the following article: http://github.com/guides/keeping-a-git-fork-in-sync-with-the-forked-repo, where they mention essentially pulling in changes from two repos at the same time by creating the following alias: pu = !"git fetch origin -v; git fetch wycats -v; git merge wycats/master" This makes sense, but, as someone new to Git, I'm curious why the commands is that versus: pu = !"git fetch origin -v; git merge origin/master; git fetch wycats -v; git merge wycats/master" or something along those lines. Basically, I'm wondering why the argument to merge is wycats/master and how it knows about origin/master automatically. Looking for a quick explanation.

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  • Simple bad merge scenario in mercurial

    - by user281180
    I have created a repository AAA and another BBB. In AAA I have created a file A with the values a1, a2, a3 and commit In BBB I have created a file B with the values b1, b2, b3, commit and export a bundle. I add the bundle in AAA and merge. I make a change in B, and write b33 in AAA and another change in B and write b23 in BBB. and commit both. I create bundle of BBB and add the bundle in AAA. I do a merge. Now I decide to revert to the revert to step 2. I no more want to have the merge of 4. changes done in B as they were bad merges. Now I want to add the bundle of 3 but I can see that it can`t see any changes anymore. Why? How can I do the merge once more?

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  • SQL Server Merge statement issue

    - by George2
    Hello everyone, I am learning and using SQL Server 2008 new Merge statement, merge statement will compare/operate source table and destination table row by row ("operate" I mean operations performed for when matched or not-matched conditions). My question is whether the whole merge process will be one transaction or each row comparison/operation will be one transaction? Appreciate if any document to prove it. thanks in advance, George

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  • How can I merge two lists and sort them working in 'linear' time?

    - by Sergio Tapia
    I have this, and it works: # E. Given two lists sorted in increasing order, create and return a merged # list of all the elements in sorted order. You may modify the passed in lists. # Ideally, the solution should work in "linear" time, making a single # pass of both lists. def linear_merge(list1, list2): finalList = [] for item in list1: finalList.append(item) for item in list2: finalList.append(item) finalList.sort() return finalList # +++your code here+++ return But, I'd really like to learn this stuff well. :) What does 'linear' time mean?

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  • Could I use Revert after abrupting Merge?

    - by John
    Hallo all, Just now I tried to upload a modified working copy to its branch in the following steps: 1. Update 2. Commit Then I attempted to Merge the changes in the trunk to this branch. However during editing of the conflicts I realized there were so many conflicting codes that I could not address completely today, then I gived up the Merge, and the working copy got an exclamation mark immediately. Thru Check for modifications I found that many many files had been modified or had conflicts. It seems that the Merge has been somehow wrongly carried out. My question: could I return to the state before the Merge simply using Revert? Thanks a lot in advance, John

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  • TFS 2010 Source Branches Never The Same

    - by Lukasz
    I have my root branch lets call it Alpha and one branch that was branched from that root lets call it Beta. I made some changes in the Beta branch and merged them back to Alpha. In theory now Alpha and Beta should be identical branches and when I do a diff they are identical. If I attempt to merge Alpha with Beta again without making any changes the changes I originally merged from Beta to Alpha will merge again from Alpha to Beta. Completing that merge and checking in the branches are the same. Now I can merge again. I can do this over and over again with no end. I was just wondering if anyone has ran into this problem before and how it can be fix. At first I thought it was harmless but when I make more changes in the Beta branch and merge the new changes as well as the original changes get merges overriding changes to these files making a mess. Thanks!

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  • Is this a bad version of the Merge Sort algorithm?

    - by SebKom
    merge1(int low, int high, int S[], U[]) { int k = (high - low + 1)/2 for q (from low to high) U[q] = S[q] int j = low int p = low int i = low + k while (j <= low + k - 1) and (i <= high) do { if ( U[j] <= U[i] ) { S[p] := U[j] j := j+1 } else { S[p] := U[i] i := i+1 } p := p+1 } if (j <= low + k - 1) { for q from p to high do { S[q] := U[j] j := j+1 } } } merge_sort1(int low, int high, int S[], U[]) { if low < high { int k := (high - low + 1)/2 merge_sort1(low, low+k-1, S, U) merge_sort1(low+k, high, S, U) merge1(low, high, S, U) } } I am really sorry for the terrible formating, as you can tell I am not a regular visitor here. So, basically, this is on my lecture notes. I find it quite confusing in general but I understand the biggest part of it. What I don't understand is the need of the "if (j <= low + k - 1)" part. It looks like it checks if there are any elements "left" in the left part. Is that even possible when mergesorting?

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  • Merge Sort issue when removing the array copy step

    - by Ime Prezime
    I've been having an issue that I couldn't debug for quite some time. I am trying to implement a MergeSort algorithm with no additional steps of array copying by following Robert Sedgewick's algorithm in "Algorithm's in C++" book. Short description of the algorithm: The recursive program is set up to sort b, leaving results in a. Thus, the recursive calls are written to leave their result in b, and we use the basic merge program to merge those files from b into a. In this way, all the data movement is done during the course of the merges. The problem is that I cannot find any logical errors but the sorting isn't done properly. Data gets overwritten somewhere and I cannot determine what logical error causes this. The data is sorted when the program is finished but it is not the same data any more. For example, Input array: { A, Z, W, B, G, C } produces the array: { A, G, W, W, Z, Z }. I can obviously see that it must be a logical error somewhere, but I have been trying to debug this for a pretty long time and I think a fresh set of eyes could maybe see what I'm missing cause I really can't find anything wrong. My code: static const int M = 5; void insertion(char** a, int l, int r) { int i,j; char * temp; for (i = 1; i < r + 1; i++) { temp = a[i]; j = i; while (j > 0 && strcmp(a[j-1], temp) > 0) { a[j] = a[j-1]; j = j - 1; } a[j] = temp; } } //merging a and b into c void merge(char ** c,char ** a, int N, char ** b, int M) { for (int i = 0, j = 0, k = 0; k < N+M; k++) { if (i == N) { c[k] = b[j++]; continue; } if (j == M) { c[k] = a[i++]; continue; } c[k] = strcmp(a[i], b[j]) < 0 ? a[i++] : b[j++]; } } void mergesortAux(char ** a, char ** b, int l, int r) { if(r - l <= M) { insertion(a, l, r); return; } int m = (l + r)/2; mergesortAux(b, a, l, m); //merge sort left mergesortAux(b, a, m+1, r); //merge sort right merge(a+l, b+l, m-l+1, b+m+1, r-m); //merge } void mergesort(char ** a,int l, int r, int size) { static char ** aux = (char**)malloc(size * sizeof(char*)); for(int i = l; i < size; i++) aux[i] = a[i]; mergesortAux(a, aux, l, r); free(aux); }

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  • How to merge after reverting in mercurial?

    - by user281180
    I have had a bad merge. Now I want to start the merge all over again. I did a revert just before the merge. Now when im trying to add the bundle , im having the message and it can no more locate the changes. What is wrong? Why isn`t it finding any change? c:\Documents and Settings\Shamima\Desktop\New Folder\test_rev94_to_tip_hg\test_rev94_to_tip.hg searching for changes no changes found [command completed successfully Tue Apr 13 16:10:37 2010]

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  • How to merge arrays with same key and different value in PHP?

    - by Martin
    Hi guys, I have arrays similarly to these: 0 => Array ( [0] => Finance / Shopping / Food, [1] => 47 ) 1 => Array ( [0] => Finance / Shopping / Food, [1] => 25 ) 2 => Array ( [0] => Finance / Shopping / Electronic, [1] => 190 ) I need to create one array with [0] as a key and [1] as value. The tricky part is that if the [0] is same it add [1] to existing value. So the result I want is: array ([Finance / Shopping / Food]=> 72, [Finance / Shopping / Electronic] => 190); thanks

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  • Merge arrow in clear case

    - by cheiav
    Hi, I have to merge all objects from a sub branch to main branch recursively. I would like to merge manually by check in the code from sub branch to main branch instead of using merge command in clear case. So after the check in into the main branch I would like to draw arrow recursively to all my objects. ic from sub branch to main branch I have used this command cleartool mkhlink -unidir Merge <sub branch path>>@@/main/<<sub branch>> <<main brach path>>@@/main/LATEST But when I dit it, it is drawing the arrow for the directory only not for all contains of the directory. Please suggest how to draw the arrow recursively from sub branch to main branch objects. Thanks in advance

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  • How can I merge several hashes into one hash in Perl?

    - by Nick
    In Perl, how do I get this: $VAR1 = { '999' => { '998' => [ '908', '906', '0', '998', '907' ] } }; $VAR1 = { '999' => { '991' => [ '913', '920', '918', '998', '916', '919', '917', '915', '912', '914' ] } }; $VAR1 = { '999' => { '996' => [] } }; $VAR1 = { '999' => { '995' => [] } }; $VAR1 = { '999' => { '994' => [] } }; $VAR1 = { '999' => { '993' => [] } }; $VAR1 = { '999' => { '997' => [ '986', '987', '990', '984', '989', '988' ] } }; $VAR1 = { '995' => { '101' => [] } }; $VAR1 = { '995' => { '102' => [] } }; $VAR1 = { '995' => { '103' => [] } }; $VAR1 = { '995' => { '104' => [] } }; $VAR1 = { '995' => { '105' => [] } }; $VAR1 = { '995' => { '106' => [] } }; $VAR1 = { '995' => { '107' => [] } }; $VAR1 = { '994' => { '910' => [] } }; $VAR1 = { '993' => { '909' => [] } }; $VAR1 = { '993' => { '904' => [] } }; $VAR1 = { '994' => { '985' => [] } }; $VAR1 = { '994' => { '983' => [] } }; $VAR1 = { '993' => { '902' => [] } }; $VAR1 = { '999' => { '992' => [ '905' ] } }; to this: $VAR1 = { '999:' => [ { '992' => [ '905' ] }, { '993' => [ { '909' => [] }, { '904' => [] }, { '902' => [] } ] }, { '994' => [ { '910' => [] }, { '985' => [] }, { '983' => [] } ] }, { '995' => [ { '101' => [] }, { '102' => [] }, { '103' => [] }, { '104' => [] }, { '105' => [] }, { '106' => [] }, { '107' => [] } ] }, { '996' => [] }, { '997' => [ '986', '987', '990', '984', '989', '988' ] }, { '998' => [ '908', '906', '0', '998', '907' ] }, { '991' => [ '913', '920', '918', '998', '916', '919', '917', '915', '912', '914' ] } ]};

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  • T-SQL - Left Outer Joins - Fileters in the where clause versus the on clause.

    - by Greg Potter
    I am trying to compare two tables to find rows in each table that is not in the other. Table 1 has a groupby column to create 2 sets of data within table one. groupby number ----------- ----------- 1 1 1 2 2 1 2 2 2 4 Table 2 has only one column. number ----------- 1 3 4 So Table 1 has the values 1,2,4 in group 2 and Table 2 has the values 1,3,4. I expect the following result when joining for Group 2: `Table 1 LEFT OUTER Join Table 2` T1_Groupby T1_Number T2_Number ----------- ----------- ----------- 2 2 NULL `Table 2 LEFT OUTER Join Table 1` T1_Groupby T1_Number T2_Number ----------- ----------- ----------- NULL NULL 3 The only way I can get this to work is if I put a where clause for the first join: PRINT 'Table 1 LEFT OUTER Join Table 2, with WHERE clause' select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table1 LEFT OUTER join table2 --****************************** on table1.number = table2.number --****************************** WHERE table1.groupby = 2 AND table2.number IS NULL and a filter in the ON for the second: PRINT 'Table 2 LEFT OUTER Join Table 1, with ON clause' select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table2 LEFT OUTER join table1 --****************************** on table2.number = table1.number AND table1.groupby = 2 --****************************** WHERE table1.number IS NULL Can anyone come up with a way of not using the filter in the on clause but in the where clause? The context of this is I have a staging area in a database and I want to identify new records and records that have been deleted. The groupby field is the equivalent of a batchid for an extract and I am comparing the latest extract in a temp table to a the batch from yesterday stored in a partioneds table, which also has all the previously extracted batches as well. Code to create table 1 and 2: create table table1 (number int, groupby int) create table table2 (number int) insert into table1 (number, groupby) values (1, 1) insert into table1 (number, groupby) values (2, 1) insert into table1 (number, groupby) values (1, 2) insert into table2 (number) values (1) insert into table1 (number, groupby) values (2, 2) insert into table2 (number) values (3) insert into table1 (number, groupby) values (4, 2) insert into table2 (number) values (4)

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  • SQL different joins not making any difference to result

    - by Chrissi
    I'm trying to write a quick (ha!) program to organise some of my financial information. What I ideally want is a query that will return all records with financial information in them from TableA. There should be one row for each month, but in instances where there were no transactions for a month there will be no record. I get results like this: SELECT Period,Year,TotalValue FROM TableA WHERE Year='1997' Result: Period Year TotalValue 1 1997 298.16 2 1997 435.25 4 1997 338.37 8 1997 336.07 9 1997 578.97 11 1997 361.23 By joining on a table (well a View in this instance) which just contains a field Period with values from 1 to 12, I expect to get something like this: SELECT p.Period,a.Year,a.TotalValue FROM Periods AS p LEFT JOIN TableA AS a ON p.Period = a.Period WHERE Year='1997' Result: Period Year TotalValue 1 1997 298.16 2 1997 435.25 3 NULL NULL 4 1997 338.37 5 NULL NULL 6 NULL NULL 7 NULL NULL 8 1997 336.07 9 1997 578.97 10 NULL NULL 11 1997 361.23 12 NULL NULL What I'm actually getting though is the same result no matter how I join it (except CROSS JOIN which goes nuts, but it's really not what I wanted anyway, it was just to see if different joins are even doing anything). LEFT JOIN, RIGHT JOIN, INNER JOIN all fail to provide the NULL records I am expecting. Is there something obvious that I'm doing wrong in the JOIN? Does it matter that I'm joining onto a View?

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  • Using the JPA Criteria API, can you do a fetch join that results in only one join?

    - by Shaun
    Using JPA 2.0. It seems that by default (no explicit fetch), @OneToOne(fetch = FetchType.EAGER) fields are fetched in 1 + N queries, where N is the number of results containing an Entity that defines the relationship to a distinct related entity. Using the Criteria API, I might try to avoid that as follows: CriteriaBuilder builder = entityManager.getCriteriaBuilder(); CriteriaQuery<MyEntity> query = builder.createQuery(MyEntity.class); Root<MyEntity> root = query.from(MyEntity.class); Join<MyEntity, RelatedEntity> join = root.join("relatedEntity"); root.fetch("relatedEntity"); query.select(root).where(builder.equals(join.get("id"), 3)); The above should ideally be equivalent to the following: SELECT m FROM MyEntity m JOIN FETCH myEntity.relatedEntity r WHERE r.id = 3 However, the criteria query results in the root table needlessly being joined to the related entity table twice; once for the fetch, and once for the where predicate. The resulting SQL looks something like this: SELECT myentity.id, myentity.attribute, relatedentity2.id, relatedentity2.attribute FROM my_entity myentity INNER JOIN related_entity relatedentity1 ON myentity.related_id = relatedentity1.id INNER JOIN related_entity relatedentity2 ON myentity.related_id = relatedentity2.id WHERE relatedentity1.id = 3 Alas, if I only do the fetch, then I don't have an expression to use in the where clause. Am I missing something, or is this a limitation of the Criteria API? If it's the latter, is this being remedied in JPA 2.1 or are there any vendor-specific enhancements? Otherwise, it seems better to just give up compile-time type checking (I realize my example doesn't use the metamodel) and use dynamic JPQL TypedQueries.

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  • Join our webcast: Discover What’s New in Oracle Data Integrator and Oracle GoldenGate

    - by Irem Radzik
    Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman","serif"; mso-fareast-font-family:"Times New Roman";} Data integration team has organized a series of webcasts for this summer. We are kicking it off this Thursday June 30th at 10am PT with a product update webcast: Discover What’s New in Oracle Data Integrator and Oracle GoldenGate. In this webcast you will hear from product management about the new patch updates to both GoldenGate 11g R1 and ODI 11gR1. Jeff Pollock, Sr. Director of Product Management for ODI will talk about the new features in Oracle Data Integrator 11.1.1.5, including the data lineage integration with OBI EE, enhanced web services to support flexible architectures as well as capabilities for efficient object execution such as Load Plans. Jeff will discuss support for complex files and performance enhancements. Chris McAllister, Sr. Director of Product Management for Oracle GoldenGate will cover the new features of Oracle GoldenGate 11.1.1.1 such as increased data security by supporting Oracle Database Advanced Security option, deeper integration with Oracle Database, and the expanded list of heterogeneous databases GoldenGate supports . Chris will also talk about the new Oracle GoldenGate 11gR1 release for HP NonStop platform and will provide information on our strategic direction for product development. Join us this Thursday at 10am PT/ 1pm ET to hear directly from Data Integration Product Management . You can register here for the June 30th webcast as well as for the upcoming ones in our summer webcast series.

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  • Join us for Live Oracle VM and Oracle Linux Cloud Events in Europe

    - by Monica Kumar
    Join us for a series of live events and discover how Oracle VM and Oracle Linux offer an integrated and optimized infrastructure for quickly deploying a private cloud environment at lower cost. As one of the most widely deployed operating systems today, Oracle Linux delivers higher performance, better reliability, and stability, at a lower cost for your cloud environments. Oracle VM is an application-driven server virtualization solution fully integrated and certified with Oracle applications to deliver rapid application deployment and simplified management. With Oracle VM, you have peace of mind that the entire Oracle stack deployed is fully certified by Oracle. Register now for any of the upcoming events, and meet with Oracle experts to discuss how we can help in enabling your private cloud. Nov 20: Foundation for the Cloud: Oracle Linux and Oracle VM (Belgium) Nov 21: Oracle Linux & Oracle VM Enabling Private Cloud (Germany) Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 Nov 28: Realize Substantial Savings and Increased Efficiency with Oracle Linux and Oracle VM (Luxembourg) Nov 29: Foundation for the Cloud: Oracle Linux and Oracle VM (Netherlands)Dec 5: MySQL Tech Tour, including Oracle Linux and Oracle VM (France) Hope to see you at one of these events!

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  • Join the Geek+ Community on Google+ and Share Your Random Geekery

    - by The Geek
    It turns out that Google+ recently added a new feature that allows you to create your own community inside of Google+, where anybody that’s a member can post images, links, or start a discussion. We’ve created the Geek+ Community, so stop by and join in the fun. You’ll notice that there’s only a few members right now, but we’re hoping that we can get every How-To Geek reader to participate in the geeky discussion. You’re welcome to: Post random geeky stuff that you find. Yell at us for articles that you don’t like, or tell us how we can do things better. Participate in discussions with other HTG readers. Post up your own Geek Trivia. We might even publish it over here on How-To Geek. Ask others for advice. Just read everything that the other readers post. Lots of other things we can’t think of right now. Note: If you want tech support, you should post on our regular forum. Secure Yourself by Using Two-Step Verification on These 16 Web Services How to Fix a Stuck Pixel on an LCD Monitor How to Factory Reset Your Android Phone or Tablet When It Won’t Boot

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  • Introducing Identity Management 11g R2: Join the webcast on July 19th, 2012 at 6:00 PM GMT

    - by Cinzia Mascanzoni
    Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Join Oracle and customer executives for the launch of Oracle Identity Management 11g R2, the breakthrough technology that dramatically expands the reach of identity management to cloud and mobile environments.. Register now for the event.

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  • JOIN THE ORACLE Fusion Middleware Summer Camps

    - by mseika
    JOIN THE ORACLE Fusion Middleware Summer Camps For Specialized partners who are working on following projects & opportunities, we offer these advanced summer camps: - BPM Suite 11 - ADF 11g - WebCenter Portal - WebLogic 12c - SOA Suite 11g - ADF for BPM Suite 11 - WebCenter Sites 11g All training sessions will be from HQ product management and our PTS team. The sessions will take place in July in Lisbon Portugal and Munich Germany. . Participation is limited to two people per company and bootcamp. Registration is handled by first come first serve, please pay attention to the skill requirements, the pre-requisitions and the follow up! We will not accept people onto the training who do not match the criteria! Lisbon: Monday, July 9th 11:00AM - Friday July 13th 16:00 PM (Lisbon time) - ADF 11g advanced training by Grant Ronald and Frank Nimphius - WebCenter Portal advanced training by Stefan Krantz and Angelo Santagata - WebLogic 12c training by Cosmin Tudor Munich: Monday, July 16th 11:00 AM - Wednesday July 18th 16:00 PM (CET) - ADF for BPM Suite 11g advanced training by David Read - WebCenter Sites 11g advanced training by Product Management & PTS Cost: Free of charge, cancelation or no-show fee 2.000€ Bootcamps are limited to 20 persons first come first serve For details and registration please visit Lisbon registration page: & Munich registration page Quotes summer camps 2011 “From zero to hero with this BPM workshop” Steven Boon, Ordina Linkedin “This is the training that prepares for real projects and POCs” Jon Petter Hjulstad, eVita – blog & twitter SOA & BPM Partner Community registration Please first login at http://partner.oracle.com and then visit: http://www.oracle.com/goto/emea/soa. If you have any questions please contact the Oracle Partner Business Center. If you have questions please feel free to contact us any time! Best regards Jürgen KressOracle EMEA SOA & BPM Partner Adoption EMEATel. +49 89 1430 1479E-Mail: [email protected]

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  • Join our team at Microsoft

    - by Daniel Moth
    If you are looking for a SDE or SDET job at Microsoft, keep on reading. Back in January I posted a Dev Lead opening on our team, which was quickly filled internally (by Maria Blees). Our team is part of the recently announced Microsoft Technical Computing group. Specifically, we are working on new debugger functionality, integrated with Visual Studio (we are starting work on the next version), aimed to address HPC and GPGPU scenarios (and continuing the Parallel Debugging scenarios we started addressing with VS2010). We now have many more openings on our debugger team. We posted three of those on the careers website: Software Development Engineer Software Development Engineer II Software Development Engineer in Test II (don't let the word "Test" fool you: An SDET on our team is no different than a developer in any way, including the skills required) Please do read the contents of the links above. Specifically, note that for both positions you need to be as proficient in writing C++ code as you are with managed code (WPF experience is a plus). If you think you have what it takes, you wish to join a quality and schedule driven project, and want to contribute features to a product that has global impact, then send me your resume and I'll pass it on to the hiring managers. Comments about this post welcome at the original blog.

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