Search Results

Search found 24334 results on 974 pages for 'directory loop'.

Page 9/974 | < Previous Page | 5 6 7 8 9 10 11 12 13 14 15 16  | Next Page >

  • Windows Share authentication from Active Directory Linux login

    - by Kenny
    Hi, I'm using Active Directory to log into RHEL. To do this, I followed the steps outlined here: http://www.markwilson.co.uk/blog/2007/05/using-active-directory-to-authenticate-users-on-a-linux-computer.htm I'd like to be able to read data from Windows Servers shared folders without being prompted for a password. On Windows I log into an AD domain, and when I access windows file shares on a server on the LAN (also part of the AD domain) my I can just access them with no authentication step. I've used SMBclient on Linux to access these shares, but it asks for my password. I would like to be able to script access to the data on the shares, but I can't if there's a password prompt in the way. Well, I could, but it's not how I want to do it. Now, since I'm logged in using my active directory username & password, can't I just access the shares without jumping that extra hoop? I know I can mount the share using something like: //192.168.0.5/share /mnt/windows cifs auto,username=steve,password=secret,rw 0 0 but access will depend who is logged in... each user logging in should have their own unique AD access privelages. Thanks for reading!

    Read the article

  • Minimum permissions needed to create a user Home Folder in Windows Active Directory

    - by Jim
    We would like the Help Desk to have the responsibility of creating User Home folders instead of our 2nd level support. The help desk global group is already an Account Operator, so in Active Directory they are able to edit all User Attributes just fine. The problem is figuring out the minimum level of permissions needed on the File Server to create the home share, with out giving them access to everyone home share. So if they open AD Users and Computer, open the properties for a user, and enter \home\users\%username% in the profile tab and then click OK, they get the following error. The \home\users\username home folder was not created because you do not have create access on the server. The user account has been updated with the new home folder value but you must create the directory manually after obtaining the required access right. Right now I have given the Helpdesk group Full Control on the root folder only (no files or subdirectories) The directory is actually created, but the permissions on the newly created folder only show administrators full control, and no permissions for the configured user account. It sure sounds like I'd have to make the helpdesk local admins on the file servers, which is what I'd like to avoid. Especially since the file servers are a large cluster hosting much much more than the entire orgs home share structure.

    Read the article

  • Active Directory Child Domain Replication Problems

    - by MikeR
    Hi, I've recently inherited an Active Directory (all DCs Windows 2003) which has been configured with several child domains that are used as test environments for out CRM software. Two of these child domains have been used for testing using dates in the future (2015), throwing them well outside of the Kerberos tolerance for time, and they're flooding my event logs with replication errors such as the following: Description: The attempt to establish a replication link for the following writable directory partition failed. Directory partition: CN=Schema,CN=Configuration,DC=ad,DC=xxxxxxx,DC=com Source domain controller: CN=NTDS Settings,CN=TESTDC001,CN=Servers,CN=SiteName,CN=Sites,CN=Configuration,DC=ad,DC=xxxxxxx,DC=com Source domain controller address: 38e95b2a-35af-4174-84ba-9ab039528cce._msdcs.ad.xxxxxxx.com Intersite transport (if any): This domain controller will be unable to replicate with the source domain controller until this problem is corrected. User Action Verify if the source domain controller is accessible or network connectivity is available. Additional Data Error value: 5 Access is denied. I'd also like to upgrade to Windows 2008 at some point, but wouldn't want to attempt any schema updates while I'm not 100% confident on the replication. I'm guessing my only real solution will be to get rid of these child domains. The child domains are operating as stand alone domains, the DC is up and running and authenticating test users fine. I'm guessing the best solution to this would be to delete the domains (although I'd be happily told otherwise). The clock forwarding appears to have been happening for several years, so I'm assuming I can't just put the clock right (I'm guessing scope for this would be 180days, the same as the tombstone lifetime) With the replication errors would I be able to dcpromo the child domains DC, select it as the last domain controller in the domain and the child domain would be deleted? Or would I be better off treating the domain as an orphaned domain and use Microsoft's instructions to clear up as such. Any advice would be much appreciated.

    Read the article

  • Allowing XP Home Clients To Access Active Directory Printers

    - by Sean M
    My school's network is based on Active Directory on Windows Server 2003 servers. Most of the computers in the school are members of the domain. However, we also acquired a passel of netbooks that are running Windows XP Home (as netbooks tend to), and we're trying to make those useful. The netbooks are made available to students by check-out, so none of them are dedicated to a specific user. I only want to allow the netbooks to do two significant network activities: to access the Internet (this is working acceptably well so far), and to print to one or more printers on the network. That second one is where trouble starts. I'm trying to find a way to allow the XP Home clients to access those Active Directory printers. All the solutions that I can come up with right now are expensive, ugly, or both - for example, changing the OS on the netbooks (even with imaging, that would take a lot of my time) or making sure that the user account on each netbook has a matching account in Active Directory with permissions for printing (invites security/maintainability disaster). Are there any elegant solutions? Failing that, what's the best ugly solution for allowing my students to print from the netbooks?

    Read the article

  • Sudoers file allow sudo on specific file for active directory group

    - by tubaguy50035
    I have active directory sign in working on an Ubuntu 12.04 box. When the user signs in, I have a script that runs that needs sudo permission (since it modifies the samba config file). How would I specify this in my sudoer's file? I've tried: %DOMAIN\\AD+Programmers ALL=NOPASSWD: /usr/local/bin/createSambaShare.php I've found various resources on the internet stating that this is how it would be done, but I'm not sure that I have the first part right. What are they using as the DOMAIN? The workgroup or the realm? I use Samba + winbind for active directory integration. Here's my smb.conf: [global] security = ads netbios name = hostname realm = COMPANYNAME.COM password server = passwordserver workgroup = COMPANYNAME idmap uid = 1000-10000 idmap gid = 1000-10000 winbind separator = + winbind enum users = no winbind enum groups = no winbind use default domain = yes template homedir = /home/%D/%U template shell = /bin/bash client use spnego = yes domain master = no EDIT: The users that should have access to run that script are all part of the Programmers group which has an Active Directory Domain Services Folder of Company.com/Staff/Security Groups (not sure if that matters or not).

    Read the article

  • IIS / Virtual Directory authentication.

    - by Chris L
    I have an IIS(v6)/Windows 2003/.Net 3.5(app code, libraries etc.) server hosting a website at www.mywebsite.com mapped to E:\Inetpub\wwwroot\mywebsite, we also have a virtual directory (VirtDir) mapped out to E:\Inetpub\wwwroot\mywebsite\files (although in theory this could be in a different directory or a separate machine) where we store a customer's files(a bunch of .pdf & .xls). Currently to access a file you can enter into the url something like: www.mywebsite.com/VirtDir/Customer/myFile.pdf and get access to the file. The problem is the user doesn't have to log into www.mywebsite.com to get access to the file, we would prefer them to log in first. We would like the user to login via the mywebsite and if valid, let them download files from the virtual directory. The www.mywebsite.com and VirtDir are separate sites on the same farm. Allow Anon Access, and Integrated Windows Authentication both enabled. I'm more of a developer and less of a Sys Admin, but hopefully I'm in the right spot, any help would be appreciated.

    Read the article

  • Extracting information from active directory

    - by Nop at NaDa
    I work in the IT support department of a branch of a huge company. I have to take care of a database with all the users, computers, etc. I'm trying to find a way to automatically update the database as much as possible, but the IT infrastructure guys doesn't give me enough privileges to use Active Directory in order to dump the users, nor they have the time to give me the information that I need. Some days ago I found Active Directory explorer from Sysinternals that allows me to browse through Active Directory, and I found all the information that I need there (username, real name, date when it was created, privileges, company, etc.). Unfortunately I'm unable to export the data to a human readable format. I'm just able to take a snapshot of the whole database in a machine-readable format. Doing the snapshot takes hours and I'm afraid that the infrastructure guys won't like me doing entire snapshots on a regular basis. Do you know of any tool (command-line is preferable) that would allow me to retrieve the values of the keys or export it to XML, CSV, etc?

    Read the article

  • Dumping active directory

    - by Nop at NaDa
    I work in the IT support department of a branch of a huge company. I have to take care of a database with all the users, computers, etc. I'm trying to find a way to automatically update the database as much as possible, but the IT infrastructure guys doesn't give me enough privileges to use Active Directory in order to dump the users, nor they have the time to give me the information that I need. Some days ago I found Active Directory explorer from Sysinternals that allows me to browse through Active Directory, and I found all the information that I need there (username, real name, date when it was created, privileges, company, etc.). Unfortunately I'm unable to export the data to a human readable format. I'm just able to take a snapshot of the whole database in a machine-readable format. Doing the snapshot takes hours and I'm afraid that the infrastructure guys won't like me doing entire snapshots on a regular basis. Do you know of any tool (command-line is preferable) that would allow me to retrieve the values of the keys or export it to XML, CSV, etc?

    Read the article

  • Directory service unavailiable, new hardware same settings

    - by Alex
    I'm working on a project with 2 sites connected by a VPN. Site 1 has the main server and there is a secondary server at site 2 which I am trying to replace. The current setup works perfectly however I can't for the life of me get the replacement server at site 2 up and running. I'm trying to replace like for like just upgraded hardware. I have installed the OS (all Server 2003 Standard SP2) and used exactly the same settings as the old server. I have setup Active Directory, DNS Server, DHCP Server and WINS Server configured. I have used all the same settings as the old server (except IP address and name). I can access the active directory but I can't do anything; add, edit, delete all returns "the directory service is unavaliable". No-one can login on any of the computers on site 2 and the internet is down. Plugging the old server back in and connecting it to the network rectifies the issue (so both new and old are connected at site 2), everyone can login and the internet is back (curious since the modem connects direct to the switch, and even with the new server online I can connect to the router via IP but not the net). I really don't have much experience but I've been roped into doing this because my company is too cheap to hire a real network admin. Any suggestions of where I can start to troubleshoot this, its driving me crazy and I only have a day before all the users are back on site.

    Read the article

  • Fireing Android Dialogs from another thread without Message Loop

    - by Jox
    In a SurfaceView, I'm dispatching new thread that draws on canvas within standard "LockCanvas-Draw-unlockCanvasAndPost" loop. (note that thread doesn't contains message loop). How to show Android standard Dialog from that thread? As thread doesn't have msg loop, following code doesn't work: Builder builder = new AlertDialog.Builder(this); builder.setTitle("Alert"); builder.setMessage("Stackoverflow!"); builder.setNegativeButton("cancel", null); builder.show();

    Read the article

  • PHP: Infinity loop and Time Limit!

    - by Jonathan
    Hi, I have a piece of code that fetches data by giving it an ID. If I give it an ID of 1230 for example, the code fetches an article data with an ID of 1230 from a web site (external) and insert it into a DB. Now, the problem is that I need to fetch all the articles, lets say from ID 00001 to 99999. If a do a 'for' loop, after 60 seconds the PHP internal time limit stops the loop. If a use some kind of header("Location: code.php?id=00001") or header("Location: code.php?id=".$ID) and increase $ID++ and then redirect to the same page the browser stops me because of the infinite loop or redirection problem. Please HELP!

    Read the article

  • Flash AS 3 Loader OnComplete Inside a Loop

    - by meengla
    Hi, I think I posted my question as an answer elsewhere (http://stackoverflow.com/questions/2338317/how-to-get-associated-urlrequest-from-event-complete-fired-by-urlloader/2776515#2776515) . Sorry. Here is my question again: Hi, How can I make your function work for loader object in a loop? Thanks! Meengla Here is my existing (rough) code; I always get the mylabel from the last element of the array. var _loader = new Loader(); for (j = 0; j < 5; j++) { //mylabel variable is correct setup in the loop _loader.contentLoaderInfo.addEventListener(Event.COMPLETE, function(e:Event):void { doneLoad(e, mylabel); }); _loader.load(new URLRequest(encodeURI(recAC[j].url))); }//for loop

    Read the article

  • Nested loop with dependent bounds trip count

    - by aaa
    hello. just out of curiosity I tried to do the following, which turned out to be not so obvious to me; Suppose I have nested loops with runtime bounds, for example: t = 0 // trip count for l in 0:N for k in 0:N for j in max(l,k):N for i in k:j+1 t += 1 t is loop trip count is there a general algorithm/way (better than N^4 obviously) to calculate loop trip count? I am working on the assumption that the iteration bounds depend only on constant or previous loop variables.

    Read the article

  • Flash AS3 for loop query

    - by jimbo
    I was hoping for a way that I could save on code by creating a loop for a few lines of code. Let me explain a little, with-out loop: icon1.button.iconLoad.load(new URLRequest("icons/icon1.jpg")); icon2.button.iconLoad.load(new URLRequest("icons/icon2.jpg")); icon3.button.iconLoad.load(new URLRequest("icons/icon3.jpg")); icon4.button.iconLoad.load(new URLRequest("icons/icon4.jpg")); etc... But with a loop could I have something like: for (var i:uint = 0; i < 4; i++) { icon+i+.button.iconLoad.load(new URLRequest("icons/icon"+i+"jpg")); } Any ideas welcome...

    Read the article

  • Javascript Replace Child/Loop issue

    - by Charles John Thompson III
    I have this really bizarre issue where I have a forloop that is supposed to replace all divs with the class of "original" to text inputs with a class of "new". When I run the loop, it only replaces every-other div with an input, but if I run the loop to just replace the class of the div and not change the tag to input, it does every single div, and doesn't only do every-other. Here is my loop code, and a link to the live version: live version here function divChange() { var divs = document.getElementsByTagName("div"); for (var i=0; i<divs.length; i++) { if (divs[i].className == 'original') { var textInput = document.createElement('input'); textInput.className = 'new'; textInput.type = 'text'; textInput.value = divs[i].innerHTML; var parent = divs[i].parentNode; parent.replaceChild(textInput, divs[i]); } } }

    Read the article

  • for loop vs while loop

    - by Atul
    We can use for loop and while loop for same purpose. in what means they effect our code if I use for instead of while? same question arises between if-else and switch-case? how to decide what to use? for example which one you would prefer? This code: int main() { int n; cin>>n; for(int i=0;i<n;i++) { do_something(); } return 0; } Or this code: int main() { int n,i=0; cin>>n; while(i<n) { do_something(); i++; } return 0; } if using for or while loop does not effect the code by any means then may I know What was the need to make 2 solution for same problem?

    Read the article

  • can i changed for loop index variable inside the loop in Matlab???

    - by shawana
    hi every one. i need to change my loop variable inside the iterationa as ihave to acces array elements in the loop which is changing w.r.t size inside the loop. ashort piece of code is that: que=[]; que=[2,3,4]; global len; len=size(que,2) x=4; for i=1:len if x<=10 que(x)= 5; len=size(que,2) x=x+1; end end que array should print like: 2 3 4 5 5 5 5 5 5 5 but it is printed in such a way: 2 3 4 5 5 5. how shuould it be accomplished in matlab? in visual c++ it happen correctly and print whole array of 10 elements which increases at run time plz reply me if have any ideaabout this as i m new into matlab

    Read the article

  • jquery for loop on click image swap

    - by user2939914
    I've created a for loop of images. I would like each image to swap with another image on click individually. Here's the jQuery I've written so far: for ( var i = 1; i < 50; i++) { $('article').append('<div class="ps-block" id="' + i + '"><img src="img/bw/' + i + 'bw.png"></div>'); } $('img').click(function() { var imgid = $(this).attr('id'); $(this).attr("src", "img/color/" + imgid + ".png"); }); I also attempted to use this code inside the for loop after the append, but i ends up returning 50 every time you click since the loop has already ran: $('img[src="img/bw/' + i + 'bw.png"]').click(function() { $(this).attr("src", "img/color/" + this.id + ".png"); }); Thanks!

    Read the article

  • How John Got 15x Improvement Without Really Trying

    - by rchrd
    The following article was published on a Sun Microsystems website a number of years ago by John Feo. It is still useful and worth preserving. So I'm republishing it here.  How I Got 15x Improvement Without Really Trying John Feo, Sun Microsystems Taking ten "personal" program codes used in scientific and engineering research, the author was able to get from 2 to 15 times performance improvement easily by applying some simple general optimization techniques. Introduction Scientific research based on computer simulation depends on the simulation for advancement. The research can advance only as fast as the computational codes can execute. The codes' efficiency determines both the rate and quality of results. In the same amount of time, a faster program can generate more results and can carry out a more detailed simulation of physical phenomena than a slower program. Highly optimized programs help science advance quickly and insure that monies supporting scientific research are used as effectively as possible. Scientific computer codes divide into three broad categories: ISV, community, and personal. ISV codes are large, mature production codes developed and sold commercially. The codes improve slowly over time both in methods and capabilities, and they are well tuned for most vendor platforms. Since the codes are mature and complex, there are few opportunities to improve their performance solely through code optimization. Improvements of 10% to 15% are typical. Examples of ISV codes are DYNA3D, Gaussian, and Nastran. Community codes are non-commercial production codes used by a particular research field. Generally, they are developed and distributed by a single academic or research institution with assistance from the community. Most users just run the codes, but some develop new methods and extensions that feed back into the general release. The codes are available on most vendor platforms. Since these codes are younger than ISV codes, there are more opportunities to optimize the source code. Improvements of 50% are not unusual. Examples of community codes are AMBER, CHARM, BLAST, and FASTA. Personal codes are those written by single users or small research groups for their own use. These codes are not distributed, but may be passed from professor-to-student or student-to-student over several years. They form the primordial ocean of applications from which community and ISV codes emerge. Government research grants pay for the development of most personal codes. This paper reports on the nature and performance of this class of codes. Over the last year, I have looked at over two dozen personal codes from more than a dozen research institutions. The codes cover a variety of scientific fields, including astronomy, atmospheric sciences, bioinformatics, biology, chemistry, geology, and physics. The sources range from a few hundred lines to more than ten thousand lines, and are written in Fortran, Fortran 90, C, and C++. For the most part, the codes are modular, documented, and written in a clear, straightforward manner. They do not use complex language features, advanced data structures, programming tricks, or libraries. I had little trouble understanding what the codes did or how data structures were used. Most came with a makefile. Surprisingly, only one of the applications is parallel. All developers have access to parallel machines, so availability is not an issue. Several tried to parallelize their applications, but stopped after encountering difficulties. Lack of education and a perception that parallelism is difficult prevented most from trying. I parallelized several of the codes using OpenMP, and did not judge any of the codes as difficult to parallelize. Even more surprising than the lack of parallelism is the inefficiency of the codes. I was able to get large improvements in performance in a matter of a few days applying simple optimization techniques. Table 1 lists ten representative codes [names and affiliation are omitted to preserve anonymity]. Improvements on one processor range from 2x to 15.5x with a simple average of 4.75x. I did not use sophisticated performance tools or drill deep into the program's execution character as one would do when tuning ISV or community codes. Using only a profiler and source line timers, I identified inefficient sections of code and improved their performance by inspection. The changes were at a high level. I am sure there is another factor of 2 or 3 in each code, and more if the codes are parallelized. The study’s results show that personal scientific codes are running many times slower than they should and that the problem is pervasive. Computational scientists are not sloppy programmers; however, few are trained in the art of computer programming or code optimization. I found that most have a working knowledge of some programming language and standard software engineering practices; but they do not know, or think about, how to make their programs run faster. They simply do not know the standard techniques used to make codes run faster. In fact, they do not even perceive that such techniques exist. The case studies described in this paper show that applying simple, well known techniques can significantly increase the performance of personal codes. It is important that the scientific community and the Government agencies that support scientific research find ways to better educate academic scientific programmers. The inefficiency of their codes is so bad that it is retarding both the quality and progress of scientific research. # cacheperformance redundantoperations loopstructures performanceimprovement 1 x x 15.5 2 x 2.8 3 x x 2.5 4 x 2.1 5 x x 2.0 6 x 5.0 7 x 5.8 8 x 6.3 9 2.2 10 x x 3.3 Table 1 — Area of improvement and performance gains of 10 codes The remainder of the paper is organized as follows: sections 2, 3, and 4 discuss the three most common sources of inefficiencies in the codes studied. These are cache performance, redundant operations, and loop structures. Each section includes several examples. The last section summaries the work and suggests a possible solution to the issues raised. Optimizing cache performance Commodity microprocessor systems use caches to increase memory bandwidth and reduce memory latencies. Typical latencies from processor to L1, L2, local, and remote memory are 3, 10, 50, and 200 cycles, respectively. Moreover, bandwidth falls off dramatically as memory distances increase. Programs that do not use cache effectively run many times slower than programs that do. When optimizing for cache, the biggest performance gains are achieved by accessing data in cache order and reusing data to amortize the overhead of cache misses. Secondary considerations are prefetching, associativity, and replacement; however, the understanding and analysis required to optimize for the latter are probably beyond the capabilities of the non-expert. Much can be gained simply by accessing data in the correct order and maximizing data reuse. 6 out of the 10 codes studied here benefited from such high level optimizations. Array Accesses The most important cache optimization is the most basic: accessing Fortran array elements in column order and C array elements in row order. Four of the ten codes—1, 2, 4, and 10—got it wrong. Compilers will restructure nested loops to optimize cache performance, but may not do so if the loop structure is too complex, or the loop body includes conditionals, complex addressing, or function calls. In code 1, the compiler failed to invert a key loop because of complex addressing do I = 0, 1010, delta_x IM = I - delta_x IP = I + delta_x do J = 5, 995, delta_x JM = J - delta_x JP = J + delta_x T1 = CA1(IP, J) + CA1(I, JP) T2 = CA1(IM, J) + CA1(I, JM) S1 = T1 + T2 - 4 * CA1(I, J) CA(I, J) = CA1(I, J) + D * S1 end do end do In code 2, the culprit is conditionals do I = 1, N do J = 1, N If (IFLAG(I,J) .EQ. 0) then T1 = Value(I, J-1) T2 = Value(I-1, J) T3 = Value(I, J) T4 = Value(I+1, J) T5 = Value(I, J+1) Value(I,J) = 0.25 * (T1 + T2 + T5 + T4) Delta = ABS(T3 - Value(I,J)) If (Delta .GT. MaxDelta) MaxDelta = Delta endif enddo enddo I fixed both programs by inverting the loops by hand. Code 10 has three-dimensional arrays and triply nested loops. The structure of the most computationally intensive loops is too complex to invert automatically or by hand. The only practical solution is to transpose the arrays so that the dimension accessed by the innermost loop is in cache order. The arrays can be transposed at construction or prior to entering a computationally intensive section of code. The former requires all array references to be modified, while the latter is cost effective only if the cost of the transpose is amortized over many accesses. I used the second approach to optimize code 10. Code 5 has four-dimensional arrays and loops are nested four deep. For all of the reasons cited above the compiler is not able to restructure three key loops. Assume C arrays and let the four dimensions of the arrays be i, j, k, and l. In the original code, the index structure of the three loops is L1: for i L2: for i L3: for i for l for l for j for k for j for k for j for k for l So only L3 accesses array elements in cache order. L1 is a very complex loop—much too complex to invert. I brought the loop into cache alignment by transposing the second and fourth dimensions of the arrays. Since the code uses a macro to compute all array indexes, I effected the transpose at construction and changed the macro appropriately. The dimensions of the new arrays are now: i, l, k, and j. L3 is a simple loop and easily inverted. L2 has a loop-carried scalar dependence in k. By promoting the scalar name that carries the dependence to an array, I was able to invert the third and fourth subloops aligning the loop with cache. Code 5 is by far the most difficult of the four codes to optimize for array accesses; but the knowledge required to fix the problems is no more than that required for the other codes. I would judge this code at the limits of, but not beyond, the capabilities of appropriately trained computational scientists. Array Strides When a cache miss occurs, a line (64 bytes) rather than just one word is loaded into the cache. If data is accessed stride 1, than the cost of the miss is amortized over 8 words. Any stride other than one reduces the cost savings. Two of the ten codes studied suffered from non-unit strides. The codes represent two important classes of "strided" codes. Code 1 employs a multi-grid algorithm to reduce time to convergence. The grids are every tenth, fifth, second, and unit element. Since time to convergence is inversely proportional to the distance between elements, coarse grids converge quickly providing good starting values for finer grids. The better starting values further reduce the time to convergence. The downside is that grids of every nth element, n > 1, introduce non-unit strides into the computation. In the original code, much of the savings of the multi-grid algorithm were lost due to this problem. I eliminated the problem by compressing (copying) coarse grids into continuous memory, and rewriting the computation as a function of the compressed grid. On convergence, I copied the final values of the compressed grid back to the original grid. The savings gained from unit stride access of the compressed grid more than paid for the cost of copying. Using compressed grids, the loop from code 1 included in the previous section becomes do j = 1, GZ do i = 1, GZ T1 = CA(i+0, j-1) + CA(i-1, j+0) T4 = CA1(i+1, j+0) + CA1(i+0, j+1) S1 = T1 + T4 - 4 * CA1(i+0, j+0) CA(i+0, j+0) = CA1(i+0, j+0) + DD * S1 enddo enddo where CA and CA1 are compressed arrays of size GZ. Code 7 traverses a list of objects selecting objects for later processing. The labels of the selected objects are stored in an array. The selection step has unit stride, but the processing steps have irregular stride. A fix is to save the parameters of the selected objects in temporary arrays as they are selected, and pass the temporary arrays to the processing functions. The fix is practical if the same parameters are used in selection as in processing, or if processing comprises a series of distinct steps which use overlapping subsets of the parameters. Both conditions are true for code 7, so I achieved significant improvement by copying parameters to temporary arrays during selection. Data reuse In the previous sections, we optimized for spatial locality. It is also important to optimize for temporal locality. Once read, a datum should be used as much as possible before it is forced from cache. Loop fusion and loop unrolling are two techniques that increase temporal locality. Unfortunately, both techniques increase register pressure—as loop bodies become larger, the number of registers required to hold temporary values grows. Once register spilling occurs, any gains evaporate quickly. For multiprocessors with small register sets or small caches, the sweet spot can be very small. In the ten codes presented here, I found no opportunities for loop fusion and only two opportunities for loop unrolling (codes 1 and 3). In code 1, unrolling the outer and inner loop one iteration increases the number of result values computed by the loop body from 1 to 4, do J = 1, GZ-2, 2 do I = 1, GZ-2, 2 T1 = CA1(i+0, j-1) + CA1(i-1, j+0) T2 = CA1(i+1, j-1) + CA1(i+0, j+0) T3 = CA1(i+0, j+0) + CA1(i-1, j+1) T4 = CA1(i+1, j+0) + CA1(i+0, j+1) T5 = CA1(i+2, j+0) + CA1(i+1, j+1) T6 = CA1(i+1, j+1) + CA1(i+0, j+2) T7 = CA1(i+2, j+1) + CA1(i+1, j+2) S1 = T1 + T4 - 4 * CA1(i+0, j+0) S2 = T2 + T5 - 4 * CA1(i+1, j+0) S3 = T3 + T6 - 4 * CA1(i+0, j+1) S4 = T4 + T7 - 4 * CA1(i+1, j+1) CA(i+0, j+0) = CA1(i+0, j+0) + DD * S1 CA(i+1, j+0) = CA1(i+1, j+0) + DD * S2 CA(i+0, j+1) = CA1(i+0, j+1) + DD * S3 CA(i+1, j+1) = CA1(i+1, j+1) + DD * S4 enddo enddo The loop body executes 12 reads, whereas as the rolled loop shown in the previous section executes 20 reads to compute the same four values. In code 3, two loops are unrolled 8 times and one loop is unrolled 4 times. Here is the before for (k = 0; k < NK[u]; k++) { sum = 0.0; for (y = 0; y < NY; y++) { sum += W[y][u][k] * delta[y]; } backprop[i++]=sum; } and after code for (k = 0; k < KK - 8; k+=8) { sum0 = 0.0; sum1 = 0.0; sum2 = 0.0; sum3 = 0.0; sum4 = 0.0; sum5 = 0.0; sum6 = 0.0; sum7 = 0.0; for (y = 0; y < NY; y++) { sum0 += W[y][0][k+0] * delta[y]; sum1 += W[y][0][k+1] * delta[y]; sum2 += W[y][0][k+2] * delta[y]; sum3 += W[y][0][k+3] * delta[y]; sum4 += W[y][0][k+4] * delta[y]; sum5 += W[y][0][k+5] * delta[y]; sum6 += W[y][0][k+6] * delta[y]; sum7 += W[y][0][k+7] * delta[y]; } backprop[k+0] = sum0; backprop[k+1] = sum1; backprop[k+2] = sum2; backprop[k+3] = sum3; backprop[k+4] = sum4; backprop[k+5] = sum5; backprop[k+6] = sum6; backprop[k+7] = sum7; } for one of the loops unrolled 8 times. Optimizing for temporal locality is the most difficult optimization considered in this paper. The concepts are not difficult, but the sweet spot is small. Identifying where the program can benefit from loop unrolling or loop fusion is not trivial. Moreover, it takes some effort to get it right. Still, educating scientific programmers about temporal locality and teaching them how to optimize for it will pay dividends. Reducing instruction count Execution time is a function of instruction count. Reduce the count and you usually reduce the time. The best solution is to use a more efficient algorithm; that is, an algorithm whose order of complexity is smaller, that converges quicker, or is more accurate. Optimizing source code without changing the algorithm yields smaller, but still significant, gains. This paper considers only the latter because the intent is to study how much better codes can run if written by programmers schooled in basic code optimization techniques. The ten codes studied benefited from three types of "instruction reducing" optimizations. The two most prevalent were hoisting invariant memory and data operations out of inner loops. The third was eliminating unnecessary data copying. The nature of these inefficiencies is language dependent. Memory operations The semantics of C make it difficult for the compiler to determine all the invariant memory operations in a loop. The problem is particularly acute for loops in functions since the compiler may not know the values of the function's parameters at every call site when compiling the function. Most compilers support pragmas to help resolve ambiguities; however, these pragmas are not comprehensive and there is no standard syntax. To guarantee that invariant memory operations are not executed repetitively, the user has little choice but to hoist the operations by hand. The problem is not as severe in Fortran programs because in the absence of equivalence statements, it is a violation of the language's semantics for two names to share memory. Codes 3 and 5 are C programs. In both cases, the compiler did not hoist all invariant memory operations from inner loops. Consider the following loop from code 3 for (y = 0; y < NY; y++) { i = 0; for (u = 0; u < NU; u++) { for (k = 0; k < NK[u]; k++) { dW[y][u][k] += delta[y] * I1[i++]; } } } Since dW[y][u] can point to the same memory space as delta for one or more values of y and u, assignment to dW[y][u][k] may change the value of delta[y]. In reality, dW and delta do not overlap in memory, so I rewrote the loop as for (y = 0; y < NY; y++) { i = 0; Dy = delta[y]; for (u = 0; u < NU; u++) { for (k = 0; k < NK[u]; k++) { dW[y][u][k] += Dy * I1[i++]; } } } Failure to hoist invariant memory operations may be due to complex address calculations. If the compiler can not determine that the address calculation is invariant, then it can hoist neither the calculation nor the associated memory operations. As noted above, code 5 uses a macro to address four-dimensional arrays #define MAT4D(a,q,i,j,k) (double *)((a)->data + (q)*(a)->strides[0] + (i)*(a)->strides[3] + (j)*(a)->strides[2] + (k)*(a)->strides[1]) The macro is too complex for the compiler to understand and so, it does not identify any subexpressions as loop invariant. The simplest way to eliminate the address calculation from the innermost loop (over i) is to define a0 = MAT4D(a,q,0,j,k) before the loop and then replace all instances of *MAT4D(a,q,i,j,k) in the loop with a0[i] A similar problem appears in code 6, a Fortran program. The key loop in this program is do n1 = 1, nh nx1 = (n1 - 1) / nz + 1 nz1 = n1 - nz * (nx1 - 1) do n2 = 1, nh nx2 = (n2 - 1) / nz + 1 nz2 = n2 - nz * (nx2 - 1) ndx = nx2 - nx1 ndy = nz2 - nz1 gxx = grn(1,ndx,ndy) gyy = grn(2,ndx,ndy) gxy = grn(3,ndx,ndy) balance(n1,1) = balance(n1,1) + (force(n2,1) * gxx + force(n2,2) * gxy) * h1 balance(n1,2) = balance(n1,2) + (force(n2,1) * gxy + force(n2,2) * gyy)*h1 end do end do The programmer has written this loop well—there are no loop invariant operations with respect to n1 and n2. However, the loop resides within an iterative loop over time and the index calculations are independent with respect to time. Trading space for time, I precomputed the index values prior to the entering the time loop and stored the values in two arrays. I then replaced the index calculations with reads of the arrays. Data operations Ways to reduce data operations can appear in many forms. Implementing a more efficient algorithm produces the biggest gains. The closest I came to an algorithm change was in code 4. This code computes the inner product of K-vectors A(i) and B(j), 0 = i < N, 0 = j < M, for most values of i and j. Since the program computes most of the NM possible inner products, it is more efficient to compute all the inner products in one triply-nested loop rather than one at a time when needed. The savings accrue from reading A(i) once for all B(j) vectors and from loop unrolling. for (i = 0; i < N; i+=8) { for (j = 0; j < M; j++) { sum0 = 0.0; sum1 = 0.0; sum2 = 0.0; sum3 = 0.0; sum4 = 0.0; sum5 = 0.0; sum6 = 0.0; sum7 = 0.0; for (k = 0; k < K; k++) { sum0 += A[i+0][k] * B[j][k]; sum1 += A[i+1][k] * B[j][k]; sum2 += A[i+2][k] * B[j][k]; sum3 += A[i+3][k] * B[j][k]; sum4 += A[i+4][k] * B[j][k]; sum5 += A[i+5][k] * B[j][k]; sum6 += A[i+6][k] * B[j][k]; sum7 += A[i+7][k] * B[j][k]; } C[i+0][j] = sum0; C[i+1][j] = sum1; C[i+2][j] = sum2; C[i+3][j] = sum3; C[i+4][j] = sum4; C[i+5][j] = sum5; C[i+6][j] = sum6; C[i+7][j] = sum7; }} This change requires knowledge of a typical run; i.e., that most inner products are computed. The reasons for the change, however, derive from basic optimization concepts. It is the type of change easily made at development time by a knowledgeable programmer. In code 5, we have the data version of the index optimization in code 6. Here a very expensive computation is a function of the loop indices and so cannot be hoisted out of the loop; however, the computation is invariant with respect to an outer iterative loop over time. We can compute its value for each iteration of the computation loop prior to entering the time loop and save the values in an array. The increase in memory required to store the values is small in comparison to the large savings in time. The main loop in Code 8 is doubly nested. The inner loop includes a series of guarded computations; some are a function of the inner loop index but not the outer loop index while others are a function of the outer loop index but not the inner loop index for (j = 0; j < N; j++) { for (i = 0; i < M; i++) { r = i * hrmax; R = A[j]; temp = (PRM[3] == 0.0) ? 1.0 : pow(r, PRM[3]); high = temp * kcoeff * B[j] * PRM[2] * PRM[4]; low = high * PRM[6] * PRM[6] / (1.0 + pow(PRM[4] * PRM[6], 2.0)); kap = (R > PRM[6]) ? high * R * R / (1.0 + pow(PRM[4]*r, 2.0) : low * pow(R/PRM[6], PRM[5]); < rest of loop omitted > }} Note that the value of temp is invariant to j. Thus, we can hoist the computation for temp out of the loop and save its values in an array. for (i = 0; i < M; i++) { r = i * hrmax; TEMP[i] = pow(r, PRM[3]); } [N.B. – the case for PRM[3] = 0 is omitted and will be reintroduced later.] We now hoist out of the inner loop the computations invariant to i. Since the conditional guarding the value of kap is invariant to i, it behooves us to hoist the computation out of the inner loop, thereby executing the guard once rather than M times. The final version of the code is for (j = 0; j < N; j++) { R = rig[j] / 1000.; tmp1 = kcoeff * par[2] * beta[j] * par[4]; tmp2 = 1.0 + (par[4] * par[4] * par[6] * par[6]); tmp3 = 1.0 + (par[4] * par[4] * R * R); tmp4 = par[6] * par[6] / tmp2; tmp5 = R * R / tmp3; tmp6 = pow(R / par[6], par[5]); if ((par[3] == 0.0) && (R > par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * tmp5; } else if ((par[3] == 0.0) && (R <= par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * tmp4 * tmp6; } else if ((par[3] != 0.0) && (R > par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * TEMP[i] * tmp5; } else if ((par[3] != 0.0) && (R <= par[6])) { for (i = 1; i <= imax1; i++) KAP[i] = tmp1 * TEMP[i] * tmp4 * tmp6; } for (i = 0; i < M; i++) { kap = KAP[i]; r = i * hrmax; < rest of loop omitted > } } Maybe not the prettiest piece of code, but certainly much more efficient than the original loop, Copy operations Several programs unnecessarily copy data from one data structure to another. This problem occurs in both Fortran and C programs, although it manifests itself differently in the two languages. Code 1 declares two arrays—one for old values and one for new values. At the end of each iteration, the array of new values is copied to the array of old values to reset the data structures for the next iteration. This problem occurs in Fortran programs not included in this study and in both Fortran 77 and Fortran 90 code. Introducing pointers to the arrays and swapping pointer values is an obvious way to eliminate the copying; but pointers is not a feature that many Fortran programmers know well or are comfortable using. An easy solution not involving pointers is to extend the dimension of the value array by 1 and use the last dimension to differentiate between arrays at different times. For example, if the data space is N x N, declare the array (N, N, 2). Then store the problem’s initial values in (_, _, 2) and define the scalar names new = 2 and old = 1. At the start of each iteration, swap old and new to reset the arrays. The old–new copy problem did not appear in any C program. In programs that had new and old values, the code swapped pointers to reset data structures. Where unnecessary coping did occur is in structure assignment and parameter passing. Structures in C are handled much like scalars. Assignment causes the data space of the right-hand name to be copied to the data space of the left-hand name. Similarly, when a structure is passed to a function, the data space of the actual parameter is copied to the data space of the formal parameter. If the structure is large and the assignment or function call is in an inner loop, then copying costs can grow quite large. While none of the ten programs considered here manifested this problem, it did occur in programs not included in the study. A simple fix is always to refer to structures via pointers. Optimizing loop structures Since scientific programs spend almost all their time in loops, efficient loops are the key to good performance. Conditionals, function calls, little instruction level parallelism, and large numbers of temporary values make it difficult for the compiler to generate tightly packed, highly efficient code. Conditionals and function calls introduce jumps that disrupt code flow. Users should eliminate or isolate conditionls to their own loops as much as possible. Often logical expressions can be substituted for if-then-else statements. For example, code 2 includes the following snippet MaxDelta = 0.0 do J = 1, N do I = 1, M < code omitted > Delta = abs(OldValue ? NewValue) if (Delta > MaxDelta) MaxDelta = Delta enddo enddo if (MaxDelta .gt. 0.001) goto 200 Since the only use of MaxDelta is to control the jump to 200 and all that matters is whether or not it is greater than 0.001, I made MaxDelta a boolean and rewrote the snippet as MaxDelta = .false. do J = 1, N do I = 1, M < code omitted > Delta = abs(OldValue ? NewValue) MaxDelta = MaxDelta .or. (Delta .gt. 0.001) enddo enddo if (MaxDelta) goto 200 thereby, eliminating the conditional expression from the inner loop. A microprocessor can execute many instructions per instruction cycle. Typically, it can execute one or more memory, floating point, integer, and jump operations. To be executed simultaneously, the operations must be independent. Thick loops tend to have more instruction level parallelism than thin loops. Moreover, they reduce memory traffice by maximizing data reuse. Loop unrolling and loop fusion are two techniques to increase the size of loop bodies. Several of the codes studied benefitted from loop unrolling, but none benefitted from loop fusion. This observation is not too surpising since it is the general tendency of programmers to write thick loops. As loops become thicker, the number of temporary values grows, increasing register pressure. If registers spill, then memory traffic increases and code flow is disrupted. A thick loop with many temporary values may execute slower than an equivalent series of thin loops. The biggest gain will be achieved if the thick loop can be split into a series of independent loops eliminating the need to write and read temporary arrays. I found such an occasion in code 10 where I split the loop do i = 1, n do j = 1, m A24(j,i)= S24(j,i) * T24(j,i) + S25(j,i) * U25(j,i) B24(j,i)= S24(j,i) * T25(j,i) + S25(j,i) * U24(j,i) A25(j,i)= S24(j,i) * C24(j,i) + S25(j,i) * V24(j,i) B25(j,i)= S24(j,i) * U25(j,i) + S25(j,i) * V25(j,i) C24(j,i)= S26(j,i) * T26(j,i) + S27(j,i) * U26(j,i) D24(j,i)= S26(j,i) * T27(j,i) + S27(j,i) * V26(j,i) C25(j,i)= S27(j,i) * S28(j,i) + S26(j,i) * U28(j,i) D25(j,i)= S27(j,i) * T28(j,i) + S26(j,i) * V28(j,i) end do end do into two disjoint loops do i = 1, n do j = 1, m A24(j,i)= S24(j,i) * T24(j,i) + S25(j,i) * U25(j,i) B24(j,i)= S24(j,i) * T25(j,i) + S25(j,i) * U24(j,i) A25(j,i)= S24(j,i) * C24(j,i) + S25(j,i) * V24(j,i) B25(j,i)= S24(j,i) * U25(j,i) + S25(j,i) * V25(j,i) end do end do do i = 1, n do j = 1, m C24(j,i)= S26(j,i) * T26(j,i) + S27(j,i) * U26(j,i) D24(j,i)= S26(j,i) * T27(j,i) + S27(j,i) * V26(j,i) C25(j,i)= S27(j,i) * S28(j,i) + S26(j,i) * U28(j,i) D25(j,i)= S27(j,i) * T28(j,i) + S26(j,i) * V28(j,i) end do end do Conclusions Over the course of the last year, I have had the opportunity to work with over two dozen academic scientific programmers at leading research universities. Their research interests span a broad range of scientific fields. Except for two programs that relied almost exclusively on library routines (matrix multiply and fast Fourier transform), I was able to improve significantly the single processor performance of all codes. Improvements range from 2x to 15.5x with a simple average of 4.75x. Changes to the source code were at a very high level. I did not use sophisticated techniques or programming tools to discover inefficiencies or effect the changes. Only one code was parallel despite the availability of parallel systems to all developers. Clearly, we have a problem—personal scientific research codes are highly inefficient and not running parallel. The developers are unaware of simple optimization techniques to make programs run faster. They lack education in the art of code optimization and parallel programming. I do not believe we can fix the problem by publishing additional books or training manuals. To date, the developers in questions have not studied the books or manual available, and are unlikely to do so in the future. Short courses are a possible solution, but I believe they are too concentrated to be much use. The general concepts can be taught in a three or four day course, but that is not enough time for students to practice what they learn and acquire the experience to apply and extend the concepts to their codes. Practice is the key to becoming proficient at optimization. I recommend that graduate students be required to take a semester length course in optimization and parallel programming. We would never give someone access to state-of-the-art scientific equipment costing hundreds of thousands of dollars without first requiring them to demonstrate that they know how to use the equipment. Yet the criterion for time on state-of-the-art supercomputers is at most an interesting project. Requestors are never asked to demonstrate that they know how to use the system, or can use the system effectively. A semester course would teach them the required skills. Government agencies that fund academic scientific research pay for most of the computer systems supporting scientific research as well as the development of most personal scientific codes. These agencies should require graduate schools to offer a course in optimization and parallel programming as a requirement for funding. About the Author John Feo received his Ph.D. in Computer Science from The University of Texas at Austin in 1986. After graduate school, Dr. Feo worked at Lawrence Livermore National Laboratory where he was the Group Leader of the Computer Research Group and principal investigator of the Sisal Language Project. In 1997, Dr. Feo joined Tera Computer Company where he was project manager for the MTA, and oversaw the programming and evaluation of the MTA at the San Diego Supercomputer Center. In 2000, Dr. Feo joined Sun Microsystems as an HPC application specialist. He works with university research groups to optimize and parallelize scientific codes. Dr. Feo has published over two dozen research articles in the areas of parallel parallel programming, parallel programming languages, and application performance.

    Read the article

  • Samba with Active Directory - shares are readonly, NT_STATUS_MEDIA_WRITE_PROTECTED

    - by froh42
    I've set a samba server that seems to work, all shares are seemingly exported as readonly, however. The machine is called "lx". When I'm on lx I can run the following command: froh@lx:~$ smbclient //lx/export -UAdministrator Enter Administrator's password: Domain=[CUSTOMER] OS=[Unix] Server=[Samba 3.5.4] smb: \> mkdir wrzlbrmpf NT_STATUS_MEDIA_WRITE_PROTECTED making remote directory \wrzlbrmpf smb: \> ls . D 0 Fri Dec 3 19:04:20 2010 .. D 0 Sun Nov 28 01:32:37 2010 zork D 0 Fri Dec 3 18:53:33 2010 bar D 0 Sun Nov 28 23:52:43 2010 ork 1 Fri Dec 3 18:53:02 2010 foo 1 Sun Nov 28 23:52:41 2010 gaga D 0 Fri Dec 3 19:04:20 2010 How can I troubleshoot this? What I did: First I set up a fresh install of Ubuntu 10.10 x64. Second I got kerberos working with the following krb5.conf file: [libdefaults] ticket_lifetime = 24000 clock_skew = 300 default_realm = CUSTOMER.LOCAL [realms] CUSTOMER.LOCAL = { kdc = SB4.customer.local:88 admin_server = SB4.customer.local:464 default_domain = CUSTOMER.LOCAL } [domain_realm] .customer.local = CUSTOMER.LOCAL customer.local = CUSTOMER.LOCAL #[login] # krb4_convert = true # krb4_get_tickets = false I also added winbind to group, passwd and shadow in nsswitch.conf. Seemingly Kerberos works: root@lx:~# net ads testjoin Join is OK root@lx:~# wbinfo -a 'Administrator%MYSECRETPASSWORD' plaintext password authentication succeeded challenge/response password authentication succeeded wbinfo -u and wbinfo -g also spit out a list of users and a list of groups respectiveley. I noted that domain accounts did NOT include a domain and they are in german (as on the SBS 2003 that is the domain server). So I get a "Domänenbenutzer" in wbinfo -u's output not a "CUSTOMER+Domain User" or something similar. I'm not sure anymore what I did to the PAM configuration, but here is what I currently have: root@lx:/etc/pam.d# cat samba @include common-auth @include common-account @include common-session-noninteractive root@lx:/etc/pam.d# grep -ve '^#' common-auth auth [success=3 default=ignore] pam_krb5.so minimum_uid=1000 auth [success=2 default=ignore] pam_unix.so nullok_secure try_first_pass auth [success=1 default=ignore] pam_winbind.so krb5_auth krb5_ccache_type=FILE cached_login try_first_pass auth requisite pam_deny.so auth required pam_permit.so root@lx:/etc/pam.d# grep -ve '^#' common-account account [success=2 new_authtok_reqd=done default=ignore] pam_unix.so account [success=1 new_authtok_reqd=done default=ignore] pam_winbind.so account requisite pam_deny.so account required pam_permit.so account required pam_krb5.so minimum_uid=1000 root@lx:/etc/pam.d# grep -ve '^#' common-session-noninteractive session [default=1] pam_permit.so session requisite pam_deny.so session required pam_permit.so session optional pam_krb5.so minimum_uid=1000 session required pam_unix.so session optional pam_winbind.so At some point I joined the linux box into the AD domain. After (manually) creating a home directory on the linux box I can log in using the Adminstrator user with the password taken from AD. Now I run samba with the following setup: [global] netbios name = LX realm = CUSTOMER.LOCAL workgroup = CUSTOMER security = ADS encrypt passwords = yes password server = 192.168.20.244 #IP des Domain Controllers os level = 0 socket options = TCP_NODELAY SO_RCVBUF=16384 SO_SNDBUF=16384 idmap uid = 10000-20000 idmap gid = 10000-20000 winbind enum users = Yes winbind enum groups = Yes preferred master = no winbind separator = + dns proxy = no wins proxy = no # client NTLMv2 auth = Yes log level = 2 logfile = /var/log/samba/log.smbd.%U template homedir = /home/%U template shell = /bin/bash [export] path = /mnt/sdc1/export read only = No public = Yes Currently I don't care whether export is exported to everyone or just one user, I want to see somebody WRITING to that directory before I start fiddling with the authentication settings. (Who may access it). As mentioned, accessing the share from smbclient results in this NT_STATUS_MEDIA_WRITE_PROTECTED . Accessing it from windows shows ACLs that look correct (The user may write) - but it does not work, I can only read files not write. The directory to be exported looks like this: root@lx:/etc/pam.d# ls -ld /mnt/ drwxr-xr-x 5 root root 4096 2010-11-28 01:29 /mnt/ root@lx:/etc/pam.d# ls -ld /mnt/sdc1/ drwxr-xr-x 4 froh froh 4096 2010-11-28 01:32 /mnt/sdc1/ root@lx:/etc/pam.d# ls -ld /mnt/sdc1/export/ drwxrwxrwx+ 5 administrator domänen-admins 4096 2010-12-03 19:04 /mnt/sdc1/export/ root@lx:/etc/pam.d# getfacl /mnt/ getfacl: Entferne führende '/' von absoluten Pfadnamen # file: mnt/ # owner: root # group: root user::rwx group::r-x other::r-x root@lx:/etc/pam.d# getfacl /mnt/sdc1/ getfacl: Entferne führende '/' von absoluten Pfadnamen # file: mnt/sdc1/ # owner: froh # group: froh user::rwx group::r-x other::r-x root@lx:/etc/pam.d# getfacl /mnt/sdc1/export/ getfacl: Entferne führende '/' von absoluten Pfadnamen # file: mnt/sdc1/export/ # owner: administrator # group: domänen-admins user::rwx group::rwx group:domänen-admins:rwx mask::rwx other::rwx default:user::rwx default:group::rwx default:group:domänen-admins:rwx default:mask::rwx default:other::rwx My, oh my what am I overlooking? What am I to blind to see?

    Read the article

  • Samba with Active Directory - shares are readonly, NT_STATUS_MEDIA_WRITE_PROTECTED

    - by froh42
    I've set a samba server that seems to work, all shares are seemingly exported as readonly, however. The machine is called "lx". When I'm on lx I can run the following command: froh@lx:~$ smbclient //lx/export -UAdministrator Enter Administrator's password: Domain=[CUSTOMER] OS=[Unix] Server=[Samba 3.5.4] smb: \> mkdir wrzlbrmpf NT_STATUS_MEDIA_WRITE_PROTECTED making remote directory \wrzlbrmpf smb: \> ls . D 0 Fri Dec 3 19:04:20 2010 .. D 0 Sun Nov 28 01:32:37 2010 zork D 0 Fri Dec 3 18:53:33 2010 bar D 0 Sun Nov 28 23:52:43 2010 ork 1 Fri Dec 3 18:53:02 2010 foo 1 Sun Nov 28 23:52:41 2010 gaga D 0 Fri Dec 3 19:04:20 2010 How can I troubleshoot this? What I did: First I set up a fresh install of Ubuntu 10.10 x64. Second I got kerberos working with the following krb5.conf file: [libdefaults] ticket_lifetime = 24000 clock_skew = 300 default_realm = CUSTOMER.LOCAL [realms] CUSTOMER.LOCAL = { kdc = SB4.customer.local:88 admin_server = SB4.customer.local:464 default_domain = CUSTOMER.LOCAL } [domain_realm] .customer.local = CUSTOMER.LOCAL customer.local = CUSTOMER.LOCAL #[login] # krb4_convert = true # krb4_get_tickets = false I also added winbind to group, passwd and shadow in nsswitch.conf. Seemingly Kerberos works: root@lx:~# net ads testjoin Join is OK root@lx:~# wbinfo -a 'Administrator%MYSECRETPASSWORD' plaintext password authentication succeeded challenge/response password authentication succeeded wbinfo -u and wbinfo -g also spit out a list of users and a list of groups respectiveley. I noted that domain accounts did NOT include a domain and they are in german (as on the SBS 2003 that is the domain server). So I get a "Domänenbenutzer" in wbinfo -u's output not a "CUSTOMER+Domain User" or something similar. I'm not sure anymore what I did to the PAM configuration, but here is what I currently have: root@lx:/etc/pam.d# cat samba @include common-auth @include common-account @include common-session-noninteractive root@lx:/etc/pam.d# grep -ve '^#' common-auth auth [success=3 default=ignore] pam_krb5.so minimum_uid=1000 auth [success=2 default=ignore] pam_unix.so nullok_secure try_first_pass auth [success=1 default=ignore] pam_winbind.so krb5_auth krb5_ccache_type=FILE cached_login try_first_pass auth requisite pam_deny.so auth required pam_permit.so root@lx:/etc/pam.d# grep -ve '^#' common-account account [success=2 new_authtok_reqd=done default=ignore] pam_unix.so account [success=1 new_authtok_reqd=done default=ignore] pam_winbind.so account requisite pam_deny.so account required pam_permit.so account required pam_krb5.so minimum_uid=1000 root@lx:/etc/pam.d# grep -ve '^#' common-session-noninteractive session [default=1] pam_permit.so session requisite pam_deny.so session required pam_permit.so session optional pam_krb5.so minimum_uid=1000 session required pam_unix.so session optional pam_winbind.so At some point I joined the linux box into the AD domain. After (manually) creating a home directory on the linux box I can log in using the Adminstrator user with the password taken from AD. Now I run samba with the following setup: [global] netbios name = LX realm = CUSTOMER.LOCAL workgroup = CUSTOMER security = ADS encrypt passwords = yes password server = 192.168.20.244 #IP des Domain Controllers os level = 0 socket options = TCP_NODELAY SO_RCVBUF=16384 SO_SNDBUF=16384 idmap uid = 10000-20000 idmap gid = 10000-20000 winbind enum users = Yes winbind enum groups = Yes preferred master = no winbind separator = + dns proxy = no wins proxy = no # client NTLMv2 auth = Yes log level = 2 logfile = /var/log/samba/log.smbd.%U template homedir = /home/%U template shell = /bin/bash [export] path = /mnt/sdc1/export read only = No public = Yes Currently I don't care whether export is exported to everyone or just one user, I want to see somebody WRITING to that directory before I start fiddling with the authentication settings. (Who may access it). As mentioned, accessing the share from smbclient results in this NT_STATUS_MEDIA_WRITE_PROTECTED . Accessing it from windows shows ACLs that look correct (The user may write) - but it does not work, I can only read files not write. The directory to be exported looks like this: root@lx:/etc/pam.d# ls -ld /mnt/ drwxr-xr-x 5 root root 4096 2010-11-28 01:29 /mnt/ root@lx:/etc/pam.d# ls -ld /mnt/sdc1/ drwxr-xr-x 4 froh froh 4096 2010-11-28 01:32 /mnt/sdc1/ root@lx:/etc/pam.d# ls -ld /mnt/sdc1/export/ drwxrwxrwx+ 5 administrator domänen-admins 4096 2010-12-03 19:04 /mnt/sdc1/export/ root@lx:/etc/pam.d# getfacl /mnt/ getfacl: Entferne führende '/' von absoluten Pfadnamen # file: mnt/ # owner: root # group: root user::rwx group::r-x other::r-x root@lx:/etc/pam.d# getfacl /mnt/sdc1/ getfacl: Entferne führende '/' von absoluten Pfadnamen # file: mnt/sdc1/ # owner: froh # group: froh user::rwx group::r-x other::r-x root@lx:/etc/pam.d# getfacl /mnt/sdc1/export/ getfacl: Entferne führende '/' von absoluten Pfadnamen # file: mnt/sdc1/export/ # owner: administrator # group: domänen-admins user::rwx group::rwx group:domänen-admins:rwx mask::rwx other::rwx default:user::rwx default:group::rwx default:group:domänen-admins:rwx default:mask::rwx default:other::rwx My, oh my what am I overlooking? What am I to blind to see?

    Read the article

  • Win7 System folder contains infinitely looping SYSTEM(!) directory

    - by Matt
    My Windows 7 Enterprise computer has been crashing fairly frequently recently, so I decided to boot up in safe mode and run the TrendMicro client I have installed. It froze about 10 minutes into the full system scan, so in the spirit of http://whathaveyoutried.com, I started scanning each folder individually. When I got to ProgramData, the AV failed with an uncaught exception. I then went down a level and tried scanning Application Data, which failed as well. Imagine my surprise when I open the folder just to see the same folder again! As far as I can tell, this folder loop continues indefinitely. (If you are trying to recreate this, keep in mind that ProgramData is a hidden folder.) I'm actually a bit concerned that these are system folders, as this is a brand-new computer with a clean installation. I guess I have three questions: Has anyone else seen/experienced this before? I'm running Win7 SP1. How do I fix this? I've run CHKDSK \F with no success (although it was incredibly slow). What are the ramifications of an infinitely recursive directory? Theoretically speaking, each link takes up memory, so shouldn't I have no space available on my hard drive? (I've got about 180GB left.) I noticed that the tree view on the left only shows the "linked folder" icon on the deeper folders--does this mean anything special? (I've circled the icons or lack thereof in red.) How can the OS even resolve this aberration? And above all, what would happen if I were to select "Expand all folders"??? :P Matt

    Read the article

  • filezilla Command: MLSD Response: 500 Error: Failed to retrieve directory listing

    - by solomongaby
    Hello, Recently our network was moved to the corporate network and behind the company firewall. Since the i could not access the FTP Servers using Filezilla. I can connect to them but cannot receive the directory listing. Command: MLSD Response: 500 Error: Failed to retrieve directory listing I can access the server using other softwares ( GnomeCommander ) but i would like to use Filezilla for its advanced features. I tried active and pasive mode but that doesnt work. I trace the problem to the MSLD command. I think if i could make filezilla use LIST command it will work. Any ideeas. Thanks.

    Read the article

< Previous Page | 5 6 7 8 9 10 11 12 13 14 15 16  | Next Page >