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  • Is it possible to use 2 versions of jQuery on the same page?

    - by Ben McCormack
    NOTE: I know similar questions have already been asked here and here, but I'm looking for additional clarification as to how to make this work. I'm adding functionality to an existing web site that is already using an older version of the jQuery library (1.1.3.1). I've been writing my added functionality against the newest version of the jQuery library (1.4.2). I've tested the website using only the newer version of jQuery and it breaks functionality, so now I'm looking at using both versions on the same page. How is this possible? What do I need to do in my code to specify that I'm using one version of jQuery instead of another? For example, I'll put <script> tags for both versions of jQuery in the header of my page, but what do I need to do so that I know for sure in my calling code that I'm calling one version of the library or another? Maybe something like this: //Put some code here to specify a variable that will be using the newer //version of jquery: var $NEW = jQuery.theNewestVersion(); //Now when I use $NEW, I'll know it's the newest version and won't //conflict with the older version. $NEW('#personName').text('Ben'); //And when I use the original $ in code, or simply 'jquery', I'll know //it's the older version. $('#personName').doSomethingWithTheOlderVersion();

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  • Criticize my code, please

    - by Micky
    Hey, I was applying for a position, and they asked me to complete a coding problem for them. I did so and submitted it, but I later found out I was rejected from the position. Anyways, I have an eclectic programming background so I'm not sure if my code is grossly wrong or if I just didn't have the best solution out there. I would like to post my code and get some feedback about it. Before I do, here's a description of a problem: You are given a sorted array of integers, say, {1, 2, 4, 4, 5, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 11, 13 }. Now you are supposed to write a program (in C or C++, but I chose C) that prompts the user for an element to search for. The program will then search for the element. If it is found, then it should return the first index the entry was found at and the number of instances of that element. If the element is not found, then it should return "not found" or something similar. Here's a simple run of it (with the array I just put up): Enter a number to search for: 4 4 was found at index 2. There are 2 instances for 4 in the array. Enter a number to search for: -4. -4 is not in the array. They made a comment that my code should scale well with large arrays (so I wrote up a binary search). Anyways, my code basically runs as follows: Prompts user for input. Then it checks if it is within bounds (bigger than a[0] in the array and smaller than the largest element of the array). If so, then I perform a binary search. If the element is found, then I wrote two while loops. One while loop will count to the left of the element found, and the second while loop will count to the right of the element found. The loops terminate when the adjacent elements do not match with the desired value. EX: 4, 4, 4, 4, 4 The bold 4 is the value the binary search landed on. One loop will check to the left of it, and another loop will check to the right of it. Their sum will be the total number of instances of the the number four. Anyways, I don't know if there are any advanced techniques that I am missing or if I just don't have the CS background and made a big error. Any constructive critiques would be appreciated! #include <stdio.h> #include <stdlib.h> #include <string.h> #include <stddef.h> /* function prototype */ int get_num_of_ints( const int* arr, size_t r, int N, size_t* first, size_t* count ); int main() { int N; /* input variable */ int arr[]={1,1,2,3,3,4,4,4,4,5,5,7,7,7,7,8,8,8,9,11,12,12}; /* array of sorted integers */ size_t r = sizeof(arr)/sizeof(arr[0]); /* right bound */ size_t first; /* first match index */ size_t count; /* total number of matches */ /* prompts the user to enter input */ printf( "\nPlease input the integer you would like to find.\n" ); scanf( "%d", &N ); int a = get_num_of_ints( arr, r, N, &first, &count ); /* If the function returns -1 then the value is not found. Else it is returned */ if( a == -1) printf( "%d has not been found.\n", N ); else if(a >= 0){ printf( "The first matching index is %d.\n", first ); printf( "The total number of instances is %d.\n", count ); } return 0; } /* function definition */ int get_num_of_ints( const int* arr, size_t r, int N, size_t* first, size_t* count ) { int lo=0; /* lower bound for search */ int m=0; /* middle value obtained */ int hi=r-1; /* upper bound for search */ int w=r-1; /* used as a fixed upper bound to calculate the number of right instances of a particular value. */ /* binary search to find if a value exists */ /* first check if the element is out of bounds */ if( N < arr[0] || arr[hi] < N ){ m = -1; } else{ /* binary search to find a value, if it exists, within given parameters */ while(lo <= hi){ m = (hi + lo)/2; if(arr[m] < N) lo = m+1; else if(arr[m] > N) hi = m-1; else if(arr[m]==N){ m=m; break; } } if (lo > hi) /* if it doesn't we assign it -1 */ m = -1; } /* If the value is found, then we compute the left and right instances of it */ if( m >= 0 ){ int j = m-1; /* starting with the first term to the left */ int L = 0; /* total number of left instances */ /* while loop computes total number of left instances */ while( j >= 0 && arr[j] == arr[m] ){ L++; j--; } /* There are six possible outcomes of this. Depending on the outcome, we must assign the first index variable accordingly */ if( j > 0 && L > 0 ) *first=j+1; else if( j==0 && L==0) *first=m; else if( j > 0 && L==0 ) *first=m; else if(j < 0 && L==0 ) *first=m; else if( j < 0 && L > 0 ) *first=0; else if( j=0 && L > 0 ) *first=j+1; int h = m + 1; /* starting with the first term to the right */ int R = 0; /* total number of right instances */ /* while loop computes total number of right instances */ /* we fixed w earlier so that it's value does not change */ while( arr[h]==arr[m] && h <= w ){ R++; h++; } *count = (R + L + 1); /* total number of instances stored as value of count */ return *first; /* first instance index stored here */ } /* if value does not exist, then we return a negative value */ else if( m==-1) return -1; }

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  • Installing Rails on Mountain Lion

    - by Jordan Medlock
    I was wondering if you could help me find why I cannot install Ruby on Rails on my MBP with OS X Mountain Lion. It's a weird problem and I'll give you as much info as I can. I've installed ruby and it's working at version 1.9.3 And I've installed ruby gems and it's worked for every other gem I've tried to install. It's version is 1.8.24 When I run $ sudo gem install rails it replies with the message: Successfully installed rails-3.2.8 1 gem installed Although when I ask it rails -v it returns: `Rails is not currently installed on this system. To get the latest version, simply type: $ sudo gem install rails You can then rerun your "rails" command.` What should I do? The rails bash file (/usr/bin/rails) contains: #!/usr/bin/ruby # Stub rails command to load rails from Gems or print an error if not installed. require 'rubygems' version = ">= 0" if ARGV.first =~ /^_(.*)_$/ and Gem::Version.correct? $1 then version = $1 ARGV.shift end begin gem 'railties', version or raise rescue Exception puts 'Rails is not currently installed on this system. To get the latest version, simply type:' puts puts ' $ sudo gem install rails' puts puts 'You can then rerun your "rails" command.' exit 0 end load Gem.bin_path('railties', 'rails', version) That must mean that the gem files aren't there or are old or corrupted How can I check that?

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  • The Most Common and Least Used 4-Digit PIN Numbers [Security Analysis Report]

    - by Asian Angel
    How ‘secure’ is your 4-digit PIN number? Is your PIN number a far too common one or is it a bit more unique in comparison to others? The folks over at the Data Genetics blog have put together an interesting analysis report that looks at the most common and least used 4-digit PIN numbers chosen by people. Numerically based (0-9) 4-digit PIN numbers only allow for a total of 10,000 possible combinations, so it stands to reason that some combinations are going to be far more common than others. The question is whether or not your personal PIN number choices are among the commonly used ones or ‘stand out’ as being more unique. Note 1: Data Genetics used data condensed from released, exposed, & discovered password tables and security breaches to generate the analysis report. Note 2: The updates section at the bottom has some interesting tidbits concerning peoples’ use of dates and certain words for PIN number generation. The analysis makes for very interesting reading, so browse on over to get an idea of where you stand with regards to your personal PIN number choices. 8 Deadly Commands You Should Never Run on Linux 14 Special Google Searches That Show Instant Answers How To Create a Customized Windows 7 Installation Disc With Integrated Updates

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  • SQL SERVER – What is Denali?

    - by pinaldave
    I see following question quite common on Twitter or in my email box. “What is Denali?” Denali is code name of SQL Server 2011. Here is the list of the code name of other versions of SQL Server. In 1988, Microsoft released its first version of SQL Server. It was developed jointly by Microsoft and Sybase for the OS/2 platform. 1993 – SQL Server 4.21 for Windows NT 1995 – SQL Server 6.0, codenamed SQL95 1996 – SQL Server 6.5, codenamed Hydra 1999 – SQL Server 7.0, codenamed Sphinx 1999 – SQL Server 7.0 OLAP, codenamed Plato 2000 – SQL Server 2000 32-bit, codenamed Shiloh (version 8.0) 2003 – SQL Server 2000 64-bit, codenamed Liberty 2005 – SQL Server 2005, codenamed Yukon (version 9.0) 2008 – SQL Server 2008, codenamed Katmai (version 10.0) 2010 – SQL Server 2008 R2, Codenamed Kilimanjaro (aka KJ) Next – SQL Server 2011, Codenamed Denali Any guesses what should be the next version after 2011 should be codenamed? Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • code metrics for .net code

    - by user20358
    While the code metrics tool gives a pretty good analysis of the code being analyzed, I was wondering if there was any such benchmark on acceptable standards for the following as well: Maximum number of types per assembly Maximum number of such types that can be accessible Maximum number of parameters per method Acceptable RFC count Acceptable Afferent coupling count Acceptable Efferent coupling count Any other metrics to judge the quality of .Net code by? Thanks for your time.

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  • Make mex compiler of matlab working on mint?

    - by Erogol
    Mex compiler of matlab does not work with following error Warning: You are using gcc version "4.7.2-2ubuntu1)". The version currently supported with MEX is "4.4.6". For a list of currently supported compilers see: http://www.mathworks.com/support/compilers/current_release/ /home/krm/matlab/bin/mex: 1: eval: g++: not found mex: compile of ' "fv_cache/fv_cache.cc"' failed. it is obvious that I need preceding version of gcc but this specific version is not included in software manager of mint. I installed gcc-4.4 but it does not recognized by Matlab. I also removed latest version from my computer and set gcc as a environment variable points to gcc-4.4 but again does not work. Is there any other way around to solve that issue? Maybe a interface or something.

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  • Oracle Solaris Cluster 4.2 Event and its SNMP Interface

    - by user12609115
    Background The cluster event SNMP interface was first introduced in Oracle Solaris Cluster 3.2 release. The details of the SNMP interface are described in the Oracle Solaris Cluster System Administration Guide and the Cluster 3.2 SNMP blog. Prior to the Oracle Solaris Cluster 4.2 release, when the event SNMP interface was enabled, it would take effect on WARNING or higher severity events. The events with WARNING or higher severity are usually for the status change of a cluster component from ONLINE to OFFLINE. The interface worked like an alert/alarm interface when some components in the cluster were out of service (changed to OFFLINE). The consumers of this interface could not get notification for all status changes and configuration changes in the cluster. Cluster Event and its SNMP Interface in Oracle Solaris Cluster 4.2 The user model of the cluster event SNMP interface is the same as what was provided in the previous releases. The cluster event SNMP interface is not enabled by default on a freshly installed cluster; you can enable it by using the cluster event SNMP administration commands on any cluster nodes. Usually, you only need to enable it on one of the cluster nodes or a subset of the cluster nodes because all cluster nodes get the same cluster events. When it is enabled, it is responsible for two basic tasks. • Logs up to 100 most recent NOTICE or higher severity events to the MIB. • Sends SNMP traps to the hosts that are configured to receive the above events. The changes in the Oracle Solaris Cluster 4.2 release are1) Introduction of the NOTICE severity for the cluster configuration and status change events.The NOTICE severity is introduced for the cluster event in the 4.2 release. It is the severity between the INFO and WARNING severity. Now all severities for the cluster events are (from low to high) • INFO (not exposed to the SNMP interface) • NOTICE (newly introduced in the 4.2 release) • WARNING • ERROR • CRITICAL • FATAL In the 4.2 release, the cluster event system is enhanced to make sure at least one event with the NOTICE or a higher severity will be generated when there is a configuration or status change from a cluster component instance. In other words, the cluster events from a cluster with the NOTICE or higher severities will cover all status and configuration changes in the cluster (include all component instances). The cluster component instance here refers to an instance of the following cluster componentsnode, quorum, resource group, resource, network interface, device group, disk, zone cluster and geo cluster heartbeat. For example, pnode1 is an instance of the cluster node component, and oracleRG is an instance of the cluster resource group. With the introduction of the NOTICE severity event, when the cluster event SNMP interface is enabled, the consumers of the SNMP interface will get notification for all status and configuration changes in the cluster. A thrid-party system management platform with the cluster SNMP interface integration can generate alarms and clear alarms programmatically, because it can get notifications for the status change from ONLINE to OFFLINE and also from OFFLINE to ONLINE. 2) Customization for the cluster event SNMP interface • The number of events logged to the MIB is 100. When the number of events stored in the MIB reaches 100 and a new qualified event arrives, the oldest event will be removed before storing the new event to the MIB (FIFO, first in, first out). The 100 is the default and minimum value for the number of events stored in the MIB. It can be changed by setting the log_number property value using the clsnmpmib command. The maximum number that can be set for the property is 500. • The cluster event SNMP interface takes effect on the NOTICE or high severity events. The NOTICE severity is also the default and lowest event severity for the SNMP interface. The SNMP interface can be configured to take effect on other higher severity events, such as WARNING or higher severity events by setting the min_severity property to the WARNING. When the min_severity property is set to the WARNING, the cluster event SNMP interface would behave the same as the previous releases (prior to the 4.2 release). Examples, • Set the number of events stored in the MIB to 200 # clsnmpmib set -p log_number=200 event • Set the interface to take effect on WARNING or higher severity events. # clsnmpmib set -p min_severity=WARNING event Administering the Cluster Event SNMP Interface Oracle Solaris Cluster provides the following three commands to administer the SNMP interface. • clsnmpmib: administer the SNMP interface, and the MIB configuration. • clsnmphost: administer hosts for the SNMP traps • clsnmpuser: administer SNMP users (specific for SNMP v3 protocol) Only clsnmpmib is changed in the 4.2 release to support the aforementioned customization of the SNMP interface. Here are some simple examples using the commands. Examples: 1. Enable the cluster event SNMP interface on the local node # clsnmpmib enable event 2. Display the status of the cluster event SNMP interface on the local node # clsnmpmib show -v 3. Configure my_host to receive the cluster event SNMP traps. # clsnmphost add my_host Cluster Event SNMP Interface uses the common agent container SNMP adaptor, which is based on the JDMK SNMP implementation as its SNMP agent infrastructure. By default, the port number for the SNMP MIB is 11161, and the port number for the SNMP traps is 11162. The port numbers can be changed by using the cacaoadm. For example, # cacaoadm list-params Print all changeable parameters. The output includes the snmp-adaptor-port and snmp-adaptor-trap-port properties. # cacaoadm set-param snmp-adaptor-port=1161 Set the SNMP MIB port number to 1161. # cacaoadm set-param snmp-adaptor-trap-port=1162 Set the SNMP trap port number to 1162. The cluster event SNMP MIB is defined in sun-cluster-event-mib.mib, which is located in the /usr/cluster/lib/mibdirectory. Its OID is 1.3.6.1.4.1.42.2.80, that can be used to walk through the MIB data. Again, for more detail information about the cluster event SNMP interface, please see the Oracle Solaris Cluster 4.2 System Administration Guide. - Leland Chen 

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  • USB mouse does not work on boot

    - by Uku Loskit
    My problem is pretty much a duplicate of the one described in USB mouse late to load , but the solution there has not worked for me. I'm running the same OS and experiencing the exact same issue. It disappears after 10 seconds or so. Booting with the options specified in the other question did not fix it :/ Thanks in advance. sheepz@sheepz-desktop:~$ dmesg | egrep "hci|usb" [ 0.188000] usbcore: registered new interface driver usbfs [ 0.188000] usbcore: registered new interface driver hub [ 0.188000] usbcore: registered new device driver usb [ 0.358613] ehci_hcd: USB 2.0 'Enhanced' Host Controller (EHCI) Driver [ 0.358627] ohci_hcd: USB 1.1 'Open' Host Controller (OHCI) Driver [ 0.358637] uhci_hcd: USB Universal Host Controller Interface driver [ 0.358683] uhci_hcd 0000:00:1d.0: PCI INT A -> GSI 23 (level, low) -> IRQ 23 [ 0.358691] uhci_hcd 0000:00:1d.0: setting latency timer to 64 [ 0.358695] uhci_hcd 0000:00:1d.0: UHCI Host Controller [ 0.358726] uhci_hcd 0000:00:1d.0: new USB bus registered, assigned bus number 1 [ 0.358758] uhci_hcd 0000:00:1d.0: irq 23, io base 0x0000e100 [ 0.358927] uhci_hcd 0000:00:1d.1: PCI INT B -> GSI 19 (level, low) -> IRQ 19 [ 0.358932] uhci_hcd 0000:00:1d.1: setting latency timer to 64 [ 0.358935] uhci_hcd 0000:00:1d.1: UHCI Host Controller [ 0.358964] uhci_hcd 0000:00:1d.1: new USB bus registered, assigned bus number 2 [ 0.358991] uhci_hcd 0000:00:1d.1: irq 19, io base 0x0000e200 [ 0.359132] uhci_hcd 0000:00:1d.2: PCI INT C -> GSI 18 (level, low) -> IRQ 18 [ 0.359137] uhci_hcd 0000:00:1d.2: setting latency timer to 64 [ 0.359139] uhci_hcd 0000:00:1d.2: UHCI Host Controller [ 0.359165] uhci_hcd 0000:00:1d.2: new USB bus registered, assigned bus number 3 [ 0.359193] uhci_hcd 0000:00:1d.2: irq 18, io base 0x0000e300 [ 0.359327] uhci_hcd 0000:00:1d.3: PCI INT D -> GSI 16 (level, low) -> IRQ 16 [ 0.359332] uhci_hcd 0000:00:1d.3: setting latency timer to 64 [ 0.359334] uhci_hcd 0000:00:1d.3: UHCI Host Controller [ 0.359360] uhci_hcd 0000:00:1d.3: new USB bus registered, assigned bus number 4 [ 0.359387] uhci_hcd 0000:00:1d.3: irq 16, io base 0x0000e400 [ 0.731933] usb 1-1: new full speed USB device using uhci_hcd and address 2 [ 1.023859] usb 1-2: new full speed USB device using uhci_hcd and address 3 [ 16.136175] usb 1-2: device descriptor read/64, error -110 [ 31.352481] usb 1-2: device descriptor read/64, error -110 [ 31.568485] usb 1-2: new full speed USB device using uhci_hcd and address 4 [ 46.680794] usb 1-2: device descriptor read/64, error -110 [ 61.903555] usb 1-2: device descriptor read/64, error -110 [ 62.119671] usb 1-2: new full speed USB device using uhci_hcd and address 5 [ 72.541078] usb 1-2: device not accepting address 5, error -110 [ 72.653194] usb 1-2: new full speed USB device using uhci_hcd and address 6 [ 83.066637] usb 1-2: device not accepting address 6, error -110 [ 83.178615] usb 3-1: new low speed USB device using uhci_hcd and address 2 [ 83.562546] usbcore: registered new interface driver hiddev [ 83.578827] input: Logitech USB-PS/2 Optical Mouse as /devices/pci0000:00/0000:00:1d.2/usb3/3-1/3-1:1.0/input/input3 [ 83.579016] generic-usb 0003:046D:C01D.0001: input,hidraw0: USB HID v1.10 Mouse [Logitech USB-PS/2 Optical Mouse] on usb-0000:00:1d.2-1/input0 [ 83.579244] usbcore: registered new interface driver usbhid [ 83.579246] usbhid: USB HID core driver [114025.224407] usb 3-1: USB disconnect, address 2 sheepz@sheepz-desktop:~$ dmesg | egrep "hci|usb" [ 0.188000] usbcore: registered new interface driver usbfs [ 0.188000] usbcore: registered new interface driver hub [ 0.188000] usbcore: registered new device driver usb [ 0.358613] ehci_hcd: USB 2.0 'Enhanced' Host Controller (EHCI) Driver [ 0.358627] ohci_hcd: USB 1.1 'Open' Host Controller (OHCI) Driver [ 0.358637] uhci_hcd: USB Universal Host Controller Interface driver [ 0.358683] uhci_hcd 0000:00:1d.0: PCI INT A -> GSI 23 (level, low) -> IRQ 23 [ 0.358691] uhci_hcd 0000:00:1d.0: setting latency timer to 64 [ 0.358695] uhci_hcd 0000:00:1d.0: UHCI Host Controller [ 0.358726] uhci_hcd 0000:00:1d.0: new USB bus registered, assigned bus number 1 [ 0.358758] uhci_hcd 0000:00:1d.0: irq 23, io base 0x0000e100 [ 0.358927] uhci_hcd 0000:00:1d.1: PCI INT B -> GSI 19 (level, low) -> IRQ 19 [ 0.358932] uhci_hcd 0000:00:1d.1: setting latency timer to 64 [ 0.358935] uhci_hcd 0000:00:1d.1: UHCI Host Controller [ 0.358964] uhci_hcd 0000:00:1d.1: new USB bus registered, assigned bus number 2 [ 0.358991] uhci_hcd 0000:00:1d.1: irq 19, io base 0x0000e200 [ 0.359132] uhci_hcd 0000:00:1d.2: PCI INT C -> GSI 18 (level, low) -> IRQ 18 [ 0.359137] uhci_hcd 0000:00:1d.2: setting latency timer to 64 [ 0.359139] uhci_hcd 0000:00:1d.2: UHCI Host Controller [ 0.359165] uhci_hcd 0000:00:1d.2: new USB bus registered, assigned bus number 3 [ 0.359193] uhci_hcd 0000:00:1d.2: irq 18, io base 0x0000e300 [ 0.359327] uhci_hcd 0000:00:1d.3: PCI INT D -> GSI 16 (level, low) -> IRQ 16 [ 0.359332] uhci_hcd 0000:00:1d.3: setting latency timer to 64 [ 0.359334] uhci_hcd 0000:00:1d.3: UHCI Host Controller [ 0.359360] uhci_hcd 0000:00:1d.3: new USB bus registered, assigned bus number 4 [ 0.359387] uhci_hcd 0000:00:1d.3: irq 16, io base 0x0000e400 [ 0.731933] usb 1-1: new full speed USB device using uhci_hcd and address 2 [ 1.023859] usb 1-2: new full speed USB device using uhci_hcd and address 3 [ 16.136175] usb 1-2: device descriptor read/64, error -110 [ 31.352481] usb 1-2: device descriptor read/64, error -110 [ 31.568485] usb 1-2: new full speed USB device using uhci_hcd and address 4 [ 46.680794] usb 1-2: device descriptor read/64, error -110 [ 61.903555] usb 1-2: device descriptor read/64, error -110 [ 62.119671] usb 1-2: new full speed USB device using uhci_hcd and address 5 [ 72.541078] usb 1-2: device not accepting address 5, error -110 [ 72.653194] usb 1-2: new full speed USB device using uhci_hcd and address 6 [ 83.066637] usb 1-2: device not accepting address 6, error -110 [ 83.178615] usb 3-1: new low speed USB device using uhci_hcd and address 2 [ 83.562546] usbcore: registered new interface driver hiddev [ 83.578827] input: Logitech USB-PS/2 Optical Mouse as /devices/pci0000:00/0000:00:1d.2/usb3/3-1/3-1:1.0/input/input3 [ 83.579016] generic-usb 0003:046D:C01D.0001: input,hidraw0: USB HID v1.10 Mouse [Logitech USB-PS/2 Optical Mouse] on usb-0000:00:1d.2-1/input0 [ 83.579244] usbcore: registered new interface driver usbhid [ 83.579246] usbhid: USB HID core driver

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  • TakeWhile and SkipWhile method in LINQ

    - by vik20000in
     In my last post I talked about how to use the take and the Skip keyword to filter out the number of records that we are fetching. But there is only problem with the take and skip statement. The problem lies in the dependency where by the number of records to be fetched has to be passed to it. Many a times the number of records to be fetched is also based on the query itself. For example if we want to continue fetching records till a certain condition is met on the record set. Let’s say we want to fetch records from the array of number till we get 7. For this kind of query LINQ has exposed the TakeWhile Method.     int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };     var firstNumbersLessThan6 = numbers.TakeWhile(n => n < 7);   In the same way we can also use the SkipWhile statement. The skip while statement will skip all the records that do not match certain condition provided. In the example below we are skiping all those number which are not divisible by 3. Remember we could have done this with where clause also, but SkipWhile method can be useful in many other situation and hence the example and the keyword.     int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };     var allButFirst3Numbers = numbers.SkipWhile(n => n % 3 != 0); Vikram

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  • 3D Dice using Maya - for integration with iOS game app

    - by Anil
    My designer is building a 3D design of a dice using Maya. I want to integrate this in my iOS app so that the user can spin the dice and get a number. Then they play the game using that number. So, I have two questions: 1) How can I make the dice spin and stop at a random position so that a number is presented to the user? and 2) Once it stops spinning how can I detect the number that is displayed to the user (programmatically)? Many thanks. -Anil

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  • choosing Database and Its Design for Rails

    - by Gaurav Shah
    I am having a difficulty in deciding the database & its structure. Let us say the problem is like this. For my product I have various customers( each is an educational institute) Each customer have their own sub-clients ( Institution have students) Each student record will have some basic information like "name" & "Number" . There are also additional information that a customer(institution) might want to ask sub-client(student) like "email" or "semester" I have come up with two solutions : 1. Mysql _insititution__ id-|- Description| __Student__ id-|-instituition_id-|-Name-|-Number| __student_additional_details__ student_id -|- field_name -|- Value Student_additional_details will have multiple records for each student depending upon number of questions asked from institution. 2.MongoDb _insititution___ id-|- Description| _Student__ id-|-instituition_id-|-Name-|-Number|-otherfield1 -|- otherfield2 with mongo the structure itself can be dynamic so student table seems really good in mongo . But the problem comes when I have to relate student with institution . So which one is a better design ? Or some other idea ?

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  • SQL SERVER – Removing Leading Zeros From Column in Table

    - by pinaldave
    Some questions surprises me and make me write code which I have never explored before. Today was similar experience as well. I have always received the question regarding how to reserve leading zeroes in SQL Server while displaying them on the SSMS or another application. I have written articles on this subject over here. SQL SERVER – Pad Ride Side of Number with 0 – Fixed Width Number Display SQL SERVER – UDF – Pad Ride Side of Number with 0 – Fixed Width Number Display SQL SERVER – Preserve Leading Zero While Coping to Excel from SSMS Today I received a very different question where the user wanted to remove leading zero and white space. I am using the same sample sent by user in this example. USE tempdb GO -- Create sample table CREATE TABLE Table1 (Col1 VARCHAR(100)) INSERT INTO Table1 (Col1) SELECT '0001' UNION ALL SELECT '000100' UNION ALL SELECT '100100' UNION ALL SELECT '000 0001' UNION ALL SELECT '00.001' UNION ALL SELECT '01.001' GO -- Original data SELECT * FROM Table1 GO -- Remove leading zeros SELECT SUBSTRING(Col1, PATINDEX('%[^0 ]%', Col1 + ' '), LEN(Col1)) FROM Table1 GO -- Clean up DROP TABLE Table1 GO Here is the resultset of above script. It will remove any leading zero or space and will display the number accordingly. This problem is a very generic problem and I am confident there are alternate solutions to this problem as well. If you have an alternate solution or can suggest a sample data which does not satisfy the SUBSTRING solution proposed, I will be glad to include them in follow up blog post with due credit. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Function, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Windows Azure Platform Training Kit - June Update

    Microsoft released an update to its Azure training kit. Here is what is new in the kit: Introduction to Windows Azure - VS2010 version Introduction To SQL Azure - VS2010 version Introduction to the Windows Azure Platform AppFabric Service Bus - VS2010 version Introduction to Dallas - VS2010 version Introduction to the Windows Azure Platform AppFabric Access Control Service - VS2010 version Web Services and Identity in the Cloud Exploring Windows Azure Storage VS2010...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Login failed for user 'sa' because the account is currently locked out. The system administrator can

    - by cabhilash
    Login failed for user 'sa' because the account is currently locked out. The system administrator can unlock it. (Microsoft SQL Server, Error: 18486) SQL server has local password policies. If policy is enabled which locks down the account after X number of failed attempts then the account is automatically locked down.This error with 'sa' account is very common. sa is default administartor login available with SQL server. So there are chances that an ousider has tried to bruteforce your system. (This can cause even if a legitimate tries to access the account with wrong password.Sometimes a user would have changed the password without informing others. So the other users would try to lo) You can unlock the account with the following options (use another admin account or connect via windows authentication) Alter account & unlock ALTER LOGIN sa WITH PASSWORD='password' UNLOCK Use another account Almost everyone is aware of the sa account. This can be the potential security risk. Even if you provide strong password hackers can lock the account by providing the wrong password. ( You can provide extra security by installing firewall or changing the default port but these measures are not always practical). As a best practice you can disable the sa account and use another account with same privileges.ALTER LOGIN sa DISABLE You can edit the lock-ot options using gpedit.msc( in command prompt type gpedit.msc and press enter). Navigate to Account Lokout policy as shown in the figure The Following options are available Account lockout threshold This security setting determines the number of failed logon attempts that causes a user account to be locked out. A locked-out account cannot be used until it is reset by an administrator or until the lockout duration for the account has expired. You can set a value between 0 and 999 failed logon attempts. If you set the value to 0, the account will never be locked out. Failed password attempts against workstations or member servers that have been locked using either CTRL+ALT+DELETE or password-protected screen savers count as failed logon attempts. Account lockout duration This security setting determines the number of minutes a locked-out account remains locked out before automatically becoming unlocked. The available range is from 0 minutes through 99,999 minutes. If you set the account lockout duration to 0, the account will be locked out until an administrator explicitly unlocks it. If an account lockout threshold is defined, the account lockout duration must be greater than or equal to the reset time. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified. Reset account lockout counter after This security setting determines the number of minutes that must elapse after a failed logon attempt before the failed logon attempt counter is reset to 0 bad logon attempts. The available range is 1 minute to 99,999 minutes. If an account lockout threshold is defined, this reset time must be less than or equal to the Account lockout duration. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified.When creating SQL user you can set CHECK_POLICY=on which will enforce the windows password policy on the account. The following policies will be applied Define the Enforce password history policy setting so that several previous passwords are remembered. With this policy setting, users cannot use the same password when their password expires.  Define the Maximum password age policy setting so that passwords expire as often as necessary for your environment, typically, every 30 to 90 days. With this policy setting, if an attacker cracks a password, the attacker only has access to the network until the password expires.  Define the Minimum password age policy setting so that passwords cannot be changed until they are more than a certain number of days old. This policy setting works in combination with the Enforce password historypolicy setting. If a minimum password age is defined, users cannot repeatedly change their passwords to get around the Enforce password history policy setting and then use their original password. Users must wait the specified number of days to change their passwords.  Define a Minimum password length policy setting so that passwords must consist of at least a specified number of characters. Long passwords--seven or more characters--are usually stronger than short ones. With this policy setting, users cannot use blank passwords, and they have to create passwords that are a certain number of characters long.  Enable the Password must meet complexity requirements policy setting. This policy setting checks all new passwords to ensure that they meet basic strong password requirements.  Password must meet the following complexity requirement, when they are changed or created: Not contain the user's entire Account Name or entire Full Name. The Account Name and Full Name are parsed for delimiters: commas, periods, dashes or hyphens, underscores, spaces, pound signs, and tabs. If any of these delimiters are found, the Account Name or Full Name are split and all sections are verified not to be included in the password. There is no check for any character or any three characters in succession. Contain characters from three of the following five categories:  English uppercase characters (A through Z) English lowercase characters (a through z) Base 10 digits (0 through 9) Non-alphabetic characters (for example, !, $, #, %) A catch-all category of any Unicode character that does not fall under the previous four categories. This fifth category can be regionally specific.

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  • Oracle Linux and Oracle VM pricing guide

    - by wcoekaer
    A few days ago someone showed me a pricing guide from a Linux vendor and I was a bit surprised at the complexity of it. Especially when you look at larger servers (4 or 8 sockets) and when adding virtual machine use into the mix. I think we have a very compelling and simple pricing model for both Oracle Linux and Oracle VM. Let me see if I can explain it in 1 page, not 10 pages. This pricing information is publicly available on the Oracle store, I am using the current public list prices. Also keep in mind that this is for customers using non-oracle x86 servers. When a customer purchases an Oracle x86 server, the annual systems support includes full use (all you can eat) of Oracle Linux, Oracle VM and Oracle Solaris (no matter how many VMs you run on that server, in case you deploy guests on a hypervisor). This support level is the equivalent of premier support in the list below. Let's start with Oracle VM (x86) : Oracle VM support subscriptions are per physical server on which you deploy the Oracle VM Server product. (1) Oracle VM Premier Limited - 1- or 2 socket server : $599 per server per year (2) Oracle VM Premier - more than 2 socket server (4, or 8 or whatever more) : $1199 per server per year The above includes the use of Oracle VM Manager and Oracle Enterprise Manager Cloud Control's Virtualization management pack (including self service cloud portal, etc..) 24x7 support, access to bugfixes, updates and new releases. It also includes all options, live migrate, dynamic resource scheduling, high availability, dynamic power management, etc If you want to play with the product, or even use the product without access to support services, the product is freely downloadable from edelivery. Next, Oracle Linux : Oracle Linux support subscriptions are per physical server. If you plan to run Oracle Linux as a guest on Oracle VM, VMWare or Hyper-v, you only have to pay for a single subscription per system, we do not charge per guest or per number of guests. In other words, you can run any number of Oracle Linux guests per physical server and count it as just a single subscription. (1) Oracle Linux Network Support - any number of sockets per server : $119 per server per year Network support does not offer support services. It provides access to the Unbreakable Linux Network and also offers full indemnification for Oracle Linux. (2) Oracle Linux Basic Limited Support - 1- or 2 socket servers : $499 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management. It includes ocfs2 as a clustered filesystem. (3) Oracle Linux Basic Support - more than 2 socket server (4, or 8 or more) : $1199 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management. It includes ocfs2 as a clustered filesystem (4) Oracle Linux Premier Limited Support - 1- or 2 socket servers : $1399 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management, XFS filesystem support. It also offers Oracle Lifetime support, backporting of patches for critical customers in previous versions of package and ksplice zero-downtime updates. (5) Oracle Linux Premier Support - more than 2 socket servers : $2299 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management, XFS filesystem support. It also offers Oracle Lifetime support, backporting of patches for critical customers in previous versions of package and ksplice zero-downtime updates. (6) Freely available Oracle Linux - any number of sockets You can freely download Oracle Linux, install it on any number of servers and use it for any reason, without support, without right to use of these extra features like Oracle Clusterware or ksplice, without indemnification. However, you do have full access to all errata as well. Need support? then use options (1)..(5) So that's it. Count number of 2 socket boxes, more than 2 socket boxes, decide on basic or premier support level and you are done. You don't have to worry about different levels based on how many virtual instance you deploy or want to deploy. A very simple menu of choices. We offer, inclusive, Linux OS clusterware, Linux OS Management, provisioning and monitoring, cluster filesystem (ocfs), high performance filesystem (xfs), dtrace, ksplice, ofed (infiniband stack for high performance networking). No separate add-on menus. NOTE : socket/cpu can have any number of cores. So whether you have a 4,6,8,10 or 12 core CPU doesn't matter, we count the number of physical CPUs.

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  • Oracle SOA Suite for healthcare integration Dashboard By Nitesh Jain

    - by JuergenKress
    Oracle SOA Suite Healthcare came up with a new way of monitoring where user can configure a dashboard and follow the dynamic runtime changes. Oracle SOA Suite for healthcare integration dashboards display information about the current health of the endpoints in a healthcare integration application. You can create and configure multiple dashboards as needed to monitor the status and volume metrics for the endpoints you have defined. The Dashboards reflects changes that occur in the runtime repository, such as purging runtime instance data, new messages processed, and new error messages. You can display data for various time periods, and you can manually refresh the data in real time or set the dashboard to automatically refresh at set intervals. Dashboard shows the following information: Status: The current status of the endpoint, such as Running, Idle, Disabled, or Errors. Messages Sent: The number of messages sent by the endpoint in the specified time period. Messages Received: The number of messages received by the endpoint in the specified time period. Errors: The number of messages with errors for the endpoint in the given time period. Last Sent: The date and time the last message was sent from the endpoint. Last Received: The date and time the last message was received from the endpoint. Last Error: The date and time of the last error for the endpoint. It also shows the detailed view of a specific Endpoint. The document type. The number of messages received per second. The total number of message processed in the specified time period. The average size of each message. For more information please visit Nitesh Jain blog SOA & BPM Partner Community For regular information on Oracle SOA Suite become a member in the SOA & BPM Partner Community for registration please visit  www.oracle.com/goto/emea/soa (OPN account required) If you need support with your account please contact the Oracle Partner Business Center. Blog Twitter LinkedIn Mix Forum Technorati Tags: SOA Suite,SOA heathcare,soa health,SOA Community,Oracle SOA,Oracle BPM,Community,OPN,Jürgen Kress

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  • Are CK Metrics still considered useful? Is there an open source tool to help?

    - by DeveloperDon
    Chidamber & Kemerer proposed several metrics for object oriented code. Among them, depth of inheritance tree, weighted number of methods, number of member functions, number of children, and coupling between objects. Using a base of code, they tried to correlated these metrics to the defect density and maintenance effort using covariant analysis. Are these metrics actionable in projects? Perhaps they can guide refactoring. For example weighted number of methods might show which God classes needed to be broken into more cohesive classes that address a single concern. Is there approach superseded by a better method, and is there a tool that can identify problem code, particularly in moderately large project being handed off to a new developer or team?

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  • Strange Windows DIR command behavior

    - by user1167442
    I discovered this quite by accident while looking for a file with a number in the name. When I type: dir *number* (where number represents any number from 0 to 9 and with no spaces between the asterisks and the number) at the cmd.exe command prompt, it returns various files do not appear in any to fit the search criteria. What's weird, is that depending on the directory, some numbers will work and not others. An example is, in a directory associated with a website, I type the following: dir *4* and what is returned is: Directory of C:\Ampps\www\includes\pages 04/30/2012 03:55 PM 153 inventory_list_retrieve.php 06/18/2012 11:17 AM 6,756 ix.html 06/19/2012 01:47 PM 257,501 jquery.1.7.1.js 3 File(s) 264,410 bytes 0 Dir(s) 362,280,906,752 bytes free That just doesn't make any sense to me. Any clue? The question is posed on stackOverflow because the DIR command is often combined with FOR in batch programs. The strange DIR behavior would seem to make batch programs potentially unreliable if they use the DIR command.

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  • Inside the Concurrent Collections: ConcurrentDictionary

    - by Simon Cooper
    Using locks to implement a thread-safe collection is rather like using a sledgehammer - unsubtle, easy to understand, and tends to make any other tool redundant. Unlike the previous two collections I looked at, ConcurrentStack and ConcurrentQueue, ConcurrentDictionary uses locks quite heavily. However, it is careful to wield locks only where necessary to ensure that concurrency is maximised. This will, by necessity, be a higher-level look than my other posts in this series, as there is quite a lot of code and logic in ConcurrentDictionary. Therefore, I do recommend that you have ConcurrentDictionary open in a decompiler to have a look at all the details that I skip over. The problem with locks There's several things to bear in mind when using locks, as encapsulated by the lock keyword in C# and the System.Threading.Monitor class in .NET (if you're unsure as to what lock does in C#, I briefly covered it in my first post in the series): Locks block threads The most obvious problem is that threads waiting on a lock can't do any work at all. No preparatory work, no 'optimistic' work like in ConcurrentQueue and ConcurrentStack, nothing. It sits there, waiting to be unblocked. This is bad if you're trying to maximise concurrency. Locks are slow Whereas most of the methods on the Interlocked class can be compiled down to a single CPU instruction, ensuring atomicity at the hardware level, taking out a lock requires some heavy lifting by the CLR and the operating system. There's quite a bit of work required to take out a lock, block other threads, and wake them up again. If locks are used heavily, this impacts performance. Deadlocks When using locks there's always the possibility of a deadlock - two threads, each holding a lock, each trying to aquire the other's lock. Fortunately, this can be avoided with careful programming and structured lock-taking, as we'll see. So, it's important to minimise where locks are used to maximise the concurrency and performance of the collection. Implementation As you might expect, ConcurrentDictionary is similar in basic implementation to the non-concurrent Dictionary, which I studied in a previous post. I'll be using some concepts introduced there, so I recommend you have a quick read of it. So, if you were implementing a thread-safe dictionary, what would you do? The naive implementation is to simply have a single lock around all methods accessing the dictionary. This would work, but doesn't allow much concurrency. Fortunately, the bucketing used by Dictionary allows a simple but effective improvement to this - one lock per bucket. This allows different threads modifying different buckets to do so in parallel. Any thread making changes to the contents of a bucket takes the lock for that bucket, ensuring those changes are thread-safe. The method that maps each bucket to a lock is the GetBucketAndLockNo method: private void GetBucketAndLockNo( int hashcode, out int bucketNo, out int lockNo, int bucketCount) { // the bucket number is the hashcode (without the initial sign bit) // modulo the number of buckets bucketNo = (hashcode & 0x7fffffff) % bucketCount; // and the lock number is the bucket number modulo the number of locks lockNo = bucketNo % m_locks.Length; } However, this does require some changes to how the buckets are implemented. The 'implicit' linked list within a single backing array used by the non-concurrent Dictionary adds a dependency between separate buckets, as every bucket uses the same backing array. Instead, ConcurrentDictionary uses a strict linked list on each bucket: This ensures that each bucket is entirely separate from all other buckets; adding or removing an item from a bucket is independent to any changes to other buckets. Modifying the dictionary All the operations on the dictionary follow the same basic pattern: void AlterBucket(TKey key, ...) { int bucketNo, lockNo; 1: GetBucketAndLockNo( key.GetHashCode(), out bucketNo, out lockNo, m_buckets.Length); 2: lock (m_locks[lockNo]) { 3: Node headNode = m_buckets[bucketNo]; 4: Mutate the node linked list as appropriate } } For example, when adding another entry to the dictionary, you would iterate through the linked list to check whether the key exists already, and add the new entry as the head node. When removing items, you would find the entry to remove (if it exists), and remove the node from the linked list. Adding, updating, and removing items all follow this pattern. Performance issues There is a problem we have to address at this point. If the number of buckets in the dictionary is fixed in the constructor, then the performance will degrade from O(1) to O(n) when a large number of items are added to the dictionary. As more and more items get added to the linked lists in each bucket, the lookup operations will spend most of their time traversing a linear linked list. To fix this, the buckets array has to be resized once the number of items in each bucket has gone over a certain limit. (In ConcurrentDictionary this limit is when the size of the largest bucket is greater than the number of buckets for each lock. This check is done at the end of the TryAddInternal method.) Resizing the bucket array and re-hashing everything affects every bucket in the collection. Therefore, this operation needs to take out every lock in the collection. Taking out mutiple locks at once inevitably summons the spectre of the deadlock; two threads each hold a lock, and each trying to acquire the other lock. How can we eliminate this? Simple - ensure that threads never try to 'swap' locks in this fashion. When taking out multiple locks, always take them out in the same order, and always take out all the locks you need before starting to release them. In ConcurrentDictionary, this is controlled by the AcquireLocks, AcquireAllLocks and ReleaseLocks methods. Locks are always taken out and released in the order they are in the m_locks array, and locks are all released right at the end of the method in a finally block. At this point, it's worth pointing out that the locks array is never re-assigned, even when the buckets array is increased in size. The number of locks is fixed in the constructor by the concurrencyLevel parameter. This simplifies programming the locks; you don't have to check if the locks array has changed or been re-assigned before taking out a lock object. And you can be sure that when a thread takes out a lock, another thread isn't going to re-assign the lock array. This would create a new series of lock objects, thus allowing another thread to ignore the existing locks (and any threads controlling them), breaking thread-safety. Consequences of growing the array Just because we're using locks doesn't mean that race conditions aren't a problem. We can see this by looking at the GrowTable method. The operation of this method can be boiled down to: private void GrowTable(Node[] buckets) { try { 1: Acquire first lock in the locks array // this causes any other thread trying to take out // all the locks to block because the first lock in the array // is always the one taken out first // check if another thread has already resized the buckets array // while we were waiting to acquire the first lock 2: if (buckets != m_buckets) return; 3: Calculate the new size of the backing array 4: Node[] array = new array[size]; 5: Acquire all the remaining locks 6: Re-hash the contents of the existing buckets into array 7: m_buckets = array; } finally { 8: Release all locks } } As you can see, there's already a check for a race condition at step 2, for the case when the GrowTable method is called twice in quick succession on two separate threads. One will successfully resize the buckets array (blocking the second in the meantime), when the second thread is unblocked it'll see that the array has already been resized & exit without doing anything. There is another case we need to consider; looking back at the AlterBucket method above, consider the following situation: Thread 1 calls AlterBucket; step 1 is executed to get the bucket and lock numbers. Thread 2 calls GrowTable and executes steps 1-5; thread 1 is blocked when it tries to take out the lock in step 2. Thread 2 re-hashes everything, re-assigns the buckets array, and releases all the locks (steps 6-8). Thread 1 is unblocked and continues executing, but the calculated bucket and lock numbers are no longer valid. Between calculating the correct bucket and lock number and taking out the lock, another thread has changed where everything is. Not exactly thread-safe. Well, a similar problem was solved in ConcurrentStack and ConcurrentQueue by storing a local copy of the state, doing the necessary calculations, then checking if that state is still valid. We can use a similar idea here: void AlterBucket(TKey key, ...) { while (true) { Node[] buckets = m_buckets; int bucketNo, lockNo; GetBucketAndLockNo( key.GetHashCode(), out bucketNo, out lockNo, buckets.Length); lock (m_locks[lockNo]) { // if the state has changed, go back to the start if (buckets != m_buckets) continue; Node headNode = m_buckets[bucketNo]; Mutate the node linked list as appropriate } break; } } TryGetValue and GetEnumerator And so, finally, we get onto TryGetValue and GetEnumerator. I've left these to the end because, well, they don't actually use any locks. How can this be? Whenever you change a bucket, you need to take out the corresponding lock, yes? Indeed you do. However, it is important to note that TryGetValue and GetEnumerator don't actually change anything. Just as immutable objects are, by definition, thread-safe, read-only operations don't need to take out a lock because they don't change anything. All lockless methods can happily iterate through the buckets and linked lists without worrying about locking anything. However, this does put restrictions on how the other methods operate. Because there could be another thread in the middle of reading the dictionary at any time (even if a lock is taken out), the dictionary has to be in a valid state at all times. Every change to state has to be made visible to other threads in a single atomic operation (all relevant variables are marked volatile to help with this). This restriction ensures that whatever the reading threads are doing, they never read the dictionary in an invalid state (eg items that should be in the collection temporarily removed from the linked list, or reading a node that has had it's key & value removed before the node itself has been removed from the linked list). Fortunately, all the operations needed to change the dictionary can be done in that way. Bucket resizes are made visible when the new array is assigned back to the m_buckets variable. Any additions or modifications to a node are done by creating a new node, then splicing it into the existing list using a single variable assignment. Node removals are simply done by re-assigning the node's m_next pointer. Because the dictionary can be changed by another thread during execution of the lockless methods, the GetEnumerator method is liable to return dirty reads - changes made to the dictionary after GetEnumerator was called, but before the enumeration got to that point in the dictionary. It's worth listing at this point which methods are lockless, and which take out all the locks in the dictionary to ensure they get a consistent view of the dictionary: Lockless: TryGetValue GetEnumerator The indexer getter ContainsKey Takes out every lock (lockfull?): Count IsEmpty Keys Values CopyTo ToArray Concurrent principles That covers the overall implementation of ConcurrentDictionary. I haven't even begun to scratch the surface of this sophisticated collection. That I leave to you. However, we've looked at enough to be able to extract some useful principles for concurrent programming: Partitioning When using locks, the work is partitioned into independant chunks, each with its own lock. Each partition can then be modified concurrently to other partitions. Ordered lock-taking When a method does need to control the entire collection, locks are taken and released in a fixed order to prevent deadlocks. Lockless reads Read operations that don't care about dirty reads don't take out any lock; the rest of the collection is implemented so that any reading thread always has a consistent view of the collection. That leads us to the final collection in this little series - ConcurrentBag. Lacking a non-concurrent analogy, it is quite different to any other collection in the class libraries. Prepare your thinking hats!

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  • July, the 31 Days of SQL Server DMO’s – Day 28 (sys.dm_db_stats_properties)

    - by Tamarick Hill
    The sys.dm_db_stats_properties Dynamic Management Function returns information about the statistics that are currently on your database objects. This function takes two parameters, an object_id and a stats_id. Let’s have a look at the result set from this function against the AdventureWorks2012.Sales.SalesOrderHeader table. To obtain the object_id and stats_id I will use a CROSS APPLY with the sys.stats system table. SELECT sp.* FROM sys.stats s CROSS APPLY sys.dm_db_stats_properties(s.object_id, s.Stats_id) sp WHERE sp.object_id = object_id('Sales.SalesOrderHeader') The first two columns returned by this function are the object_id and the stats_id columns. The next column, ‘last_updated’, gives you the date and the time that a particular statistic was last updated. The next column, ‘rows’, gives you the total number of rows in the table as of the last statistic update date. The ‘rows_sampled’ column gives you the number of rows that were sampled to create the statistic. The ‘steps’ column represents the number of specific value ranges from the statistic histogram. The ‘unfiltered_rows’ column represents the number of rows before any filters are applied. If a particular statistic is not filtered, the ‘unfiltered_rows’ column will always equal the ‘rows’ column. Lastly we have the ‘modification_counter’ column which represents the number of modification to the leading column in a given statistic since the last time the statistic was updated. Probably the most important column from this Dynamic Management Function is the ‘last_updated’ column. You want to always ensure that you have accurate and updated statistics on your database objects. Accurate statistics are vital for the query optimizer to generate efficient and reliable query execution plans. Without accurate and updated statistics, the performance of your SQL Server would likely suffer. For more information about this Dynamic Management Function, please see the below Books Online link: http://msdn.microsoft.com/en-us/library/jj553546.aspx Folllow me on Twitter @PrimeTimeDBA

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  • Vote of Disconfidence to Entity Framework

    - by Ricardo Peres
    A friend of mine has found the following problem with Entity Framework 4: Two simple classes and one association between them (one to many): One condition to filter out soft-deleted entities (WHERE Deleted = 0): 100 records in the database; A simple query: 1: var l = ctx.Person.Include("Address").Where(x => (x.Address.Name == "317 Oak Blvd." && x.Address.Number == 926) || (x.Address.Name == "891 White Milton Drive" && x.Address.Number == 497)); Will produce the following SQL: 1: SELECT 2: [Extent1].[Id] AS [Id], 3: [Extent1].[FullName] AS [FullName], 4: [Extent1].[AddressId] AS [AddressId], 5: [Extent202].[Id] AS [Id1], 6: [Extent202].[Name] AS [Name], 7: [Extent202].[Number] AS [Number] 8: FROM [dbo].[Person] AS [Extent1] 9: LEFT OUTER JOIN [dbo].[Address] AS [Extent2] ON ([Extent2].[Deleted] = 0) AND ([Extent1].[AddressId] = [Extent2].[Id]) 10: LEFT OUTER JOIN [dbo].[Address] AS [Extent3] ON ([Extent3].[Deleted] = 0) AND ([Extent1].[AddressId] = [Extent3].[Id]) 11: LEFT OUTER JOIN [dbo].[Address] AS [Extent4] ON ([Extent4].[Deleted] = 0) AND ([Extent1].[AddressId] = [Extent4].[Id]) 12: LEFT OUTER JOIN [dbo].[Address] AS [Extent5] ON ([Extent5].[Deleted] = 0) AND ([Extent1].[AddressId] = [Extent5].[Id]) 13: LEFT OUTER JOIN [dbo].[Address] AS [Extent6] ON ([Extent6].[Deleted] = 0) AND ([Extent1].[AddressId] = [Extent6].[Id]) 14: ... 15: WHERE ((N'317 Oak Blvd.' = [Extent2].[Name]) AND (926 = [Extent3].[Number])) 16: ... And will result in 680 MB of memory being taken! Now, Entity Framework has been historically known for producing less than optimal SQL, but 680 MB for 100 entities?! According to Microsoft, the problem will be addressed in the following version, there is a Connect issue open. There is even a whitepaper, Performance Considerations for Entity Framework 5, which talks about some of the changes and optimizations coming on version 5, but by reading it, I got even more concerned: “Once the cache contains a set number of entries (800), we start a timer that periodically (once-per-minute) sweeps the cache.” Say what?! The next version of Entity Framework will spawn timer threads?! When Code First came along, I thought it was a step in the right direction. Sure, it didn’t include some things that NHibernate did for quite some time – for example, different strategies for Id generation that do not rely on IDENTITY columns, which makes INSERT batching impossible, or support for enumerated types – but I thought these would come with the time. Now, enumerated types have, but so did… timer threads! I’m afraid Entity Framework is becoming a monster.

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  • Project Euler 8: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • How to get current connection settings

    - by Peter Larsson
    SELECT name AS Setting,         CASE             WHEN @@OPTIONS & number = number THEN 'ON'             ELSE 'OFF'         END AS Value FROM    master..spt_values WHERE   type= 'SOP'         AND number > 0

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  • Project migration tips [closed]

    - by Shirish11
    I would like to know what all things are to be considered when your client asks for Project Migration to a different language / version. Migrating an existing project to a higher version is not going to be much of a trouble. Have to take care of some system files, some settings changes and if version specific components included then search for them to comply with the newer version. I understand why you need to migrate projects to newer versions mostly for overcoming the drawbacks of current version. But I have absolutely no idea how to go about language migration. Any help on this is appreciated . Moreover is language migration a healthy idea?

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