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  • [Qt] Check octal number

    - by sterh
    Hello, I write simple application in C++/Qt. And i have a text and some octal number in it. My app splits this text by spaces. And i need to check octal numbers from text. How can i select octal numbers from this text with regular expressions? Thank you.

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  • How to check how many characters in variable, and add space between characters in that var?

    - by Camran
    I have a 'price' variable that contains some integer number from a MySQL database. I want to check how many numbers the 'price' variable contains, and add a space in the variable depending on how many numbers. See below: Example: If 'price' is 150000 I would like the output to be 150 000 (notice the space). OR, if it is 19000 I would like it to output 19 000... How would you do this the easiest way?

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  • mysql query trouble

    - by Bharanikumar
    Hi , in my database i have phone numbers with country code , which look somthing like 0044-123456 0044-123456 0014-123456 0014-123456 0024-123456 0024-123456 0034-123456 0044-123456 0044-123456 0024-123456 0034-123456 084-123456 084-123456 i want to total up the numbers by country, something like this output 0044 (2) 0024 (2) 0034 (1) 084 (2) 064 (5) Is it possible to do this with a SQL query?

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  • display 25 randomnumbers from an array

    - by qwerty
    Hello, I have in C++ an array of 100 elements, so v[1], ... ,v[100] contains numbers. How can i display, 25 random numbers from this array? So i wanna select 25 random positions from this array and display the values.. How can i do this in C++? Thanks!

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  • question about combinatorical

    - by davit-datuashvili
    here is task How many ways are there to choose from the set {1, 2, . . . , 100} three distinct numbers so that their sum is even? first of all sum of three numbers is even if only if 1.all number is even 2.two of them is odd and one is even i know that (n) = n!/(k!*(n-k)! (k) and can anybody help me to solve this problem

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  • Exceptions handling in SQL?

    - by Vineet
    Is there any way to handle exceptions in sql(ORACLE 9i)? Since I was trying to divide values of a column that contains both numbers and literals ,I need to fetch out only numbers from it ,as if it divisible by any number then its number else if contains literals it would not get divided it will generate error. how to handle those errors? Please suggest!!

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  • which delimeter to use while spliting String

    - by London
    I need to split this line string in each line, I need to get the third word(film name) but as you see the delimeter is one big blank character in some cases its small like before the numbers at the end or its big as in front of numbers at front. I tried using string split with(" ") regex, and also \t but get the out of the bounds error. 400115305 Lionel_Atwill The_Song_of_Songs_(1933_film) 7587 400115309 Brian_Aherne A_Night_to_Remember_(1943_film) 7952 Did anyone have the same problem?

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  • Using IF LARGE when there is text in column

    - by Ray
    I have an excel column of numbers and texts. I tried to use "IF LARGE" to find top 3 numbers of the column (A1 to A7), and return "Yes" to the cells right next to the top 3 (in column B). But unfortunately, the cells next to the texts also returned "Yes". This is the data: 0.2 0.3 Yes 0.5 Yes 0.1 0.8 Yes asdf Yes jklm Yes This is the code for cell B7: =IF(A7>=LARGE($A$1:$A$7,3),"Yes","") Any suggestions to fix this? thanks in advance

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  • Lotto program doesn't stop

    - by Naseyb Yaramis
    So I'm making a lotto game. You have to enter 6 lucky numbers and if they're the same as the lotto numbers then you win. Here is my code: using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace OefeningExaam { class Program { static void Main(string[] args) { Random getal = new Random(); int[] lottotrekking = new int[6]; Console.WriteLine("Geef je geluksgetallen in <tussen 1 en 42>"); Console.WriteLine("Geef je eerste getal in"); int getal1 = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("Geef je tweede getal in"); int getal2 = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("Geef je derde getal in"); int getal3 = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("Geef je vierde getal in"); int getal4 = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("Geef je vijfde getal in"); int getal5 = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("Geef je zesde getal in"); int getal6 = Convert.ToInt32(Console.ReadLine()); while (getal1 != lottotrekking[0] || getal2 != lottotrekking[1] || getal3 != lottotrekking[2] || getal4 != lottotrekking[3] || getal5 != lottotrekking[4] || getal5 != lottotrekking[4] || getal6 != lottotrekking[5]) { for (int i = 0; i < lottotrekking.Length; i++) { int cijfer = getal.Next(1, 43); lottotrekking[i] = cijfer; Console.WriteLine(lottotrekking[0] + "\t " + lottotrekking[1] + "\t " + lottotrekking[2] + "\t " + lottotrekking[3] + "\t " + lottotrekking[4] + "\t " + lottotrekking[5]); } } if (getal1 == lottotrekking[0] && getal2 == lottotrekking[1] && getal3 == lottotrekking[2] && getal4 == lottotrekking[3] && getal5 == lottotrekking[4] && getal5 == lottotrekking[4] && getal6 == lottotrekking[5]) { Console.WriteLine(lottotrekking[0] + " " + lottotrekking[1] + " " + lottotrekking[2] + " " + lottotrekking[3] + " " + lottotrekking[4] + " " + lottotrekking[5]); } Console.ReadLine(); } } } The problem is that the program just keeps going and doesn't stop. It's supposed to stop when the lucky numbers are the same as the lotto numbers.

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  • Array help in C++

    - by user267237
    I am writing a program that is declaring an array of 100 integer numbers and filling the array with random numbers using rand(). All I have so far is: #include <iostream> using namespace std; int main() { int random_integer = rand();

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  • Excel "Remove Duplicates" feature does not remove all duplicates

    - by raph82
    When the Data ToolsRemove Duplicates command is used on the following list, some duplicates are not removed: Key 46509 46509 51442 51442 73367 73367 80048 80048 88396 88396 102576 102576 108095 108095 118613 118613 127224 127224 137549 137549 140164 140164 151679 151679 152975 152975 154986 154986 167715 167715 169166 169166 170427 170427 176303 176303 187038 187038 199856 199856 206589 206589 225371 225371 229767 229767 230680 230680 245333 245333 253697 253697 256002 256002 262094 262094 273060 273060 275584 275584 285569 285569 301930 301930 304993 304993 305965 305965 307299 307299 311438 311438 314461 314461 340545 340545 344517 344517 355035 355035 358387 358387 359982 359982 363904 363904 368642 368642 378666 378666 378744 378744 382396 382396 393542 393542 394320 394320 395258 395258 402047 402047 421796 421796 427972 427972 442651 442651 442654 442654 445786 445786 447124 447124 456682 456682 467303 467303 470437 470437 476340 476340 480976 480976 481450 481450 498967 498967 516068 516068 517592 517592 522621 522621 524678 524678 529405 529405 532909 532909 535096 535096 540835 540835 553197 553197 566893 566893 567730 567730 570751 570751 572259 572259 581701 581701 589753 589753 597923 597923 602282 602282 612011 612011 643774 643774 649471 649471 654159 654159 661681 661681 680578 680578 709714 709714 728036 728036 728680 728680 740252 740252 750843 750843 763430 763430 777724 777724 777870 777870 779703 779703 787034 787034 789776 789776 793859 793859 798479 798479 812410 812410 816599 816599 820169 820169 821236 821236 833511 833511 833701 833701 845925 845925 854408 854408 882853 882853 891710 891710 906251 906251 929802 929802 982039 982039 984503 984503 984539 984539 987889 987889 997188 997188 1002869 1002869 1018493 1018493 1020735 1020735 1028780 1028780 1032460 1032460 1047306 1047306 1052136 1052136 1054482 1054482 1062633 1062633 1066509 1066509 1069500 1069500 1075339 1075339 1082282 1082282 1100206 1100206 1105407 1105407 1111409 1111409 1112175 1112175 1112407 1112407 1113400 1113400 1117243 1117243 1121732 1121732 1152461 1152461 1153727 1153727 1157252 1157252 1163384 1163384 1195324 1195324 1197373 1197373 1201828 1201828 1202786 1202786 1216328 1216328 1223541 1223541 1225848 1225848 1228595 1228595 1239281 1239281 1264378 1264378 1267348 1267348 1291510 1291510 1309891 1309891 1321106 1321106 1322886 1322886 1327690 1327690 1344145 1344145 1346250 1346250 1352350 1352350 1357920 1357920 1363386 1363386 1365046 1365046 1370383 1370383 1376666 1376666 1383834 1383834 1414472 1414472 1418967 1418967 1419811 1419811 1421587 1421587 1432639 1432639 1442854 1442854 1455236 1455236 1461896 1461896 1466376 1466376 1477670 1477670 1480154 1480154 1481801 1481801 1498209 1498209 1504776 1504776 1511904 1511904 1017854_447 1017854_447 1085496_19 1085496_19 1280601_77 1280601_77 1350662_135 1350662_135 141944_382 141944_382 1525893_483 1525893_483 455968_177 455968_177 469495_157 469495_157 641583_74 641583_74 704267_63 704267_63 709992_199 709992_199 819399_224 819399_224 832188_358 832188_358 857189_467 857189_467 898328_334 898328_334 91033_245 91033_245 960992_341 960992_341 999706_361 999706_361 BLA1048669 BLA1048669 BLA1052457 BLA1052457 BLA1055176 BLA1055176 BLA1060585 BLA1060585 BLA1067125 BLA1067125 BLA107601 BLA107601 BLA1087509 BLA1087509 BLA1097390 BLA1097390 BLA1122471 BLA1122471 BLA1126436 BLA1126436 BLA112694 BLA112694 BLA1148652 BLA1148652 BLA1149504 BLA1149504 BLA1155256 BLA1155256 BLA1222348 BLA1222348 BLA1225575 BLA1225575 BLA1227779 BLA1227779 BLA124071 BLA124071 BLA1247411 BLA1247411 BLA1258830 BLA1258830 BLA1281625 BLA1281625 BLA128165 BLA128165 BLA1305848 BLA1305848 BLA1307751 BLA1307751 BLA130926 BLA130926 BLA1334270 BLA1334270 BLA1335674 BLA1335674 BLA1340102 BLA1340102 BLA1345080 BLA1345080 BLA1353398 BLA1353398 BLA1369834 BLA1369834 BLA1424977 BLA1424977 BLA142669 BLA142669 BLA1429760 BLA1429760 BLA1443592 BLA1443592 BLA1447085 BLA1447085 BLA1453639 BLA1453639 BLA1456355 BLA1456355 BLA1458745 BLA1458745 BLA148663 BLA148663 BLA1494220 BLA1494220 BLA1522880 BLA1522880 BLA152562 BLA152562 BLA1548628 BLA1548628 BLA182644 BLA182644 BLA186707 BLA186707 BLA188073 BLA188073 BLA199127 BLA199127 BLA207148 BLA207148 BLA209939 BLA209939 BLA232119 BLA232119 BLA240725 BLA240725 BLA267275 BLA267275 BLA303829 BLA303829 BLA346044 BLA346044 BLA348268 BLA348268 BLA367863 BLA367863 BLA386897 BLA386897 BLA390799 BLA390799 BLA415724 BLA415724 BLA425291 BLA425291 BLA427093 BLA427093 BLA462355 BLA462355 BLA46821 BLA46821 BLA477637 BLA477637 BLA501111 BLA501111 BLA522126 BLA522126 BLA530906 BLA530906 BLA576525 BLA576525 BLA639361 BLA639361 BLA644530 BLA644530 BLA65754 BLA65754 BLA670003 BLA670003 BLA687525 BLA687525 BLA709367 BLA709367 BLA714281 BLA714281 BLA719328 BLA719328 BLA736457 BLA736457 BLA759669 BLA759669 BLA76056 BLA76056 BLA767666 BLA767666 BLA780853 BLA780853 BLA793029 BLA793029 BLA796639 BLA796639 BLA803743 BLA803743 BLA815298 BLA815298 BLA828243 BLA828243 BLA864030 BLA864030 BLA864110 BLA864110 BLA882892 BLA882892 BLA885665 BLA885665 BLA897478 BLA897478 BLA903861 BLA903861 BLA91627 BLA91627 BLA925523 BLA925523 BLA928243 BLA928243 BLA932395 BLA932395 BLA935341 BLA935341 BLA940705 BLA940705 BLA966386 BLA966386 A message is displayed indicating 293 duplicate values were removed and 307 unique values remain. 7 duplicates are not removed: 1280601_77 1525893_483 455968_177 469495_157 641583_74 704267_63 709992_199 The initial list consist of Excel-ordered values of: numbers between 40,000 and 1,500,000 numbers in the same range with and added _ and a 1-999 suffix numbers in the same range prefixed with BLA Each value is present exactly twice. There are no formula involved. I already checked for extra spaces. Immediately reapplying the Remove Duplicates command does not remove the remaining duplicate values. Reapplying the Remove Duplicates command after a reverse-sort seems to remove the remaining duplicates. Reverse-sorting before using the first Remove Duplicates changes the result (213 duplicate values removed and 387 unique values remained), only in the non-suffixed non-prefixed group. Can you reproduce this bug in Excel 2007 like I do? What triggers it? Could it the composite (numbers and text) data set? I'm not looking for alternatives. I want to know when not to use the feature and revert to already known (Excel 2003 era) alternatives.

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  • e2fsck extremely slow, although enough memory exists

    - by kaefert
    I've got this external USB-Disk: kaefert@blechmobil:~$ lsusb -s 2:3 Bus 002 Device 003: ID 0bc2:3320 Seagate RSS LLC As can be seen in this dmesg output, there is some problem that prevents that disk from beeing mounted: kaefert@blechmobil:~$ dmesg ... [ 113.084079] usb 2-1: new high-speed USB device number 3 using ehci_hcd [ 113.217783] usb 2-1: New USB device found, idVendor=0bc2, idProduct=3320 [ 113.217787] usb 2-1: New USB device strings: Mfr=2, Product=3, SerialNumber=1 [ 113.217790] usb 2-1: Product: Expansion Desk [ 113.217792] usb 2-1: Manufacturer: Seagate [ 113.217794] usb 2-1: SerialNumber: NA4J4N6K [ 113.435404] usbcore: registered new interface driver uas [ 113.455315] Initializing USB Mass Storage driver... [ 113.468051] scsi5 : usb-storage 2-1:1.0 [ 113.468180] usbcore: registered new interface driver usb-storage [ 113.468182] USB Mass Storage support registered. [ 114.473105] scsi 5:0:0:0: Direct-Access Seagate Expansion Desk 070B PQ: 0 ANSI: 6 [ 114.474342] sd 5:0:0:0: [sdb] 732566645 4096-byte logical blocks: (3.00 TB/2.72 TiB) [ 114.475089] sd 5:0:0:0: [sdb] Write Protect is off [ 114.475092] sd 5:0:0:0: [sdb] Mode Sense: 43 00 00 00 [ 114.475959] sd 5:0:0:0: [sdb] Write cache: enabled, read cache: enabled, doesn't support DPO or FUA [ 114.477093] sd 5:0:0:0: [sdb] 732566645 4096-byte logical blocks: (3.00 TB/2.72 TiB) [ 114.501649] sdb: sdb1 [ 114.502717] sd 5:0:0:0: [sdb] 732566645 4096-byte logical blocks: (3.00 TB/2.72 TiB) [ 114.504354] sd 5:0:0:0: [sdb] Attached SCSI disk [ 116.804408] EXT4-fs (sdb1): ext4_check_descriptors: Checksum for group 3976 failed (47397!=61519) [ 116.804413] EXT4-fs (sdb1): group descriptors corrupted! ... So I went and fired up my favorite partition manager - gparted, and told it to verify and repair the partition sdb1. This made gparted call e2fsck (version 1.42.4 (12-Jun-2012)) e2fsck -f -y -v /dev/sdb1 Although gparted called e2fsck with the "-v" option, sadly it doesn't show me the output of my e2fsck process (bugreport https://bugzilla.gnome.org/show_bug.cgi?id=467925 ) I started this whole thing on Sunday (2012-11-04_2200) evening, so about 48 hours ago, this is what htop says about it now (2012-11-06-1900): PID USER PRI NI VIRT RES SHR S CPU% MEM% TIME+ Command 3704 root 39 19 1560M 1166M 768 R 98.0 19.5 42h56:43 e2fsck -f -y -v /dev/sdb1 Now I found a few posts on the internet that discuss e2fsck running slow, for example: http://gparted-forum.surf4.info/viewtopic.php?id=13613 where they write that its a good idea to see if the disk is just that slow because maybe its damaged, and I think these outputs tell me that this is not the case in my case: kaefert@blechmobil:~$ sudo hdparm -tT /dev/sdb /dev/sdb: Timing cached reads: 3562 MB in 2.00 seconds = 1783.29 MB/sec Timing buffered disk reads: 82 MB in 3.01 seconds = 27.26 MB/sec kaefert@blechmobil:~$ sudo hdparm /dev/sdb /dev/sdb: multcount = 0 (off) readonly = 0 (off) readahead = 256 (on) geometry = 364801/255/63, sectors = 5860533160, start = 0 However, although I can read quickly from that disk, this disk speed doesn't seem to be used by e2fsck, considering tools like gkrellm or iotop or this: kaefert@blechmobil:~$ iostat -x Linux 3.2.0-2-amd64 (blechmobil) 2012-11-06 _x86_64_ (2 CPU) avg-cpu: %user %nice %system %iowait %steal %idle 14,24 47,81 14,63 0,95 0,00 22,37 Device: rrqm/s wrqm/s r/s w/s rkB/s wkB/s avgrq-sz avgqu-sz await r_await w_await svctm %util sda 0,59 8,29 2,42 5,14 43,17 160,17 53,75 0,30 39,80 8,72 54,42 3,95 2,99 sdb 137,54 5,48 9,23 0,20 587,07 22,73 129,35 0,07 7,70 7,51 16,18 2,17 2,04 Now I researched a little bit on how to find out what e2fsck is doing with all that processor time, and I found the tool strace, which gives me this: kaefert@blechmobil:~$ sudo strace -p3704 lseek(4, 41026998272, SEEK_SET) = 41026998272 write(4, "\212\354K[_\361\3nl\212\245\352\255jR\303\354\312Yv\334p\253r\217\265\3567\325\257\3766"..., 4096) = 4096 lseek(4, 48404766720, SEEK_SET) = 48404766720 read(4, "\7t\260\366\346\337\304\210\33\267j\35\377'\31f\372\252\ffU\317.y\211\360\36\240c\30`\34"..., 4096) = 4096 lseek(4, 41027002368, SEEK_SET) = 41027002368 write(4, "\232]7Ws\321\352\t\1@[+5\263\334\276{\343zZx\352\21\316`1\271[\202\350R`"..., 4096) = 4096 lseek(4, 48404770816, SEEK_SET) = 48404770816 read(4, "\17\362r\230\327\25\346//\210H\v\311\3237\323K\304\306\361a\223\311\324\272?\213\tq \370\24"..., 4096) = 4096 lseek(4, 41027006464, SEEK_SET) = 41027006464 write(4, "\367yy>x\216?=\324Z\305\351\376&\25\244\210\271\22\306}\276\237\370(\214\205G\262\360\257#"..., 4096) = 4096 lseek(4, 48404774912, SEEK_SET) = 48404774912 read(4, "\365\25\0\21|T\0\21}3t_\272\373\222k\r\177\303\1\201\261\221$\261B\232\3142\21U\316"..., 4096) = 4096 ^CProcess 3704 detached around 16 of these lines every second, so 4 read and 4 write operations every second, which I don't consider to be a lot.. And finally, my question: Will this process ever finish? If those numbers from fseek (48404774912) represent bytes, that would be something like 45 gigabytes, with this beeing a 3 terrabyte disk, which would give me 134 days to go, if the speed stays constant, and e2fsck scans the disk like this completly and only once. Do you have some advice for me? I have most of the data on that disk elsewhere, but I've put a lot of hours into sorting and merging it to this disk, so I would prefer to getting this disk up and running again, without formatting it anew. I don't think that the hardware is damaged since the disk is only a few months and since I can't see any I/O errors in the dmesg output. UPDATE: I just looked at the strace output again (2012-11-06_2300), now it looks like this: lseek(4, 1419860611072, SEEK_SET) = 1419860611072 read(4, "3#\f\2447\335\0\22A\355\374\276j\204'\207|\217V|\23\245[\7VP\251\242\276\207\317:"..., 4096) = 4096 lseek(4, 43018145792, SEEK_SET) = 43018145792 write(4, "]\206\231\342Y\204-2I\362\242\344\6R\205\361\324\177\265\317C\334V\324\260\334\275t=\10F."..., 4096) = 4096 lseek(4, 1419860615168, SEEK_SET) = 1419860615168 read(4, "\262\305\314Y\367\37x\326\245\226\226\320N\333$s\34\204\311\222\7\315\236\336\300TK\337\264\236\211n"..., 4096) = 4096 lseek(4, 43018149888, SEEK_SET) = 43018149888 write(4, "\271\224m\311\224\25!I\376\16;\377\0\223H\25Yd\201Y\342\r\203\271\24eG<\202{\373V"..., 4096) = 4096 lseek(4, 1419860619264, SEEK_SET) = 1419860619264 read(4, ";d\360\177\n\346\253\210\222|\250\352T\335M\33\260\320\261\7g\222P\344H?t\240\20\2548\310"..., 4096) = 4096 lseek(4, 43018153984, SEEK_SET) = 43018153984 write(4, "\360\252j\317\310\251G\227\335{\214`\341\267\31Y\202\360\v\374\307oq\3063\217Z\223\313\36D\211"..., 4096) = 4096 So the numbers in the lseek lines before the reads, like 1419860619264 are already a lot bigger, standing for 1.29 terabytes if those numbers are bytes, so it doesn't seem to be a linear progress on a big scale, maybe there are only some areas that need work, that have big gaps in between them. UPDATE2: Okey, big disappointment, the numbers are back to very small again (2012-11-07_0720) lseek(4, 52174548992, SEEK_SET) = 52174548992 read(4, "\374\312\22\\\325\215\213\23\0357U\222\246\370v^f(\312|f\212\362\343\375\373\342\4\204mU6"..., 4096) = 4096 lseek(4, 46603526144, SEEK_SET) = 46603526144 write(4, "\370\261\223\227\23?\4\4\217\264\320_Am\246CQ\313^\203U\253\274\204\277\2564n\227\177\267\343"..., 4096) = 4096 so either e2fsck goes over the data multiple times, or it just hops back and forth multiple times. Or my assumption that those numbers are bytes is wrong. UPDATE3: Since it's mentioned here http://forums.fedoraforum.org/showthread.php?t=282125&page=2 that you can testisk while e2fsck is running, i tried that, though not with a lot of success. When asking testdisk to display the data of my partition, this is what I get: TestDisk 6.13, Data Recovery Utility, November 2011 Christophe GRENIER <[email protected]> http://www.cgsecurity.org 1 P Linux 0 4 5 45600 40 8 732566272 Can't open filesystem. Filesystem seems damaged. And this is what strace currently gives me (2012-11-07_1030) lseek(4, 212460343296, SEEK_SET) = 212460343296 read(4, "\315Mb\265v\377Gn \24\f\205EHh\2349~\330\273\203\3375\206\10\r3=W\210\372\352"..., 4096) = 4096 lseek(4, 47347830784, SEEK_SET) = 47347830784 write(4, "]\204\223\300I\357\4\26\33+\243\312G\230\250\371*m2U\t_\215\265J \252\342Pm\360D"..., 4096) = 4096 (times are in CET)

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  • More Fun with C# Iterators and Generators

    - by James Michael Hare
    In my last post, I talked quite a bit about iterators and how they can be really powerful tools for filtering a list of items down to a subset of items.  This had both pros and cons over returning a full collection, which, in summary, were:   Pros: If traversal is only partial, does not have to visit rest of collection. If evaluation method is costly, only incurs that cost on elements visited. Adds little to no garbage collection pressure.    Cons: Very slight performance impact if you know caller will always consume all items in collection. And as we saw in the last post, that con for the cost was very, very small and only really became evident on very tight loops consuming very large lists completely.    One of the key items to note, though, is the garbage!  In the traditional (return a new collection) method, if you have a 1,000,000 element collection, and wish to transform or filter it in some way, you have to allocate space for that copy of the collection.  That is, say you have a collection of 1,000,000 items and you want to double every item in the collection.  Well, that means you have to allocate a collection to hold those 1,000,000 items to return, which is a lot especially if you are just going to use it once and toss it.   Iterators, though, don't have this problem.  Each time you visit the node, it would return the doubled value of the node (in this example) and not allocate a second collection of 1,000,000 doubled items.  Do you see the distinction?  In both cases, we're consuming 1,000,000 items.  But in one case we pass back each doubled item which is just an int (for example's sake) on the stack and in the other case, we allocate a list containing 1,000,000 items which then must be garbage collected.   So iterators in C# are pretty cool, eh?  Well, here's one more thing a C# iterator can do that a traditional "return a new collection" transformation can't!   It can return **unbounded** collections!   I know, I know, that smells a lot like an infinite loop, eh?  Yes and no.  Basically, you're relying on the caller to put the bounds on the list, and as long as the caller doesn't you keep going.  Consider this example:   public static class Fibonacci {     // returns the infinite fibonacci sequence     public static IEnumerable<int> Sequence()     {         int iteration = 0;         int first = 1;         int second = 1;         int current = 0;         while (true)         {             if (iteration++ < 2)             {                 current = 1;             }             else             {                 current = first + second;                 second = first;                 first = current;             }             yield return current;         }     } }   Whoa, you say!  Yes, that's an infinite loop!  What the heck is going on there?  Yes, that was intentional.  Would it be better to have a fibonacci sequence that returns only a specific number of items?  Perhaps, but that wouldn't give you the power to defer the execution to the caller.   The beauty of this function is it is as infinite as the sequence itself!  The fibonacci sequence is unbounded, and so is this method.  It will continue to return fibonacci numbers for as long as you ask for them.  Now that's not something you can do with a traditional method that would return a collection of ints representing each number.  In that case you would eventually run out of memory as you got to higher and higher numbers.  This method, though, never runs out of memory.   Now, that said, you do have to know when you use it that it is an infinite collection and bound it appropriately.  Fortunately, Linq provides a lot of these extension methods for you!   Let's say you only want the first 10 fibonacci numbers:       foreach(var fib in Fibonacci.Sequence().Take(10))     {         Console.WriteLine(fib);     }   Or let's say you only want the fibonacci numbers that are less than 100:       foreach(var fib in Fibonacci.Sequence().TakeWhile(f => f < 100))     {         Console.WriteLine(fib);     }   So, you see, one of the nice things about iterators is their power to work with virtually any size (even infinite) collections without adding the garbage collection overhead of making new collections.    You can also do fun things like this to make a more "fluent" interface for for loops:   // A set of integer generator extension methods public static class IntExtensions {     // Begins counting to inifity, use To() to range this.     public static IEnumerable<int> Every(this int start)     {         // deliberately avoiding condition because keeps going         // to infinity for as long as values are pulled.         for (var i = start; ; ++i)         {             yield return i;         }     }     // Begins counting to infinity by the given step value, use To() to     public static IEnumerable<int> Every(this int start, int byEvery)     {         // deliberately avoiding condition because keeps going         // to infinity for as long as values are pulled.         for (var i = start; ; i += byEvery)         {             yield return i;         }     }     // Begins counting to inifity, use To() to range this.     public static IEnumerable<int> To(this int start, int end)     {         for (var i = start; i <= end; ++i)         {             yield return i;         }     }     // Ranges the count by specifying the upper range of the count.     public static IEnumerable<int> To(this IEnumerable<int> collection, int end)     {         return collection.TakeWhile(item => item <= end);     } }   Note that there are two versions of each method.  One that starts with an int and one that starts with an IEnumerable<int>.  This is to allow more power in chaining from either an existing collection or from an int.  This lets you do things like:   // count from 1 to 30 foreach(var i in 1.To(30)) {     Console.WriteLine(i); }     // count from 1 to 10 by 2s foreach(var i in 0.Every(2).To(10)) {     Console.WriteLine(i); }     // or, if you want an infinite sequence counting by 5s until something inside breaks you out... foreach(var i in 0.Every(5)) {     if (someCondition)     {         break;     }     ... }     Yes, those are kinda play functions and not particularly useful, but they show some of the power of generators and extension methods to form a fluid interface.   So what do you think?  What are some of your favorite generators and iterators?

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  • PASS: The Budget Process

    - by Bill Graziano
    Every fiscal year PASS creates a detailed budget.  This helps us set priorities and communicate to our members what we’re going to do in the upcoming year.  You can review the current budget on the PASS Governance page.  That page currently requires you to login but I’m talking with HQ to see if there are any legal issues with opening that up. The Accounting Team The PASS accounting team is two people.  The Executive Vice-President of Finance (“EVP”) and the PASS Accounting Manager.  Sandy Cherry is the accounting manager and works at PASS HQ.  Sandy has been with PASS since we switched management companies in 2007.  Throughout this document when I talk about any actual work related to the budget that’s all Sandy :)  She’s the glue that gets us through this process.  Last year we went through 32 iterations of the budget before the Board approved so it’s a pretty busy time for her us – well, mostly her. Fiscal Year The PASS fiscal year runs from July 1st through June 30th the following year.  Right now we’re in fiscal year 2011.  Our 2010 Summit actually occurred in FY2011.  We switched to this schedule from a calendar year in 2006.  Our goal was to have the Summit occur early in our fiscal year.  That gives us the rest of the year to handle any significant financial impact from the Summit.  If registrations are down we can reduce spending.  If registrations are up we can decide how much to increase our reserves and how much to spend.  Keep in mind that the Summit is budgeted to generate 82% of our revenue this year.  How it performs has a significant impact on our financials.  The other benefit of this fiscal year is that it matches the Microsoft fiscal year.  We sign an annual sponsorship agreement with Microsoft and it’s very helpful that our fiscal years match. This year our budget process will probably start in earnest in March or April.  I’d like to be done in early June so we can publish before July 1st.  I was late publishing it this year and I’m trying not to repeat that. Our Budget Our actual budget is an Excel spreadsheet with 36 sheets.  We remove some of those when we publish it since they include salary information.  The budget is broken up into various portfolios or departments.  We have 20 portfolios.  They include chapters, marketing, virtual chapters, marketing, etc.  Ideally each portfolio is assigned to a Board member.  Each portfolio also typically has a staff person assigned to it.  Portfolios that aren’t assigned to a Board member are monitored by HQ and the ExecVP-Finance (me).  These are typically smaller portfolios such as deferred membership or Summit futures.  (More on those in a later post.)  All portfolios are reviewed by all Board members during the budget approval process, when interim financials are released internally and at year-end. The Process Our first step is to budget revenues.  The Board determines a target attendee number.  We have formulas based on historical performance that convert that to an overall attendee revenue number.  Other revenue projections (such as vendor sponsorships) come from different parts of the organization.  I hope to have another post with more details on how we project revenues. The next step is to budget expenses.  Board members fill out a sample spreadsheet with their budget for the year.  They can add line items and notes describing what the amounts are for.  Each Board portfolio typically has from 10 to 30 line items.  Any new initiatives they want to pursue needs to be budgeted.  The Summit operations budget is managed by HQ.  It includes the cost for food, electrical, internet, etc.  Most of these come from our estimate of attendees and our contract with the convention center.  During this process the Board can ask for more or less to be spent on various line items.  For example, if we weren’t happy with the Internet at the last Summit we can ask them to look into different options and/or increasing the budget.  HQ will also make adjustments to these numbers based on what they see at the events and the feedback we receive on the surveys. After we have all the initial estimates we start reviewing the entire budget.  It is sent out to the Board and we can see what each portfolio requested and what the overall profit and loss number is.  We usually start with too much in expenses and need to cut.  In years past the Board started haggling over these numbers as a group.  This past year they decided I should take a first cut and present them with a reasonable budget and a list of what I changed.  That worked well and I think we’ll continue to do that in the future. We go through a number of iterations on the budget.  If I remember correctly, we went through 32 iterations before we passed the budget.  At each iteration various revenue and expense numbers can change.  Keep in mind that the PASS budget has 200+ line items spread over 20 portfolios.  Many of these depend on other numbers.  For example, if we decide increase the projected attendees that cascades through our budget.  At each iteration we list what changed and the impact.  Ideally these discussions will take place at a face-to-face Board meeting.  Many of them also take place over the phone.  Board members explain any increase they are asking for while performing due diligence on other budget requests.  Eventually a budget emerges and is passed. Publishing After the budget is passed we create a version without the formulas and salaries for posting on the web site.  Sandy also creates some charts to help our members understand the budget.  The EVP writes a nice little letter describing some of the changes from last year’s budget.  You can see my letter and our budget on the PASS Governance page. And then, eight months later, we start all over again.

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  • c++ and c# speed compared

    - by Mack
    I was worried about C#'s speed when it deals with heavy calculations, when you need to use raw CPU power. I always thought that C++ is much faster than C# when it comes to calculations. So I did some quick tests. The first test computes prime numbers < an integer n, the second test computes some pandigital numbers. The idea for second test comes from here: Pandigital Numbers C# prime computation: using System; using System.Diagnostics; class Program { static int primes(int n) { uint i, j; int countprimes = 0; for (i = 1; i <= n; i++) { bool isprime = true; for (j = 2; j <= Math.Sqrt(i); j++) if ((i % j) == 0) { isprime = false; break; } if (isprime) countprimes++; } return countprimes; } static void Main(string[] args) { int n = int.Parse(Console.ReadLine()); Stopwatch sw = new Stopwatch(); sw.Start(); int res = primes(n); sw.Stop(); Console.WriteLine("I found {0} prime numbers between 0 and {1} in {2} msecs.", res, n, sw.ElapsedMilliseconds); Console.ReadKey(); } } C++ variant: #include <iostream> #include <ctime> int primes(unsigned long n) { unsigned long i, j; int countprimes = 0; for(i = 1; i <= n; i++) { int isprime = 1; for(j = 2; j < (i^(1/2)); j++) if(!(i%j)) { isprime = 0; break; } countprimes+= isprime; } return countprimes; } int main() { int n, res; cin>>n; unsigned int start = clock(); res = primes(n); int tprime = clock() - start; cout<<"\nI found "<<res<<" prime numbers between 1 and "<<n<<" in "<<tprime<<" msecs."; return 0; } When I ran the test trying to find primes < than 100,000, C# variant finished in 0.409 seconds and C++ variant in 5.553 seconds. When I ran them for 1,000,000 C# finished in 6.039 seconds and C++ in about 337 seconds. Pandigital test in C#: using System; using System.Diagnostics; class Program { static bool IsPandigital(int n) { int digits = 0; int count = 0; int tmp; for (; n > 0; n /= 10, ++count) { if ((tmp = digits) == (digits |= 1 << (n - ((n / 10) * 10) - 1))) return false; } return digits == (1 << count) - 1; } static void Main() { int pans = 0; Stopwatch sw = new Stopwatch(); sw.Start(); for (int i = 1; i <= 123456789; i++) { if (IsPandigital(i)) { pans++; } } sw.Stop(); Console.WriteLine("{0}pcs, {1}ms", pans, sw.ElapsedMilliseconds); Console.ReadKey(); } } Pandigital test in C++: #include <iostream> #include <ctime> using namespace std; int IsPandigital(int n) { int digits = 0; int count = 0; int tmp; for (; n > 0; n /= 10, ++count) { if ((tmp = digits) == (digits |= 1 << (n - ((n / 10) * 10) - 1))) return 0; } return digits == (1 << count) - 1; } int main() { int pans = 0; unsigned int start = clock(); for (int i = 1; i <= 123456789; i++) { if (IsPandigital(i)) { pans++; } } int ptime = clock() - start; cout<<"\nPans:"<<pans<<" time:"<<ptime; return 0; } C# variant runs in 29.906 seconds and C++ in about 36.298 seconds. I didn't touch any compiler switches and bot C# and C++ programs were compiled with debug options. Before I attempted to run the test I was worried that C# will lag well behind C++, but now it seems that there is a pretty big speed difference in C# favor. Can anybody explain this? C# is jitted and C++ is compiled native so it's normal that a C++ will be faster than a C# variant. Thanks for the answers!

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  • C++ Program performs better when piped

    - by ET1 Nerd
    I haven't done any programming in a decade. I wanted to get back into it, so I made this little pointless program as practice. The easiest way to describe what it does is with output of my --help codeblock: ./prng_bench --help ./prng_bench: usage: ./prng_bench $N $B [$T] This program will generate an N digit base(B) random number until all N digits are the same. Once a repeating N digit base(B) number is found, the following statistics are displayed: -Decimal value of all N digits. -Time & number of tries taken to randomly find. Optionally, this process is repeated T times. When running multiple repititions, averages for all N digit base(B) numbers are displayed at the end, as well as total time and total tries. My "problem" is that when the problem is "easy", say a 3 digit base 10 number, and I have it do a large number of passes the "total time" is less when piped to grep. ie: command ; command |grep took : ./prng_bench 3 10 999999 ; ./prng_bench 3 10 999999|grep took .... Pass# 999999: All 3 base(10) digits = 3 base(10). Time: 0.00005 secs. Tries: 23 It took 191.86701 secs & 99947208 tries to find 999999 repeating 3 digit base(10) numbers. An average of 0.00019 secs & 99 tries was needed to find each one. It took 159.32355 secs & 99947208 tries to find 999999 repeating 3 digit base(10) numbers. If I run the same command many times w/o grep time is always VERY close. I'm using srand(1234) for now, to test. The code between my calls to clock_gettime() for start and stop do not involve any stream manipulation, which would obviously affect time. I realize this is an exercise in futility, but I'd like to know why it behaves this way. Below is heart of the program. Here's a link to the full source on DB if anybody wants to compile and test. https://www.dropbox.com/s/6olqnnjf3unkm2m/prng_bench.cpp clock_gettime() requires -lrt. for (int pass_num=1; pass_num<=passes; pass_num++) { //Executes $passes # of times. clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &temp_time); //get time start_time = timetodouble(temp_time); //convert time to double, store as start_time for(i=1, tries=0; i!=0; tries++) { //loops until 'comparison for' fully completes. counts reps as 'tries'. <------------ for (i=0; i<Ndigits; i++) //Move forward through array. | results[i]=(rand()%base); //assign random num of base to element (digit). | /*for (i=0; i<Ndigits; i++) //---Debug Lines--------------- | std::cout<<" "<<results[i]; //---a LOT of output.---------- | std::cout << "\n"; //---Comment/decoment to disable/enable.*/ // | for (i=Ndigits-1; i>0 && results[i]==results[0]; i--); //Move through array, != element breaks & i!=0, new digits drawn. -| } //If all are equal i will be 0, nested for condition satisfied. -| clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &temp_time); //get time draw_time = (timetodouble(temp_time) - start_time); //convert time to dbl, subtract start_time, set draw_time to diff. total_time += draw_time; //add time for this pass to total. total_tries += tries; //add tries for this pass to total. /*Formated output for each pass: Pass# ---: All -- base(--) digits = -- base(10) Time: ----.---- secs. Tries: ----- (LINE) */ std::cout<<"Pass# "<<std::setw(width_pass)<<pass_num<<": All "<<Ndigits<<" base("<<base<<") digits = " <<std::setw(width_base)<<results[0]<<" base(10). Time: "<<std::setw(width_time)<<draw_time <<" secs. Tries: "<<tries<<"\n"; } if(passes==1) return 0; //No need for totals and averages of 1 pass. /* It took ----.---- secs & ------ tries to find --- repeating -- digit base(--) numbers. (LINE) An average of ---.---- secs & ---- tries was needed to find each one. (LINE)(LINE) */ std::cout<<"It took "<<total_time<<" secs & "<<total_tries<<" tries to find " <<passes<<" repeating "<<Ndigits<<" digit base("<<base<<") numbers.\n" <<"An average of "<<total_time/passes<<" secs & "<<total_tries/passes <<" tries was needed to find each one. \n\n"; return 0;

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  • Redaction in AutoVue

    - by [email protected]
    As the trend to digitize all paper assets continues, so does the push to digitize all the processes around these assets. One such process is redaction - removing sensitive or classified information from documents. While for some this may conjure up thoughts of old CIA documents filled with nothing but blacked out pages, there are actually many uses for redaction today beyond military and government. Many companies have a need to remove names, phone numbers, social security numbers, credit card numbers, etc. from documents that are being scanned in and/or released to the public or less privileged users - insurance companies, banks and legal firms are a few examples. The process of digital redaction actually isn't that far from the old paper method: Step 1. Find a folder with a big red stamp on it labeled "TOP SECRET" Step 2. Make a copy of that document, since some folks still need to access the original contents Step 3. Black out the text or pages you want to hide Step 4. Release or distribute this new 'redacted' copy So where does a solution like AutoVue come in? Well, we've really been doing all of these things for years! 1. With AutoVue's VueLink integration and iSDK, we can integrate to virtually any content management system and view documents of almost any format with a single click. Finding the document and opening it in AutoVue: CHECK! 2. With AutoVue's markup capabilities, adding filled boxes (or other shapes) around certain text is a no-brainer. You can even leverage AutoVue's powerful APIs to automate the addition of markups over certain text or pre-defined regions using our APIs. Black out the text you want to hide: CHECK! 3. With AutoVue's conversion capabilities, you can 'burn-in' the comments into a new file, either as a TIFF, JPEG or PDF document. Burning-in the redactions avoids slip-ups like the recent (well-publicized) TSA one. Through our tight integrations, the newly created copies can be directly checked into the content management system with no manual intervention. Make a copy of that document: CHECK! 4. Again, leveraging AutoVue's integrations, we can now define rules in the system based on a user's privileges. An 'authorized' user wishing to view the document from the repository will get exactly that - no redactions. An 'unauthorized' user, when requesting to view that same document, can get redirected to open the redacted copy of the same document. Release or distribute the new 'redacted' copy: CHECK! See this movie (WMV format, 2mins, 20secs, no audio) for a quick illustration of AutoVue's redaction capabilities. It shows how redactions can be added based on text searches, manual input or pre-defined templates/regions. Let us know what you think in the comments. And remember - this is all in our flagship AutoVue product - no additional software required!

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  • Grouping data in LINQ with the help of group keyword

    - by vik20000in
    While working with any kind of advanced query grouping is a very important factor. Grouping helps in executing special function like sum, max average etc to be performed on certain groups of data inside the date result set. Grouping is done with the help of the Group method. Below is an example of the basic group functionality.     int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };         var numberGroups =         from num in numbers         group num by num % 5 into numGroup         select new { Remainder = numGroup.Key, Numbers = numGroup };  In the above example we have grouped the values based on the reminder left over when divided by 5. First we are grouping the values based on the reminder when divided by 5 into the numgroup variable.  numGroup.Key gives the value of the key on which the grouping has been applied. And the numGroup itself contains all the records that are contained in that group. Below is another example to explain the same. string[] words = { "blueberry", "abacus", "banana", "apple", "cheese" };         var wordGroups =         from num in words         group num by num[0] into grp         select new { FirstLetter = grp.Key, Words = grp }; In the above example we are grouping the value with the first character of the string (num[0]). Just like the order operator the group by clause also allows us to write our own logic for the Equal comparison (That means we can group Item by ignoring case also by writing out own implementation). For this we need to pass an object that implements the IEqualityComparer<string> interface. Below is an example. public class AnagramEqualityComparer : IEqualityComparer<string> {     public bool Equals(string x, string y) {         return getCanonicalString(x) == getCanonicalString(y);     }      public int GetHashCode(string obj) {         return getCanonicalString(obj).GetHashCode();     }         private string getCanonicalString(string word) {         char[] wordChars = word.ToCharArray();         Array.Sort<char>(wordChars);         return new string(wordChars);     } }  string[] anagrams = {"from   ", " salt", " earn", "  last   ", " near "}; var orderGroups = anagrams.GroupBy(w => w.Trim(), new AnagramEqualityComparer()); Vikram  

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  • New January 2013 Release of the Ajax Control Toolkit

    - by Stephen.Walther
    I am super excited to announce the January 2013 release of the Ajax Control Toolkit! I have one word to describe this release and that word is “Charts” – we’ve added lots of great new chart controls to the Ajax Control Toolkit. You can download the new release directly from http://AjaxControlToolkit.CodePlex.com – or, just fire the following command from the Visual Studio Library Package Manager Console Window (NuGet): Install-Package AjaxControlToolkit You also can view the new chart controls by visiting the “live” Ajax Control Toolkit Sample Site. 5 New Ajax Control Toolkit Chart Controls The Ajax Control Toolkit contains five new chart controls: the AreaChart, BarChart, BubbleChart, LineChart, and PieChart controls. Here is a sample of each of the controls: AreaChart: BarChart: BubbleChart: LineChart: PieChart: We realize that people love to customize the appearance of their charts so all of the chart controls include properties such as color properties. The chart controls render the chart on the browser using SVG. The chart controls are compatible with any browser which supports SVG including Internet Explorer 9 and new and recent versions of Google Chrome, Mozilla Firefox, and Apple Safari. (If you attempt to display a chart on a browser which does not support SVG then you won’t get an error – you just won’t get anything). Updates to the HTML Sanitizer If you are using the HtmlEditorExtender on a public-facing website then it is really important that you enable the HTML Sanitizer to prevent Cross-Site Scripting (XSS) attacks. The HtmlEditorExtender uses the HTML Sanitizer by default. The HTML Sanitizer strips out any suspicious content (like JavaScript code and CSS expressions) from the HTML submitted with the HtmlEditorExtender. We followed the recommendations of OWASP and ha.ckers.org to identify suspicious content. We updated the HTML Sanitizer with this release to protect against new types of XSS attacks. The HTML Sanitizer now has over 220 unit tests. The Ajax Control Toolkit team would like to thank Gil Cohen who helped us identify and block additional XSS attacks. Change in Ajax Control Toolkit Version Format We ran out of numbers. The Ajax Control Toolkit was first released way back in 2006. In previous releases, the version of the Ajax Control Toolkit followed the format: Release Year + Date. So, the previous release was 60919 where 6 represented the 6th release year and 0919 represent September 19. Unfortunately, the AssembyVersion attribute uses a UInt16 data type which has a maximum size of 65,534. The number 70123 is bigger than 65,534 so we had to change our version format with this release. Fortunately, the AssemblyVersion attribute actually accepts four UInt16 numbers so we used another one. This release of the Ajax Control Toolkit is officially version 7.0123. This new version format should work for another 65,000 years. And yes, I realize that 7.0123 is less than 60,919, but we ran out of numbers. Summary I hope that you find the chart controls included with this latest release of the Ajax Control Toolkit useful. Let me know if you use them in applications that you build. And, let me know if you run into any issues using the new chart controls. Next month, back to improving the File Upload control – more exciting stuff.

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  • SQLAuthority News – Monthly list of Puzzles and Solutions on SQLAuthority.com

    - by pinaldave
    This month has been very interesting month for SQLAuthority.com we had multiple and various puzzles which everybody participated and lots of interesting conversation which we have shared. Let us start in latest puzzles and continue going down. There are few answers also posted on facebook as well. SQL SERVER – Puzzle Involving NULL – Resolve – Error – Operand data type void type is invalid for sum operator This puzzle involves NULL and throws an error. The challenge is to resolve the error. There are multiple ways to resolve this error. Readers has contributed various methods. Few of them even have supplied the answer why this error is showing up. NULL are very important part of the database and if one of the column has NULL the result can be totally different than the one expected. SQL SERVER – T-SQL Scripts to Find Maximum between Two Numbers I modified script provided by friend to find greatest number between two number. My script has small bug in it. However, lots of readers have suggested better scripts. Madhivanan has written blog post on the subject over here. SQL SERVER – BI Quiz Hint – Performance Tuning Cubes – Hints This quiz is hosted on my friend Jacob‘s site. I have written many hints how one can tune cubes. Now one can take part here and win exciting prizes. SQL SERVER – Solution – Generating Zero Without using Any Numbers in T-SQL Madhivanan has asked very interesting question on his blog about How to Generate Zero without using Any Numbers in T-SQL. He has demonstrated various methods how one can generate Zero. I asked the same question on blog and got many interesting answers which I have shared. SQL SERVER – Solution – Puzzle – Statistics are not Updated but are Created Once I have to accept that this was most difficult puzzle. In this puzzle I have asked even though settings are correct, why statistics of the tables are not getting updated. In this puzzle one is tested with various concepts 1) Indexes, 2) Statistics, 3) database settings etc. There are multiple ways of solving this puzzles. It was interesting as many took interest but only few got it right. SQL SERVER – Question to You – When to use Function and When to use Stored Procedure This is rather straight forward question and not the typical puzzle. The answers from readers are great however, still there is chance of more detailed answers. SQL SERVER – Selecting Domain from Email Address I wrote on selecting domains from email addresses. Madhivanan makes puzzle out of a simple question. He wrote a follow-up post over here. In his post he writes various way how one can find email addresses from list of domains. Well, this is not a puzzle but amazing Guest Post by Feodor Georgiev who has written on subject Job Interviewing the Right Way (and for the Right Reasons). An article which everyone should read. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, Readers Contribution, Readers Question, SQL, SQL Authority, SQL Puzzle, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, SQLServer, T SQL, Technology

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  • Caesar Cipher Program In C++ [migrated]

    - by xaliap81
    I am trying to write a caesar cipher program in c++. This is my codes template: int chooseKEY (){ //choose key shift from 1-26 } void encrypt (char * w, char *e, int key) { //Encryption function, *w is the text in the beginning, the *e is the encrypted text //Encryption in being made only in letters noy in numbers and punctuations // *e = *w + key } void decrypt (char * e, char *w, int key) { // Decryption function, *e is the text in the beginning, the *w is the decrypted text //Dencryption in being made only in letters no numbers and punctuations // *w = *e - key } void Caesar (char * inputFile, char * outputFile, int key, int mode) { // Read the inputfile which contains some data. If mode ==1 then the data is being //encrypted else if mode == 0 the data is being decrypted and is being written in //the output file } void main() { // call the Caesar function } The program has four functions, chooseKey function have to return an int as a shift key from 1-26. Encrypt function has three parameters, *w is the text in the beginning, *e is the encrypted text and the key is from the choosekey function.For encryption : Only letters have to be encrypted not numbers or punctuation and the letters are counting cyclic. Decrypt function has three parameters *e is the encrypted text, *w is the beginning text and the key. Caesar function has four parameters, inputfile which is the file that contains the beginning text, output file which contains the encrypted text, the key and the mode (if mode==1) encryption, (mode ==0) decryption. This is my sample code: #include <iostream> #include <fstream> using namespace std; int chooseKey() { int key_number; cout << "Give a number from 1-26: "; cin >> key_number; while(key_number<1 || key_number>26) { cout << "Your number have to be from 1-26.Retry: "; cin >> key_number; } return key_number; } void encryption(char *w, char *e, int key){ char *ptemp = w; while(*ptemp){ if(isalpha(*ptemp)){ if(*ptemp>='a'&&*ptemp<='z') { *ptemp-='a'; *ptemp+=key; *ptemp%=26; *ptemp+='A'; } } ptemp++; } w=e; } void decryption (char *e, char *w, int key){ char *ptemp = e; while(*ptemp){ if(isalpha(*ptemp)) { if(*ptemp>='A'&&*ptemp<='Z') { *ptemp-='A'; *ptemp+=26-key; *ptemp%=26; *ptemp+='a'; } } ptemp++; } e=w; } void Caesar (char *inputFile, char *outputFile, int key, int mode) { ifstream input; ofstream output; char buf, buf1; input.open(inputFile); output.open(outputFile); buf=input.get(); while(!input.eof()) { if(mode == 1){ encryption(&buf, &buf1, key); }else{ decryption(&buf1, &buf, key); } output << buf; buf=input.get(); } input.close(); output.close(); } int main(){ int key, mode; key = chooseKey(); cout << "1 or 0: "; cin >> mode; Caesar("test.txt","coded.txt",key,mode); system("pause"); } I am trying to run the code and it is being crashed (Debug Assertion Failed). I think the problem is somewhere inside Caesar function.How to call the encrypt and decrypt functions. But i don't know what exactly is. Any ideas?

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  • Using C# regex to select text based on custom tags

    - by spaceman
    I have a string in c# containing some data i need to extract based on certain conditions. The string contains many tenders in the following form : <TENDER> some words, don't know how many, may contain numbers and things like slashes (/) or whatever <DESCRIPTION> some more words and possibly other things like numbers or whatever describing the tender here </DESCRIPTION> some more words and possibly numbers and weird things </TENDER> This string doesn't contain any nested <TENDER> tags, its flat. The <DESCRIPTION> tags occur only once within the <TENDER> tags. I'm using : <TENDER>(.+?)</TENDER> as the regex to split up the tenders and it works fine. If this is wrong or stupid and you know a better way to write this please let me know as I have discovered I suck at regex. My problem that I now need to only select a tender if its description contains any word in a list of keywords (lets say for now i want to select a tender only if it contains either "concrete" or"brick" in the description). So far the regex I have come up with looks like this, but I don't know what to put in the middle. Also I have a vague suspicion that this might return me some false positives. <TENDER>(.+?)<DESCRIPTION>have no idea what to do here</DESCRIPTION>(.+?)</TENDER> If any of you regex guru's could point me in the right direction I would be most appreciative.

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  • nth ugly number

    - by Anil Katti
    Numbers whose only prime factors are 2, 3 or 5 are called ugly numbers. Example: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... 1 can be considered as 2^0. I am working on finding nth ugly number. Note that these numbers are extremely sparsely distributed as n gets large. I wrote a trivial program that computes if a given number is ugly or not. For n 500 - it became super slow. I tried using memoization - observation: ugly_number * 2, ugly_number * 3, ugly_number * 5 are all ugly. Even with that it is slow. I tried using some properties of log - since that will reduce this problem from multiplication to addition - but, not much luck yet. Thought of sharing this with you all. Any interesting ideas? Using a concept similar to "Sieve of Eratosthenes" (thanks Anon) for (int i(2), uglyCount(0); ; i++) { if (i % 2 == 0) continue; if (i % 3 == 0) continue; if (i % 5 == 0) continue; uglyCount++; if (uglyCount == n - 1) break; } i is the nth ugly number. Even this is pretty slow. I am trying to find 1500th ugly number.

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