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• Apache logs other user read permissions

  - by user2344668

  We have several developers who maintain the system and I want them to easily read the log files in /var/log/httpd without needing root access. I set the read permission for 'other' users but when I run tail on the log files I get permission denied: [root@ourserver httpd]# chmod -R go+r /var/log/httpd [root@ourserver httpd]# ls -la drwxr--r-- 13 root root 4096 Oct 25 03:31 . drwxr-xr-x. 6 root root 4096 Oct 20 03:24 .. drwxr-xr-x 2 root root 4096 Oct 20 03:24 oursite.com drwxr-xr-x 2 root root 4096 Oct 20 03:24 oursite2.com -rw-r--r-- 1 root root 0 May 7 03:46 access_log -rw-r--r-- 1 root root 3446 Oct 24 22:05 error_log [me@ourserver ~]$ tail -f /var/log/httpd/oursite.com/error.log tail: cannot open `/var/log/httpd/oursite/error.log' for reading: Permission denied Maybe I'm missing something on how permissions work but I'm not finding any easy answers on it.

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• linked list elements gone?

  - by Hristo

  I create a linked list dynamically and initialize the first node in main(), and I add to the list every time I spawn a worker process. Before the worker process exits, I print the list. Also, I print the list inside my sigchld signal handler. in main(): head = NULL; tail = NULL; // linked list to keep track of worker process dll_node_t *node; node = (dll_node_t *) malloc(sizeof(dll_node_t)); // initialize list, allocate memory append_node(node); node->pid = mainPID; // the first node is the MAIN process node->type = MAIN; in a fork()'d process: // add to list dll_node_t *node; node = (dll_node_t *) malloc(sizeof(dll_node_t)); append_node(node); node->pid = mmapFileWorkerStats->childPID; node->workerFileName = mmapFileWorkerStats->workerFileName; node->type = WORK; functions: void append_node(dll_node_t *nodeToAppend) { /* * append param node to end of list */ // if the list is empty if (head == NULL) { // create the first/head node head = nodeToAppend; nodeToAppend->prev = NULL; } else { tail->next = nodeToAppend; nodeToAppend->prev = tail; } // fix the tail to point to the new node tail = nodeToAppend; nodeToAppend->next = NULL; } finally... the signal handler: void chld_signalHandler() { dll_node_t *temp1 = head; while (temp1 != NULL) { printf("2. node's pid: %d\n", temp1->pid); temp1 = temp1->next; } int termChildPID = waitpid(-1, NULL, WNOHANG); dll_node_t *temp = head; while (temp != NULL) { if (temp->pid == termChildPID) { printf("found process: %d\n", temp->pid); } temp = temp->next; } return; } Is it true that upon the worker process exiting, the SIGCHLD signal handler is triggered? If so, that would mean that after I print the tree before exiting, the next thing I do is in the signal handler which is print the tree... which would mean i would print the tree twice? But the tree isn't the same. The node I add in the worker process doesn't exist when I print in the signal handler or at the very end of main(). Any idea why? Thanks, Hristo

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• Give the mount point of a path

  - by Charles Stewart

  The following, very non-robust shell code will give the mount point of $path: (for i in $(df|cut -c 63-99); do case $path in $i*) echo $i;; esac; done) | tail -n 1 Is there a better way to do this? Postscript This script is really awful, but has the redeeming quality that it Works On My Systems. Note that several mount points may be prefixes of $path. Examples On a Linux system: cas@txtproof:~$ path=/sys/block/hda1 cas@txtproof:~$ for i in $(df -a|cut -c 57-99); do case $path in $i*) echo $i;; esac; done| tail -1 /sys On a Mac osx system cas local$ path=/dev/fd/0 cas local$ for i in $(df -a|cut -c 63-99); do case $path in $i*) echo $i;; esac; done| tail -1 /dev Note the need to vary cut's parameters, because of the way df's output differs: indeed, awk is better. Answer It looks like munging tabular output is the only way within the shell, but df /dev/fd/impossible | tail -1 | awk '{ print $NF}' is a big improvement on what I had. Note two differences in semantics: firstly, df $path insists that $path names an existing file, the script I had above doesn't care; secondly, there are no worries about dereferncing symlinks. It's not difficult to write Python code to do the job.

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• Why is this removing all elements from my LinkedList?

  - by Brian

  Why is my remove method removing every element from my Doubly Linked List? If I take out that if/else statements then I can successfully remove middle elements, but elements at the head or tail of the list still remain. However, I added the if/else statements to take care of elements at the head and tail, unfortunately this method now removes every element in my list. What am I do wrong? public void remove(int n) { LinkEntry<E> remove_this = new LinkEntry<E>(); //if nothing comes before remove_this, set the head to equal the element after remove_this if (remove_this.previous == null) head = remove_this.next; //otherwise set the element before remove_this equal to the element after remove_this else remove_this.previous.next = remove_this.next; //if nothing comes after remove_this, set the tail equal to the element before remove_this if (remove_this.next == null) tail = remove_this.previous; //otherwise set the next element's previous pointer to the element before remove_this else remove_this.next.previous = remove_this.previous; //if remove_this is located in the middle of the list, enter this loop until it is //found, then remove it, closing the gap afterwards. int i = 0; for (remove_this = head; remove_this != null; remove_this = remove_this.next) { //if i == n, stop and delete 'remove_this' from the list if (i == n) { //set the previous element's next to the element that comes after remove_this remove_this.previous.next = remove_this.next; //set the element after remove_this' previous pointer to the element before remove_this remove_this.next.previous = remove_this.previous; break; } //if i != n, keep iterating through the list i++; } }

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• cannot delete from doubly linked list using visual studios in C

  - by Jesus Sanchez

  hello I am currently doing an assignment that is supposed to read in a file, use the information, and then print out another file. all using doubly linked list. Currently i am trying to just read in the file into a doubly linked list, print it out onto the screen and a file, and finally delete the list and close the program. The program works fine as long as I don't call the dlist_distroy function which is supposed to delete the string. as soon as I do it the program starts running and then a window pops up saying "Windows has triggered a breakpoint in tempfilter.exe. This may be due to a corruption of the heap, which indicates a bug in tempfilter.exe or any of the DLLs it has loaded. This may also be due to the user pressing F12 while tempfilter.exe has focus. The output window may have more diagnostic information." I have revised the destroy and remove functions and cant understand the problem. my program is the following main.c #include <stdlib.h> #include <stdio.h> #include <string.h> #include "dlinklist.h" #include "DlistElmt.h" #include "Dlist.h" #include "dlistdata.h" /**************************************************************************************************/ int main ( int argc, char *argv[] ) { FILE *ifp, *ofp; int hour, min; Dlist *list; DlistElmt *current=NULL, *element; float temp; list = (Dlist *)malloc(sizeof(list)); element = (DlistElmt *)malloc(sizeof(element)); if ( argc != 3 ) /* argc should be 3 for correct execution */ { /* We print argv[0] assuming it is the program name */ /* TODO: This is wrong, it should be: usage: %s inputfile outputfile */ printf( "usage: %s filename", argv[0] ); } else { // We assume argv[1] is a filename to open ifp = fopen( argv[1], "r" ); if (ifp == 0 ){ printf("Could not open file\n"); } else{ ofp = fopen(argv[2], "w"); dlist_init(list);//, (destroy)(hour, min, temp)); while (fscanf(ifp, "%d:%d %f ", &hour, &min, &temp) == 3) { current=list->tail; if(dlist_size(list)==0){ dlist_ins_prev(list, current, hour, min, temp); } else{ dlist_ins_next(list, current, hour, min, temp); } } current = list->head; while(current != NULL){ if(current==list->head){ current=current->next; } else if((current->temp > (current->prev->temp +5)) || (current->temp < (current->prev->temp -5))){ dlist_remove(list, current); current=current->next; } else current=current->next; } current = list->head; while(current != NULL){ printf("%d:%d %2.1lf\n" ,current->time, current->time2, current->temp ); fprintf(ofp, "%d:%d %2.1lf\n", current->time, current->time2, current->temp ); current = current->next; } //dlist_destroy(list); //} fclose(ifp); fclose(ofp); } } getchar(); } dlistdata.c #include <stdlib.h> #include <stdio.h> #include <string.h> #include "dlinklist.h" #include "DlistElmt.h" #include "dlistdata.h" /**************************************************************************************************/ void dlist_init(Dlist *list){ list->size = 0; list->head = NULL; list->tail = NULL; return; } void dlist_destroy(Dlist *list){ while (dlist_size(list) > 0){ dlist_remove(list, list->head); } memset(list, 0, sizeof(Dlist)); return; } int dlist_ins_next(Dlist *list, DlistElmt *element, const int time, const int time2, const float temp){ DlistElmt *new_element; if (element == NULL && dlist_size(list) != 0) return -1; if ((new_element = (DlistElmt *)malloc(sizeof(new_element))) == NULL) return -1; new_element->time = (int )time; new_element->time2 = (int )time2; new_element->temp = (float )temp; if (dlist_size(list) == 0) { list->head = new_element; list->head->prev = NULL; list->head->next = NULL; list->tail = new_element; } else { new_element->next = element->next; new_element->prev = element; if(element->next == NULL) list->tail = new_element; else element->next->prev = new_element; element->next = new_element; } list->size++; return 0; } int dlist_ins_prev(Dlist *list, DlistElmt *element, const int time, const int time2, const float temp){ DlistElmt *new_element; if (element == NULL && dlist_size(list) != 0) return -1; if ((new_element = (DlistElmt *)malloc(sizeof(new_element))) == NULL) return -1; new_element->time = (int )time; new_element->time2 = (int )time2; new_element->temp = (float )temp; if (dlist_size(list) == 0){ list->head = new_element; list->head->prev = NULL; list->head->next=NULL; list->tail = new_element; } else{ new_element->next = element; new_element->prev = element->prev; if(element->prev ==NULL) list->head = new_element; else element->prev->next = new_element; element->prev = new_element; } list->size++; return 0; } int dlist_remove(Dlist *list, DlistElmt *element){//, int time, int time2, float temp){ if (element == NULL || dlist_size(list) == 0) return -1; if (element == list->head) { list->head = element->next; if (list->head == NULL) list->tail = NULL; else element->next->prev = NULL; } else{ element->prev->next = element->next; if (element->next = NULL) list->tail = element->prev; else element->next->prev = element->prev; } free(element); list->size--; return 0; }

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• Forensic Analysis of the OOM-Killer

  - by Oddthinking

  Ubuntu's Out-Of-Memory Killer wreaked havoc on my server, quietly assassinating my applications, sendmail, apache and others. I've managed to learn what the OOM Killer is, and about its "badness" rules. While my machine is small, my applications are even smaller, and typically only half of my physical memory is in use, let alone swap-space, so I was surprised. I am trying to work out the culprit, but I don't know how to read the OOM-Killer logs. Can anyone please point me to a tutorial on how to read the data in the logs (what are ve, free and gen?), or help me parse these logs? Apr 20 20:03:27 EL135 kernel: kill_signal(13516.0): selecting to kill, queued 0, seq 1, exc 2326 0 goal 2326 0... Apr 20 20:03:27 EL135 kernel: kill_signal(13516.0): task ebb0c6f0, thg d33a1b00, sig 1 Apr 20 20:03:27 EL135 kernel: kill_signal(13516.0): selected 1, signalled 1, queued 1, seq 1, exc 2326 0 red 61795 745 Apr 20 20:03:27 EL135 kernel: kill_signal(13516.0): selecting to kill, queued 0, seq 2, exc 122 0 goal 383 0... Apr 20 20:03:27 EL135 kernel: kill_signal(13516.0): task ebb0c6f0, thg d33a1b00, sig 1 Apr 20 20:03:27 EL135 kernel: kill_signal(13516.0): selected 1, signalled 1, queued 1, seq 2, exc 383 0 red 61795 745 Apr 20 20:03:27 EL135 kernel: kill_signal(13516.0): task ebb0c6f0, thg d33a1b00, sig 2 Apr 20 20:03:27 EL135 kernel: OOM killed process watchdog (pid=14490, ve=13516) exited, free=43104 gen=24501. Apr 20 20:03:27 EL135 kernel: OOM killed process tail (pid=4457, ve=13516) exited, free=43104 gen=24502. Apr 20 20:03:27 EL135 kernel: OOM killed process ntpd (pid=10816, ve=13516) exited, free=43104 gen=24503. Apr 20 20:03:27 EL135 kernel: OOM killed process tail (pid=27401, ve=13516) exited, free=43104 gen=24504. Apr 20 20:03:27 EL135 kernel: OOM killed process tail (pid=29009, ve=13516) exited, free=43104 gen=24505. Apr 20 20:03:27 EL135 kernel: OOM killed process apache2 (pid=10557, ve=13516) exited, free=49552 gen=24506. Apr 20 20:03:27 EL135 kernel: OOM killed process apache2 (pid=24983, ve=13516) exited, free=53117 gen=24507. Apr 20 20:03:27 EL135 kernel: OOM killed process apache2 (pid=29129, ve=13516) exited, free=68493 gen=24508. Apr 20 20:03:27 EL135 kernel: OOM killed process sendmail-mta (pid=941, ve=13516) exited, free=68803 gen=24509. Apr 20 20:03:27 EL135 kernel: OOM killed process tail (pid=12418, ve=13516) exited, free=69330 gen=24510. Apr 20 20:03:27 EL135 kernel: OOM killed process python (pid=22953, ve=13516) exited, free=72275 gen=24511. Apr 20 20:03:27 EL135 kernel: OOM killed process apache2 (pid=6624, ve=13516) exited, free=76398 gen=24512. Apr 20 20:03:27 EL135 kernel: OOM killed process python (pid=23317, ve=13516) exited, free=94285 gen=24513. Apr 20 20:03:27 EL135 kernel: OOM killed process tail (pid=29030, ve=13516) exited, free=95339 gen=24514. Apr 20 20:03:28 EL135 kernel: OOM killed process apache2 (pid=20583, ve=13516) exited, free=101663 gen=24515. Apr 20 20:03:28 EL135 kernel: OOM killed process logger (pid=12894, ve=13516) exited, free=101694 gen=24516. Apr 20 20:03:28 EL135 kernel: OOM killed process bash (pid=21119, ve=13516) exited, free=101849 gen=24517. Apr 20 20:03:28 EL135 kernel: OOM killed process atd (pid=991, ve=13516) exited, free=101880 gen=24518. Apr 20 20:03:28 EL135 kernel: OOM killed process apache2 (pid=14649, ve=13516) exited, free=102748 gen=24519. Apr 20 20:03:28 EL135 kernel: OOM killed process grep (pid=21375, ve=13516) exited, free=132167 gen=24520. Apr 20 20:03:57 EL135 kernel: kill_signal(13516.0): selecting to kill, queued 0, seq 4, exc 4215 0 goal 4826 0... Apr 20 20:03:57 EL135 kernel: kill_signal(13516.0): task ede29370, thg df98b880, sig 1 Apr 20 20:03:57 EL135 kernel: kill_signal(13516.0): selected 1, signalled 1, queued 1, seq 4, exc 4826 0 red 189481 331 Apr 20 20:03:57 EL135 kernel: kill_signal(13516.0): task ede29370, thg df98b880, sig 2 Apr 20 20:04:53 EL135 kernel: kill_signal(13516.0): selecting to kill, queued 0, seq 5, exc 3564 0 goal 3564 0... Apr 20 20:04:53 EL135 kernel: kill_signal(13516.0): task c6c90110, thg cdb1a100, sig 1 Apr 20 20:04:53 EL135 kernel: kill_signal(13516.0): selected 1, signalled 1, queued 1, seq 5, exc 3564 0 red 189481 331 Apr 20 20:04:53 EL135 kernel: kill_signal(13516.0): task c6c90110, thg cdb1a100, sig 2 Apr 20 20:07:14 EL135 kernel: kill_signal(13516.0): selecting to kill, queued 0, seq 6, exc 8071 0 goal 8071 0... Apr 20 20:07:14 EL135 kernel: kill_signal(13516.0): task d7294050, thg c03f42c0, sig 1 Apr 20 20:07:14 EL135 kernel: kill_signal(13516.0): selected 1, signalled 1, queued 1, seq 6, exc 8071 0 red 189481 331 Apr 20 20:07:14 EL135 kernel: kill_signal(13516.0): task d7294050, thg c03f42c0, sig 2 Watchdog is a watchdog task, that was idle; nothing in the logs to suggest it had done anything for days. Its job is to restart one of the applications if it dies, so a bit ironic that it is the first to get killed. Tail was monitoring a few logs files. Unlikely to be consuming memory madly. The apache web-server only serves pages to a little old lady who only uses it to get to church on Sundays a couple of developers who were in bed asleep, and hadn't visited a page on the site for a few weeks. The only traffic it might have had is from the port-scanners; all the content is password-protected and not linked from anywhere, so no spiders are interested. Python is running two separate custom applications. Nothing in the logs to suggest they weren't humming along as normal. One of them was a relatively recent implementation, which makes suspect #1. It doesn't have any data-structures of any significance, and normally uses only about 8% of the total physical RAW. It hasn't misbehaved since. The grep is suspect #2, and the one I want to be guilty, because it was a once-off command. The command (which piped the output of a grep -r to another grep) had been started at least 30 minutes earlier, and the fact it was still running is suspicious. However, I wouldn't have thought grep would ever use a significant amount of memory. It took a while for the OOM killer to get to it, which suggests it wasn't going mad, but the OOM killer stopped once it was killed, suggesting it may have been a memory-hog that finally satisfied the OOM killer's blood-lust.

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• confused about variables in bash

  - by gappy

  I know that variables in bash have no type, but am confused about the value they are assigned. The following simple script works fine in bash #!/bin/bash tail -n +2 /cygdrive/c/workdir\ \(newco\,\ LLC\)/workfile.txt > \ /cygdrive/c/workdir\ \(newco\,\ LLC\)/workfile2.txt However, the following does not #!/bin/bash tmpdir=/cygdrive/c/workdir\ \(newco\,\ LLC\) tail -n +2 $tmpdir/workfile.txt > $tmpdir/workfile2.txt Is there an explanation for this behavior?

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• CNAME to another domain fails on some office networks, why?

  - by crashalpha

  Our domain "aspenfasteners.com" is hosted by Volusion. We have CNAME records "find" and "search" which point to site indexing accounts on www.picosearch.com. These addresses fail on SOME private office networks which have their own DNS. We suspect the problem comes from Volusion's own name servers, n2.volusion.com and n3.volusion.com. Volusion support on problems this technical is non-existant. We have tried an NSLOOKUP on find.aspenfasteners.com with level 2 debugging info, and we got the results below. Is it possible that the local DNS is recursing to Volusion's name servers, and that while Volusion DOES return the canonical name, they do NOT resolve the address? Can anybody with expertise in this sort of stuff PLEASE look at the NSLOOKUP below and tell me if we are right, because Volusion is giving me absolutely NO support on this topic. I need proof of where the problem lies. Thanks VERY much! Carlo find.aspenfasteners.com Server: mtl-srm-dbsv-01.fastenerwholesale.com Address: 192.168.0.44 SendRequest(), len 61 HEADER: opcode = QUERY, id = 8, rcode = NOERROR header flags: query, want recursion questions = 1, answers = 0, authority records = 0, additional = 0 QUESTIONS: find.aspenfasteners.com.fastenerwholesale.com, type = A, class = IN ------------ Got answer (138 bytes): HEADER: opcode = QUERY, id = 8, rcode = NXDOMAIN header flags: response, auth. answer, want recursion, recursion avail. questions = 1, answers = 0, authority records = 1, additional = 0 QUESTIONS: find.aspenfasteners.com.fastenerwholesale.com, type = A, class = IN AUTHORITY RECORDS: -> fastenerwholesale.com type = SOA, class = IN, dlen = 46 ttl = 3600 (1 hour) primary name server = mtl-srm-dbsv-01.fastenerwholesale.com responsible mail addr = admin.fastenerwholesale.com serial = 10219 refresh = 900 (15 mins) retry = 600 (10 mins) expire = 86400 (1 day) default TTL = 3600 (1 hour) ------------ SendRequest(), len 41 HEADER: opcode = QUERY, id = 9, rcode = NOERROR header flags: query, want recursion questions = 1, answers = 0, authority records = 0, additional = 0 QUESTIONS: find.aspenfasteners.com, type = A, class = IN ------------ Got answer (141 bytes): HEADER: opcode = QUERY, id = 9, rcode = NXDOMAIN header flags: response, auth. answer questions = 1, answers = 1, authority records = 1, additional = 1 QUESTIONS: find.aspenfasteners.com, type = A, class = IN ANSWERS: -> find.aspenfasteners.com type = CNAME, class = IN, dlen = 17 canonical name = www.picosearch.com ttl = 3600 (1 hour) AUTHORITY RECORDS: -> com type = SOA, class = IN, dlen = 43 ttl = 900 (15 mins) primary name server = ns3.volusion.com responsible mail addr = admin.volusion.com serial = 1 refresh = 900 (15 mins) retry = 600 (10 mins) expire = 86400 (1 day) default TTL = 3600 (1 hour) ADDITIONAL RECORDS: -> ns3.volusion.com type = A, class = IN, dlen = 4 internet address = 65.61.137.154 ttl = 900 (15 mins) * mtl-srm-dbsv-01.fastenerwholesale.com can't find find.aspenfasteners.com: Non-existent domain

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• Is it possible to shorten my main function in this code?

  - by AjiPorter

  Is it possible for me to shorten my main() by creating a class? If so, what part of my code would most likely be inside the class? Thanks again to those who'll answer. :) #include <iostream> #include <fstream> #include <string> #include <ctime> #include <cstdlib> #define SIZE 20 using namespace std; struct textFile { string word; struct textFile *next; }; textFile *head, *body, *tail, *temp; int main() { ifstream wordFile("WORDS.txt", ios::in); // file object constructor /* stores words in the file into an array */ string words[SIZE]; char pointer; int i; for(i = 0; i < SIZE; i++) { while(wordFile >> pointer) { if(!isalpha(pointer)) { pointer++; break; } words[i] = words[i] + pointer; } } /* stores the words in the array to a randomized linked list */ srand(time(NULL)); int index[SIZE] = {0}; // temporary array of index that will contain randomized indexes of array words int j = 0, ctr; // assigns indexes to array index while(j < SIZE) { i = rand() % SIZE; ctr = 0; for(int k = 0; k < SIZE; k++) { if(!i) break; else if(i == index[k]) { // checks if the random number has previously been stored as index ctr = 1; break; } } if(!ctr) { index[j] = i; // assigns the random number to the current index of array index j++; } } /* makes sure that there are no double zeros on the array */ ctr = 0; for(i = 0; i < SIZE; i++) { if(!index[i]) ctr++; } if(ctr > 1) { int temp[ctr-1]; for(j = 0; j < ctr-1; j++) { for(i = 0; i < SIZE; i++) { if(!index[i]) { int ctr2 = 0; for(int k = 0; k < ctr-1; k++) { if(i == temp[k]) ctr2 = 1; } if(!ctr2) temp[j] = i; } } } j = ctr - 1; while(j > 0) { i = rand() % SIZE; ctr = 0; for(int k = 0; k < SIZE; k++) { if(!i || i == index[k]) { ctr = 1; break; } } if(!ctr) { index[temp[j-1]] = i; j--; } } } head = tail = body = temp = NULL; for(j = 0; j < SIZE; j++) { body = (textFile*) malloc (sizeof(textFile)); body->word = words[index[j]]; if(head == NULL) { head = tail = body; } else { tail->next = body; tail = body; cout << tail->word << endl; } } temp = head; while(temp != NULL) { cout << temp->word << endl; temp = temp->next; } return 0; }

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• is it wasteful/bad design to use a vector/list where in most instances it will only have one element

  - by lucid

  is it wasteful/bad design to use a vector/list where in most instances it will only have one element? example: class dragon { ArrayList<head> = new ArrayList<head> Heads; tail Tail = new tail(); body Body = new body(); dragon() { theHead=new head(); Heads.add(theHead); } void nod() { for (int i=0;i<Heads.size();i++) { heads.get(i).GoUpAndDown(); } } } class firedragon extends dragon { } class icedragon extends dragon { } class lightningdragon extends dragon { } // 10 other one-headed dragon declarations here class hydra extends dragon { hydra() { anotherHead=new head(); for (int i=0;i<2;i++) { Heads.add(anotherHead); } } } class superhydra extends dragon { superhydra() { anotherHead=new head(); for (int i=0;i<4;i++) { Heads.add(anotherHead); } } }

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• Python: Seeing all files in Hex.

  - by Recursion

  I am writing a python script which looks at common computer files and examines them for similar bytes, words, double word's. Though I need/want to see the files in Hex, ande cannot really seem to get python to open a simple file in python. I have tried codecs.open with hex as the encoding, but when I operate on the file descriptor it always spits back File "main.py", line 41, in <module> main() File "main.py", line 38, in main process_file(sys.argv[1]) File "main.py", line 27, in process_file seeker(line.rstrip("\n")) File "main.py", line 15, in seeker for unit in f.read(2): File "/usr/lib/python2.6/codecs.py", line 666, in read return self.reader.read(size) File "/usr/lib/python2.6/codecs.py", line 472, in read newchars, decodedbytes = self.decode(data, self.errors) File "/usr/lib/python2.6/encodings/hex_codec.py", line 50, in decode return hex_decode(input,errors) File "/usr/lib/python2.6/encodings/hex_codec.py", line 42, in hex_decode output = binascii.a2b_hex(input) TypeError: Non-hexadecimal digit found def seeker(_file): f = codecs.open(_file, "rb", "hex") for LINE in f.read(): print LINE f.close() I really just want to see files, and operate on them as if it was in a hex editor like xxd. Also is it possible to read a file in increments of maybe a word at a time. No this is not homework.

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• viewing the updated data in a file

  - by benjamin button

  If i have file for eg: a log file created by any background process on unix, how do i view the data that getting updated each and every time. i know that i can use tail command to see the file.but let's say i have used tail -10 file.txt it will give the last 10 lines.but if lets say at one time 10 lines got added and at the next instance it has been added 2 lines. now from the tail command i can see previous 8 lines also.i don't want this.i want only those two lines which were added. In short i want to view only those lines which were appended.how can i do this?

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• Cure for puzzle piece programming habits?

  - by Recursion

  Even though I went to a decent CS school, I was still taught with the mentality of programming with puzzle pieces. By puzzle pieces I mean, looking up code segments at each step of the development process and adding them together as needed. Eventually gathering all of the pieces and having a properly working program. So as an example, if in my program the next step is to tokenize a string, I go to google and search "how do I tokenize a string in language". All instead of critically thinking about its implementation. I personally don't think its a very good way to program and I always seem to forget everything that I have searched for. So how can I get out of this puzzle piece mode of programmer that I was taught.

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• SH/BASH - Scan a log file until some text occurs, then exit. How??

  - by James

  Current working environment is OSX 10.4.11. My current script: #!/bin/sh tail -f log.txt | while read line do if echo $line | grep -q 'LOL CANDY'; then echo 'LOL MATCH FOUND' exit 0 fi done It works properly the first time, but on the second run and beyond it takes 2 occurrences of 'LOL CANDY' to appear before the script will exit, for whatever reason. And although I'm not sure it is specifically related, there is the problem of the "tail -f" staying open forever. Can someone please give me an example that will work without using tail -f? If you want you can give me a bash script, as OSX can handle sh, bash, and some other shells I think.

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• Understanding pattern matching with cons operator

  - by Mathias

  In "Programming F#" I came across a pattern-matching like this one (I simplified a bit): let rec len list = match list with | [] -> 0 | [_] -> 1 | head :: tail -> 1 + len tail;; Practically, I understand that the last match recognizes the head and tail of the list. Conceptually, I don't get why it works. As far as I understand, :: is the cons operator, which appends a value in head position of a list, but it doesn't look to me like it is being used as an operator here. Should I understand this as a "special syntax" for lists, where :: is interpreted as an operator or a "match pattern" depending on context? Or can the same idea be extended for types other than lists, with other operators?

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• Counting common Bytes, Words and Double Words.

  - by Recursion

  I am scanning over a large amount of data and looking for common trends in it. Every time I meet a recurrence of a unit, I want to increment the count of it. What is the best data structure or way to hold this data. I need to be able to search it quickly, and also have a count with each unit of data.

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• Python ValueError: not allowed to raise maximum limit

  - by Ricky Bobby

  I'm using python 2.7.2 on mac os 10.7.3 I'm doing a recursive algorithm in python with more than 50 000 recursion levels. I tried to increase the maximum recursion level to 1 000 000 but my python shell still exit after 18 000 recursion levels. I tried to increase the resources available : import resource resource.setrlimit(resource.RLIMIT_STACK, (2**29,-1)) sys.setrecursionlimit(10**6) and I get this error : Traceback (most recent call last): File "<pyshell#58>", line 1, in <module> resource.setrlimit(resource.RLIMIT_STACK,(2**29,-1)) ValueError: not allowed to raise maximum limit I don't know why I cannot raise the maximum limit ? thanks for your suggestions .

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• How can I remove sensitive data from the debug_backtrace function?

  - by RenderIn

  I am using print_r(debug_backtrace(), true) to retrieve a string representation of the debug backtrace. This works fine, as print_r handles recursion. When I tried to recursively iterate through the debug_backtrace() return array before turning it into a string it ran into recursion and never ended. Is there some simple way I can remove certain sensitive key/value pairs from the backtrace array? Perhaps some way to turn the array to a string using print_r, then back to an array with the recursive locations changed to the string RECURSION, which I could the iterate through. I don't want to execute regular expressions on the string representation if possible.

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• GPA and Resume and PDF vs Doc.

  - by Recursion

  As a recent graduate of a CS program, I am looking for my first job. My GPA was not above 3.0, but incredibly close. Should I still put my GPA on my resume, or is it best to leave it out? Also, is it best to submit a resume as a PDF or a DOC file?

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• google maps plot route between two points

  - by amarsh-anand

  I have written this innocent javascript code, which lets the user create two markers and plot the route between them. It doesnt work, instead, it gives a weird error: Uncaught TypeError: Cannot read property 'ya' of undefined Can someone suggest me whats wrong here: // called upon a click GEvent.addListener(map, "click", function(overlay,point) { if (isCreateHeadPoint) { // add the head marker headMarker = new GMarker(point,{icon:redIcon,title:'Head'}); map.addOverlay(headMarker); isCreateHeadPoint = false; } else { // add the tail marker tailMarker = new GMarker(point,{icon:greenIcon,title:'Tail'}); map.addOverlay(tailMarker); isCreateHeadPoint = true; // create a path from head to tail direction.load("from:" + headMarker.getPoint().lat()+ ", " + headMarker.getPoint().lng()+ " to:" + tailMarker.getPoint().lat() + "," + tailMarker.getPoint().lng(), { getPolyline: true, getSteps: true }); // display the path map.addOverlay(direction.getPolyline()); } });

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• For kernel/OS is C still it

  - by Recursion

  I like operating systems and would eventually like to become a OS developer mostly working on kernels. In the future will C still be the language of choice and what else should I be trying to learn.

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• Remove redundant entries, scala way

  - by andersbohn

  Edit: Added the fact, that the list is sorted, and realizing 'duplicate' is misleading, replaced that with 'redundant' in the title. I have a sorted list of entries stating a production value in a given interval. Entries stating the exact same value at a later time adds no information and can safely be left out. case class Entry(minute:Int, production:Double) val entries = List(Entry(0, 100.0), Entry(5, 100.0), Entry(10, 100.0), Entry(20, 120.0), Entry(30, 100.0), Entry(180, 0.0)) Experimenting with the scala 2.8 collection functions, so far I have this working implementation: entries.foldRight(List[Entry]()) { (entry, list) => list match { case head :: tail if (entry.production == head.production) => entry :: tail case head :: tail => entry :: list case List() => entry :: List() } } res0: List[Entry] = List(Entry(0,100.0), Entry(20,120.0), Entry(30,100.0), Entry(180,0.0)) Any comments? Am I missing out on some scala magic?

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• Remove duplicates in entries, scala way

  - by andersbohn

  I have a list of entries stating a production value in a given interval. Entries stating the exact same value at a later time adds no information and can safely be left out. case class Entry(minute:Int, production:Double) val entries = List(Entry(0, 100.0), Entry(5, 100.0), Entry(10, 100.0), Entry(20, 120.0), Entry(30, 100.0), Entry(180, 0.0)) Experimenting with the scala 2.8 collection functions, so far I have this working implementation: entries.foldRight(List[Entry]()) { (entry, list) => list match { case head :: tail if (entry.production == head.production) => entry :: tail case head :: tail => entry :: list case List() => entry :: List() } } res0: List[Entry] = List(Entry(0,100.0), Entry(20,120.0), Entry(30,100.0), Entry(180,0.0)) Any comments? Am I missing out on some scala magic?

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• Cure for puzzle piece programming habbits?

  - by Recursion

  Even though I went to a decent CS school, I was still taught with the mentality of programming with puzzle pieces. By puzzle pieces I mean, looking up code segments at each step of the development process and adding them together as needed. Eventually gathering all of the pieces and having a properly working program. So as an example, if in my program the next step is to tokenize a string, I go to google and search "how do I tokenize a string in language". All instead of critically thinking about its implementation. I personally don't think its a very good way to program and I always seem to forget everything that I have searched for. So how can I get out of this puzzle piece mode of programmer that I was taught.

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• Linked List. Insert integers in order

  - by user69514

  I have a linked list of integers. When I insert a new Node I need to insert it not at the end, but in oder... i.e. 2, 4, 5, 8, 11, 12, 33, 55, 58, 102, etc. I don't think I am inserting it in the correct position. Do see what Im doing wrong? Node newNode = new Node(someInt); Node current = head; for(int i=0; i<count; i++){ if(current == tail && tail.data < someInt){ tail.next = newNode; } if(current.data < someInt && current.next.data >= someInt){ newNode.next = current.next; current.next = newNode; } }

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