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  • fork() in perl on windows

    - by Darioush
    I'm using fork() on PERL in windows (activeperl) for a basic socket server, but apparently there are problems (it won't accept connections after a few times), is there any workaround? while($client = $bind->accept()) { $client->autoflush(); if(fork()){ $client->close(); } else { $bind->close(); new_client($client); exit(); } } is the portion of the relevant code.

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  • Fork in Perl not working inside a while loop reading from file

    - by Sag
    Hi, I'm running a while loop reading each line in a file, and then fork processes with the data of the line to a child. After N lines I want to wait for the child processes to end and continue with the next N lines, etc. It looks something like this: while ($w=<INP>) { # ignore file header if ($w=~m/^\D/) { next;} # get data from line chomp $w; @ws = split(/\s/,$w); $m = int($ws[0]); $d = int($ws[1]); $h = int($ws[2]); # only for some days in the year if (($m==3)and($d==15) or ($m==4)and($d==21) or ($m==7)and($d==18)) { die "could not fork" unless defined (my $pid = fork); unless ($pid) { some instructions here using $m, $d, $h ... } push @qpid,$pid; # when all processors are busy, wait for child processes if ($#qpid==($procs-1)) { for my $pid (@qpid) { waitpid $pid, 0; } reset 'q'; } } } close INP; This is not working. After the first round of processes I get some PID equal to 0, the @qpid array gets mixed up, and the file starts to get read at (apparently) random places, jumping back and forth. The end result is that most lines in the file get read two or three times. Any ideas? Thanks a lot in advance, S.

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  • c - fork() and wait()

    - by Joe
    Hi there, I need to use the fork() and wait() functions to complete an assignment. We are modelling non-deterministic behaviour and need the program to fork() if there is more than one possible transition. In order to try and work out how fork and wait work, I have just made a simple program. I think I understand now how the calls work and would be fine if the program only branched once because the parent process could use the exit status from the single child process to determine whether the child process reached the accept state or not. As you can see from the code that follows though, I want to be able to handle situations where there must be more than one child processes. My problem is that you seem to only be able to set the status using an _exit function once. So, as in my example the exit status that the parent process tests for shows that the first child process issued 0 as it's exit status, but has no information on the second child process. I tried simply not _exit()-ing on a reject, but then that child process would carry on, and in effect there would seem to be two parent processes. Sorry for the waffle, but I would be grateful if someone could tell me how my parent process could obtain the status information on more than one child process, or I would be happy for the parent process to only notice accept status's from the child processes, but in that case I would successfully need to exit from the child processes which have a reject status. My test code is as follows: #include <stdio.h> #include <unistd.h> #include <stdlib.h> #include <errno.h> #include <sys/wait.h> int main(void) { pid_t child_pid, wpid, pid; int status = 0; int i; int a[3] = {1, 2, 1}; for(i = 1; i < 3; i++) { printf("i = %d\n", i); pid = getpid(); printf("pid after i = %d\n", pid); if((child_pid = fork()) == 0) { printf("In child process\n"); pid = getpid(); printf("pid in child process is %d\n", pid); /* Is a child process */ if(a[i] < 2) { printf("Should be accept\n"); _exit(1); } else { printf("Should be reject\n"); _exit(0); } } } if(child_pid > 0) { /* Is the parent process */ pid = getpid(); printf("parent_pid = %d\n", pid); wpid = wait(&status); if(wpid != -1) { printf("Child's exit status was %d\n", status); if(status > 0) { printf("Accept\n"); } else { printf("Complete parent process\n"); if(a[0] < 2) { printf("Accept\n"); } else { printf("Reject\n"); } } } } return 0; } Many thanks Joe

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  • Weird behavior of fork() and execvp() in C

    - by ron
    After some remarks from my previous post , I made the following modifications : int main() { char errorStr[BUFF3]; while (1) { int i , errorFile; char *line = malloc(BUFFER); char *origLine = line; fgets(line, 128, stdin); // get a line from stdin // get complete diagnostics on the given string lineData info = runDiagnostics(line); char command[20]; sscanf(line, "%20s ", command); line = strchr(line, ' '); // here I remove the command from the line , the command is stored in "commmand" above printf("The Command is: %s\n", command); int currentCount = 0; // number of elements in the line int *argumentsCount = &currentCount; // pointer to that // get the elements separated char** arguments = separateLineGetElements(line,argumentsCount); printf("\nOutput after separating the given line from the user\n"); for (i = 0; i < *argumentsCount; i++) { printf("Argument %i is: %s\n", i, arguments[i]); } // here we call a method that would execute the commands pid_t pid ; if (-1 == (pid = fork())) { sprintf(errorStr,"fork: %s\n",strerror(errno)); write(errorFile,errorStr,strlen(errorStr + 1)); perror("fork"); exit(1); } else if (pid == 0) // fork was successful { printf("\nIn son process\n"); // if (execvp(arguments[0],arguments) < 0) // for the moment I ignore this line if (execvp(command,arguments) < 0) // execute the command { perror("execvp"); printf("ERROR: execvp failed\n"); exit(1); } } else // parent { int status = 0; pid = wait(&status); printf("Process %d returned with status %d.", pid, status); } // print each element of the line for (i = 0; i < *argumentsCount; i++) { printf("Argument %i is: %s\n", i, arguments[i]); } // free all the elements from the memory for (i = 0; i < *argumentsCount; i++) { free(arguments[i]); } free(arguments); free(origLine); } return 0; } When I enter in the Console : ls out.txt I get : The Command is: ls execvp: No such file or directory In son process ERROR: execvp failed Process 4047 returned with status 256.Argument 0 is: > Argument 1 is: out.txt So I guess that the son process is active , but from some reason the execvp fails . Why ? Regards REMARK : The ls command is just an example . I need to make this works with any given command . EDIT 1 : User input : ls > qq.out Program output : The Command is: ls Output after separating the given line from the user Argument 0 is: > Argument 1 is: qq.out In son process >: cannot access qq.out: No such file or directory Process 4885 returned with status 512.Argument 0 is: > Argument 1 is: qq.out

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  • Hibernate: Perform criteria query with Sub-Select AND Left-Outer join?

    - by Markos Fragkakis
    Can I perform a Criteria query with Sub-Select AND Left-Outer join? For example, I have A 1-many B 1-many C. With Criteria.createAlias ("b", "b", Criteria.LEFT_JOIN) I can perform Left Outer join. With Criteria.setFetchMode ("b", org.hibernate.FetchMode.DEFAULT) I can perform Join with the default fetching strategy. I assume that having set @org.hibernate.annotations.FetchMode.SUBSELECT in both A.B and B.C is enough (is it?). Question 1: Why does org.hibernate.FetchMode not have SUBSELECT option, whereas the org.hibernate.annotations.FetchMode does? Question 2: Can I perform a Criteria query with Sub-Select AND Left-Outer join?

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  • About fork system call and global variables

    - by lurks
    I have this program in C++ that forks two new processes: #include <pthread.h> #include <iostream> #include <unistd.h> #include <sys/types.h> #include <sys/wait.h> #include <cstdlib> using namespace std; int shared; void func(){ extern int shared; for (int i=0; i<10;i++) shared++; cout<<"Process "<<getpid()<<", shared " <<shared<<", &shared " <<&shared<<endl; } int main(){ extern int shared; pid_t p1,p2; int status; shared=0; if ((p1=fork())==0) {func();exit(0);}; if ((p2=fork())==0) {func();exit(0);}; for(int i=0;i<10;i++) shared++; waitpid(p1,&status,0); waitpid(p2,&status,0);; cout<<"shared variable is: "<<shared<<endl; cout<<"Process "<<getpid()<<", shared " <<shared<<", &shared " <<&shared<<endl; } The two forked processes make an increment on the shared variables and the parent process does the same. As the variable belongs to the data segment of each process, the final value is 10 because the increment is independent. However, the memory address of the shared variables is the same, you can try compiling and watching the output of the program. How can that be explained ? I cannot understand that, I thought I knew how the fork() works, but this seems very odd.. I need an explanation on why the address is the same, although they are separate variables.

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  • Can MySQL / SQL's short hand of "Using" be used without saying "Inner Join" ?

    - by Jian Lin
    The following 2 statements are to join using gifts.giftID = sentgifts.giftID: mysql> select * from gifts, sentgifts using (giftID); ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'using (giftID)' at line 1 and the second one: mysql> select * from gifts INNER JOIN sentgifts using (giftID); +--------+------------+----------------+---------------------+--------+------------+--------+------+---------------------+ | giftID | name | filename | effectiveTime | sentID | whenSent | fromID | toID | trytryWhen | +--------+------------+----------------+---------------------+--------+------------+--------+------+---------------------+ | 2 | teddy bear | bear.jpg | 2010-04-24 04:36:03 | 4 | 2010-04-24 | NULL | 111 | 2010-04-24 03:10:42 | | 6 | beer | beer_glass.png | 2010-04-24 05:18:12 | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | | 6 | beer | beer_glass.png | 2010-04-24 05:18:12 | 6 | 2010-04-24 | 11 | 222 | 2010-04-24 03:54:49 | | 6 | beer | beer_glass.png | 2010-04-24 05:18:12 | 7 | 2010-04-24 | 1 | 2 | 2010-04-24 03:58:45 | +--------+------------+----------------+---------------------+--------+------------+--------+------+---------------------+ 4 rows in set (0.00 sec) Can the first statement also use the "using" shorthand? It seems that when it is used then the word "Inner Join" must be specified... but the first statement is actually an inner join?

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  • printf anomaly after "fork()"

    - by pechenie
    OS: Linux, Language: pure C I'm moving forward in learning C progpramming in general, and C programming under UNIX in a special case :D So, I detected a strange (as for me) behaviour of the printf() function after using a fork() call. Let's take a look at simple test program: #include <stdio.h> #include <system.h> int main() { int pid; printf( "Hello, my pid is %d", getpid() ); pid = fork(); if( pid == 0 ) { printf( "\nI was forked! :D" ); sleep( 3 ); } else { waitpid( pid, NULL, 0 ); printf( "\n%d was forked!", pid ); } return 0; } In this case the output looks like: Hello, my pid is 1111 I was forked! :DHello, my pid is 1111 2222 was forked! Why the second "Hello" string occured in the child's output? Yes, it is exactly what the parent printed on it's start, with the parent's pid. But! If we place '\n' character in the end of each string we got the expected output: #include <stdio.h> #include <system.h> int main() { int pid; printf( "Hello, my pid is %d\n", getpid() ); // SIC!! pid = fork(); if( pid == 0 ) { printf( "I was forked! :D" ); //removed the '\n', no matter sleep( 3 ); } else { waitpid( pid, NULL, 0 ); printf( "\n%d was forked!", pid ); } return 0; } And the output looks like: Hello, my pid is 1111 I was forked! :D 2222 was forked! Why does it happen? Is it ... ummm ... correct behaviour? Or it's a kind of the 'bug'?

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  • fork within Cocoa application

    - by liuliu
    My problem is not the best scenario for fork(). However, this is the best func I can get. I am working on a Firefox plugin on Mac OSX. To make it robust, I need to create a new process to run my plugin. The problem is, when I forked a new process, much like this: if (fork() == 0) exit(other_main()); However, since the state is not cleaned, I cannot properly initialized my new process (call NSApplicationLoad etc.). Any ideas? BTW, I certainly don't want create a new binary and exec it.

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  • linux's fork and system resources

    - by Dmitry Bespalov
    Suppose I have following code in linux: int main() { FILE* f = fopen("file.txt", "w"); fork(); fwrite("A", 1, 1, f); fclose(f); return 0; } What I know about fork from documentation, is that it makes the copy of current process. It copies state of the memory as well, so *f should be equal in both instances. But what happens with system resources, such as a file handle? In this example I open the file with write intentions, so only one instance can write into file, right? Which of the instances will actually write into file? Who should care further about the file handle, and call fclose() ?

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  • INNER JOIN vs LEFT JOIN performance in SQL Server

    - by Ekkapop
    I've created SQL command that use INNER JOIN for 9 tables, anyway this command take a very long time (more than five minutes). So my folk suggest me to change INNER JOIN to LEFT JOIN because the performance of LEFT JOIN is better, at first time its despite what I know. After I changed, the speed of query is significantly improve. I want to know why LEFT JOIN is faster than INNER JOIN? My SQL command look like below: SELECT * FROM A INNER JOIN B ON ... INNER JOIN C ON ... INNER JOIN D and so no

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  • How sql server evaluates the multiple different joins?

    - by ziang
    Hi, i have a general question about how sql server evaluates the joins.The query is SELECT * FROM TableA INNER JOIN TableB ON TableB.id = TableA.id LEFT JOIN TABLEC ON TABLEC.id = TABLEB.id Q1: What tables is the left join based on? I know it will based on the TABLEC but what is the other one? Is it the result of the first inner join or the TABLEB specified in the left join condition? Q2: Is "LEFT JOIN TABLEC ON TABLEC.id = TABLEB.id" equivalent to "LEFT JOIN TABLEC ON TABLEB.id = TABLEC.id" Q3: Is the query equivalent to the following one? (with TABLEB.id replaced by TABLEA.id?) SELECT * FROM TableA INNER JOIN TableB ON TableB.id = TableA.id LEFT JOIN TABLEC ON TABLEC.id = TABLEA.id Thank you!

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  • Profile Picture Thumbnails, Following Projects, and Fork Collaboration

    [Do you tweet? Follow us on Twitter @matthawley and @adacole_msft] We deployed a new version of the CodePlex website last week. Profile Picture Thumbnails We have added a way to select a thumbnail from your profile picture, which will start appearing next to usernames across the site.  Managing your thumbnail is simple. From your profile page, choose Edit your profile.  On the left side, you’ll find an intuitive widget for choosing a profile picture, uploading it, and editing your thumbnail image. If you previously uploaded a profile picture, we’ve used that to generate a starter thumbnail. We welcome your suggestions and ideas for areas where seeing user thumbnails would be useful or interesting. Following Projects Based on some feedback we’ve received recently, we have taken several steps to help you discover and follow interesting and popular projects on CodePlex: The homepage now surfaces the top Projects Users are Following from the previous 7 days. When you visit any project homepage, you can see at a glance how many people follow the project. When you visit the People tab for any project, you will see both the project contributors and the 25 most recent project followers. Fork Collaboration We now support enabling collaborators on a fork based on a large number of user requests.  From the Source Code management page for your fork, you will now see the following on the right side: To add a collaborator, type in a username and click Add. All fork collaborators will have the ability to push to the fork and send/cancel pull requests.  To remove a collaborator, hover over user, and click on the X that appears: The CodePlex team values your feedback, and is frequently monitoring Twitter, our Discussions and Issue Tracker for new features or problems. If you’ve not visited the Issue Tracker recently, please take a few moments to log an idea or vote for the features you would most like to see implemented on CodePlex.

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  • 3 SQL Join Concepts to Help You Choose the Right Join

    What do SQL joins and the "teach a man to fish" Chinese proverb have in common? SQL joins, like regular expressions, are one of those commonplace programming tasks in which true success is entirely dependent upon your ability to conceptualize the outcome. Fail to do so and you'll likely wind up spending a few hours in a frustrating round of trial and error. Like regular expressions, the proliferation of online examples has actually contributed to the frustration, providing the equivalent of a day's worth of fish rather than the proverbial fishing pool. The Future of SQL Server MonitoringMonitor wherever, whenever with Red Gate's SQL Monitor. See it live in action now.

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  • Duplicate Items Using Join in NHibernate Map

    - by Colin Bowern
    I am trying to retrieve the individual detail rows without having to create an object for the parent. I have a map which joins a parent table with the detail to achieve this: Table("UdfTemplate"); Id(x => x.Id, "Template_Id"); Map(x => x.FieldCode, "Field_Code"); Map(x => x.ClientId, "Client_Id"); Join("UdfFields", join => { join.KeyColumn("Template_Id"); join.Map(x => x.Name, "COLUMN_NAME"); join.Map(x => x.Label, "DISPLAY_NAME"); join.Map(x => x.IsRequired, "MANDATORY_FLAG") .CustomType<YesNoType>(); join.Map(x => x.MaxLength, "DATA_LENGTH"); join.Map(x => x.Scale, "DATA_SCALE"); join.Map(x => x.Precision, "DATA_PRECISION"); join.Map(x => x.MinValue, "MIN_VALUE"); join.Map(x => x.MaxValue, "MAX_VALUE"); }); When I run the query in NH using: Session.CreateCriteria(typeof(UserDefinedField)) .Add(Restrictions.Eq("FieldCode", code)).List<UserDefinedField>(); I get back the first row three times as opposed to the three individual rows it should return. Looking at the SQL trace in NH Profiler the query appears to be correct. The problem feels like it is in the mapping but I am unsure how to troubleshoot that process. I am about to turn on logging to see what I can find but I thought I would post here in case someone with experience mapping joins knows where I am going wrong.

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  • Would shell command join cause out of memory?

    - by Hancy
    I have two file to join. FILE 1: a A1 a A2 a A3 ... c C1 c C2 ... FILE 2: a feature1_of_a a feature2_of_a ... a featureN_of_a ... ... c feature1_of_c c feature2_of_c ... after join, i could get File like this: A1 feature1_of_a A2 feature1_of_a A3 feature1_of_a A1 feature2_of_a A2 feature2_of_a A3 feature2_of_a ... A1 featureN_of_a A2 featureN_of_a A3 featureN_of_a ... In order to do that: i wrote shell command join -11 -21 -o1.2,2.2 file1 file2. But the problem is: number N might be huge. So if join read all feautre of a into memory at once, memory might not be enough. I don't know how join is implemented. WQould the momery become a problem? If so, is there any way to get what I want?

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  • Best way to fork SVN project with Git

    - by Jeremy Thomerson
    I have forked an SVN project using Git because I needed to add features that they didn't want. But at the same time, I wanted to be able to continue pulling in features or fixes that they added to the upstream version down into my fork (where they don't conflict). So, I have my Git project with the following branches: master - the branch I actually build and deploy from feature_* - feature branches where I work or have worked on new things, which I then merge to master when complete vendor-svn - my local-only git-svn branch that allows me to "git svn rebase" from their svn repo vendor - my local branch that i merge vendor-svn into. then i push this (vendor) branch to the public git repo (github) So, my flow is something like this: git checkout vendor-svn git svn rebase git checkout vendor git merge vendor-svn git push origin vendor Now, the question comes here: I need to review each commit that they made (preferably individually since at this point I'm about twenty commits behind them) before merging them into master. I know that I could run git checkout master; git merge vendor, but this would pull in all changes and commit them, without me being able to see if they conflict with what I need. So, what's the best way to do this? Git seems like a great tool for handling forks of projects since you can pull and push from multiple repos - I'm just not experienced with it enough to know the best way of doing this. Here's the original SVN project I'm talking about: https://appkonference.svn.sourceforge.net/svnroot/appkonference My fork is at github.com/jthomerson/AsteriskAudioKonf (sorry - I couldn't make it a link since I'm a new user here)

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  • Problem with fork exec kill when redirecting output in perl

    - by Edu
    I created a script in perl to run programs with a timeout. If the program being executed takes longer then the timeout than the script kills this program and returns the message "TIMEOUT". The script worked quite well until I decided to redirect the output of the executed program. When the stdout and stderr are being redirected, the program executed by the script is not being killed because it has a pid different than the one I got from fork. It seems perl executes a shell that executes my program in the case of redirection. I would like to have the output redirection but still be able to kill the program in the case of a timeout. Any ideas on how I could do that? A simplified code of my script is: #!/usr/bin/perl use strict; use warnings; use POSIX ":sys_wait_h"; my $timeout = 5; my $cmd = "very_long_program 1>&2 > out.txt"; my $pid = fork(); if( $pid == 0 ) { exec($cmd) or print STDERR "Couldn't exec '$cmd': $!"; exit(2); } my $time = 0; my $kid = waitpid($pid, WNOHANG); while ( $kid == 0 ) { sleep(1); $time ++; $kid = waitpid($pid, WNOHANG); print "Waited $time sec, result $kid\n"; if ($timeout > 0 && $time > $timeout) { print "TIMEOUT!\n"; #Kill process kill 9, $pid; exit(3); } } if ( $kid == -1) { print "Process did not exist\n"; exit(4); } print "Process exited with return code $?\n"; exit($?); Thanks for any help.

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  • confusing fork system call

    - by benjamin button
    Hi, i was just checking the behaviour of fork system call and i found it very confusing. i saw in a website that Unix will make an exact copy of the parent's address space and give it to the child. Therefore, the parent and child processes have separate address spaces #include <stdio.h> #include <sys/types.h> int main(void) { pid_t pid; char y='Y'; char *ptr; ptr=&y; pid = fork(); if (pid == 0) { y='Z'; printf(" *** Child process ***\n"); printf(" Address is %p\n",ptr); printf(" char value is %c\n",y); sleep(5); } else { sleep(5); printf("\n ***parent process ***\n",&y); printf(" Address is %p\n",ptr); printf(" char value is %c\n",y); } } the output of the above program is : *** Child process *** Address is 69002894 char value is Z ***parent process *** Address is 69002894 char value is Y so from the above mentioned statement it seems that child and parent have separet address spaces.this is the reason why char value is printed separately and why am i seeing the address of the variable as same in both child and parent processes.? Please help me understand this!

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  • problem with fork()

    - by john
    I'm writing a shell which forks, with the parent reading the input and the child process parsing and executing it with execvp. pseudocode of main method: do{ pid = fork(); print pid; if (p<0) { error; exit; } if (p>0) { wait for child to finish; read input; } else { call function to parse input; exit; } }while condition return; what happens is that i never seem to enter the child process (pid printed is always positive, i never enter the else). however, if i don't call the parse function and just have else exit, i do correctly enter parent and child alternatingly. full code: int main(int argc, char *argv[]){ char input[500]; pid_t p; int firstrun = 1; do{ p = fork(); printf("PID: %d", p); if (p < 0) {printf("Error forking"); exit(-1);} if (p > 0){ wait(NULL); firstrun = 0; printf("\n> "); bzero(input, 500); fflush(stdout); read(0, input, 499); input[strlen(input)-1] = '\0'; } else exit(0); else { if (parse(input) != 0 && firstrun != 1) { printf("Error parsing"); exit(-1); } exit(0); } }while(strcmp(input, "exit") != 0); return 0; }

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  • Avoiding a fork()/SIGCHLD race condition

    - by larry
    Please consider the following fork()/SIGCHLD pseudo-code. // main program excerpt for (;;) { if ( is_time_to_make_babies ) { pid = fork(); if (pid == -1) { /* fail */ } else if (pid == 0) { /* child stuff */ print "child started" exit } else { /* parent stuff */ print "parent forked new child ", pid children.add(pid); } } } // SIGCHLD handler sigchld_handler(signo) { while ( (pid = wait(status, WNOHANG)) > 0 ) { print "parent caught SIGCHLD from ", pid children.remove(pid); } } In the above example there's a race-condition. It's possible for "/* child stuff */" to finish before "/* parent stuff */" starts which can result in a child's pid being added to the list of children after it's exited, and never being removed. When the time comes for the app to close down, the parent will wait endlessly for the already-finished child to finish. One solution I can think of to counter this is to have two lists: started_children and finished_children. I'd add to started_children in the same place I'm adding to children now. But in the signal handler, instead of removing from children I'd add to finished_children. When the app closes down, the parent can simply wait until the difference between started_children and finished_children is zero. Another possible solution I can think of is using shared-memory, e.g. share the parent's list of children and let the children .add and .remove themselves? But I don't know too much about this. EDIT: Another possible solution, which was the first thing that came to mind, is to simply add a sleep(1) at the start of /* child stuff */ but that smells funny to me, which is why I left it out. I'm also not even sure it's a 100% fix. So, how would you correct this race-condition? And if there's a well-established recommended pattern for this, please let me know! Thanks.

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  • SQL SERVER – Challenge – Puzzle – Why does RIGHT JOIN Exists

    - by pinaldave
    I had interesting conversation with the attendees of the my SQL Server Performance Tuning course. I was asked if LEFT JOIN can do the same task as RIGHT JOIN by reserving the order of the tables in join, why does RIGHT JOIN exists? The definitions are as following: Left Join – select all the records from the LEFT table and then pick up any matching records from the RIGHT table   Right Join – select all the records from the RIGHT table and then pick up any matching records from the LEFT table Most of us read from LEFT to RIGHT so we are using LEFT join. Do you have any explaination why RIGHT JOIN exists or can you come up with example, where RIGHT JOIN is absolutely required and the task can not be achieved with LEFT JOIN. Other Puzzles: SQL SERVER – Puzzle – Challenge – Error While Converting Money to Decimal SQL SERVER – Challenge – Puzzle – Usage of FAST Hint Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Pinal Dave, SQL, SQL Authority, SQL Puzzle, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • mysql join with conditional

    - by Conor H
    Hi There, I am currently working on a MySQL query that contains a table: TBL:lesson_fee -fee_type_id (PRI) -lesson_type_id (PRI) -lesson_fee_amount this table contains the fees for a particular 'lesson type' and there are different 'fee names' (fee_type). Which means that there can be many entries in this table for one 'lesson type' In my query I am joining this table onto the rest of the query via the 'lesson_type' table using: lesson_fee INNER JOIN (other joins here) ON lesson_fee.lesson_type_id = lesson_type.lesson_type_id The problem with this is that it is currently returning duplicate data in the result. 1 row for every duplicate entry in the 'lesson fee' table. I am also joining the 'fee type' table using this 'fee_type_id' Is there a way of telling MySQL to say "Join the lesson_fee table rows that have lesson_fee.lesson_type_id and fee_type_id = client.fee_type_id". UPDATE: Query: SELECT lesson_booking.lesson_booking_id,lesson_fee.lesson_fee_amount FROM fee_type INNER JOIN (lesson_fee INNER JOIN (color_code INNER JOIN (employee INNER JOIN (horse_owned INNER JOIN (lesson_type INNER JOIN (timetable INNER JOIN (lesson_booking INNER JOIN CLIENT ON client.client_id = lesson_booking.client_id) ON lesson_booking.timetable_id = timetable.timetable_id) ON lesson_type.lesson_type_id = timetable.lesson_type_id) ON horse_owned.horse_owned_id = lesson_booking.horse_owned_id) ON employee.employee_id = timetable.employee_id) ON employee.color_code_id = color_code.color_code_id) ON lesson_fee.lesson_type_id = lesson_type.lesson_type_id) ON lesson_fee.fee_type_id = client.fee_type_id WHERE booking_date = '2010-04-06' ORDER BY lesson_booking_id ASC How do I keep the format(indentation) of my query?

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