scheduling

Java: waiting on synchronized block, who goes first?

╄→гoц情女王★ 提交于 2019-11-26 21:33:29
问题 This question is inspired by this other question. If multiple threads are waiting on a synchronized block, and the lock becomes available, who goes first? Is it by thread priority (and then first-come-first-served)? And do the same rules apply for notify (with multiple wait ing threads)? 回答1: According to this guy: http://tutorials.jenkov.com/java-concurrency/starvation-and-fairness.html Java issues no guarantees about the sequence. So I guess it is not based on thread priority I'll try to

Best Fit Scheduling Algorithm

橙三吉。 提交于 2019-11-26 20:17:26
I'm writing a scheduling program with a difficult programming problem. There are several events, each with multiple meeting times. I need to find an arrangement of meeting times such that each schedule contains any given event exactly once, using one of each event's multiple meeting times. Obviously I could use brute force, but that's rarely the best solution. I'm guessing this is a relatively basic computer science problem, which I'll learn about once I am able to start taking computer science classes. In the meantime, I'd prefer any links where I could read up on this, or even just a name I

Does linux schedule a process or a thread?

与世无争的帅哥 提交于 2019-11-26 19:30:34
问题 After reading this SO question I got a few doubts. Please help in understanding. Scheduling involves deciding when to run a process and for what quantum of time. Does linux kernel schedule a thread or a process? As process and thread are not differentiated inside kernel how a scheduler treats them? How quantum for each thread is decided? a. If a quantum of time (say 100us) is decided for a process is that getting shared between all the threads of the process? or b. A quantum for each thread

Quartz Java resuming a job excecutes it many times

落花浮王杯 提交于 2019-11-26 18:59:39
问题 For my application i create jobs and schedule them with CronTriggers. Each job has only one trigger and both the job name and the trigger names are the same. No jobs share a trigger. Now when i create a cron trigger like this "0/1 * * * * ?" which instructs the job to execute every second, it works just fine. The problem rises when i first pause the job by calling : scheduler.pauseJob(jobName, jobGroup); and then resuming the job after let's say 50 seconds with : scheduler.resumeJob(jobName,

How to instruct cron to execute a job every second week?

∥☆過路亽.° 提交于 2019-11-26 18:57:19
I would like to run a job through cron that will be executed every second Tuesday at given time of day. For every Tuesday is easy: 0 6 * * Tue But how to make it on "every second Tuesday" (or if you prefer - every second week)? I would not like to implement any logic in the script it self, but keep the definition only in cron. xahtep How about this, it does keep it in the crontab even if it isn't exactly defined in the first five fields: 0 6 * * Tue expr `date +\%W` \% 2 > /dev/null || /scripts/fortnightly.sh pilcrow Answer Modify your Tuesday cron logic to execute every other week since the

How does select work when multiple channels are involved?

女生的网名这么多〃 提交于 2019-11-26 18:34:53
问题 I found when using select on multiple non buffered channels like select { case <- chana: case <- chanb: } Even when both channels have data, but when processing this select, the call that falls in case chana and case chanb is not balanced. package main import ( "fmt" _ "net/http/pprof" "sync" "time" ) func main() { chana := make(chan int) chanb := make(chan int) go func() { for i := 0; i < 1000; i++ { chana <- 100 * i } }() go func() { for i := 0; i < 1000; i++ { chanb <- i } }() time.Sleep

Influence on the static scheduling overhead in OpenMP

坚强是说给别人听的谎言 提交于 2019-11-26 18:33:49
问题 I thought about which factors would influence the static scheduling overhead in OpenMP. In my opinion it is influenced by: CPU performance specific implementation of the OpenMP run-time library the number of threads But am I missing further factors? Maybe the size of the tasks, ...? And furthermore: Is the overhead linearly dependent on the number of iterations? In this case I would expect that having static scheduling and 4 cores, the overhead increases linearly with 4*i iterations. Correct

Recommended method for loading a URL via a scheduled task on Windows

[亡魂溺海] 提交于 2019-11-26 17:09:36
I have a webpage hosted on a Windows box that I need to assure gets loaded at least once/day. My current plan is to create a scheduled task that opens Internet Explorer and hits the URL: "C:\Program Files\Internet Explorer\iexplore.exe" myurl.com/script_to_run_daily.aspx This was simple to setup and works fine, but it strikes me as a hack because Internet Explorer actually has to open and hit this URL. I don't need any input back from this page, it simply stores cached data in files when it's hit. Is there a slicker way of doing this? In case it matters, this is a VB.net site. Thanks in

How does DateTimeOffset deal with daylight saving time?

夙愿已清 提交于 2019-11-26 15:51:55
问题 I am storing schedules in the database as a day of the week, hour and minute. When the data is read we create a DateTime object for the next occurrence of that day, hour and minute, but I need to modify this to be DST-aware. I am able to modify the database if necessary. I know that DateTimeOffset stores a UTC date/time and an offset. I also know from this MSDN blog entry that DateTimeOffset should be used to "Work with daylight saving times". What I'm struggling to understand is exactly how

What is an uninterruptable process?

こ雲淡風輕ζ 提交于 2019-11-26 15:34:58
Sometimes whenever I write a program in Linux and it crashes due to a bug of some sort, it will become an uninterruptable process and continue running forever until I restart my computer (even if I log out). My questions are: What causes a process to become uninterruptable? How do I stop that from happening? This is probably a dumb question, but is there any way to interrupt it without restarting my computer? An uninterruptable process is a process which happens to be in a system call (kernel function) that cannot be interrupted by a signal. To understand what that means, you need to