race-condition

问题 When I took a look at the reference of 'Launching-Jobs' in gnu.org, I didn't get this part. The shell should also call setpgid to put each of its child processes into the new process group. This is because there is a potential timing problem : each child process must be put in the process group before it begins executing a new program , and the shell depends on having all the child processes in the group before it continues executing. If both the child processes and the shell call setpgid,

问题 I plan to use redis as an unique atomic id generator. However, my concern there might be simulatoneous web requests from multiple browsers. I was wondering, what is the common practice to make the following operations atomic? get id from redis if id is not found insert id as 0 into redis else store the id in a variable increase id by one store the new id back to redis If I were in desktop app or mobile app, I would use synchronized keyword in Java to avoid race condition. However, how about

问题 Given a simplified table structure like this: CREATE TABLE t1 ( id INT, num INT, CONSTRAINT t1_pk PRIMARY KEY (id), CONSTRAINT t1_uk UNIQUE (id, num) ) Can I use a subquery like this for inserting records without causing a race condition? INSERT INTO t1 ( id, num ) VALUES ( 1, ( SELECT MAX(num) + 1 FROM t1 ) ) Or are subqueries not atomic? I'm worried about simultaneous INSERT s grabbing the same value for num and then causing a unique constraint violation. 回答1: Yes, this can most certainly

问题 I have a situation where I need to check if the other side of a fifo has opened it, however I can't use a open because otherwise the program will start doing stuff. Why I have to do this: I have a program (a monitor) that launches a server program (both created by me). The monitor uses this fifo to communicate beacause the monitor can be closed/reopened while the server is already started. The problem is when the monitor starts the server: in this case I have to wait in some way for fifos to

问题 I'll present some code first since explaining is easier that way. Assume that mutexes are correctly used with the condition variables to keep it simple: // Thread 1 while(1) { conditionVariable.wait(); // Do some work } // Thread 2 while(1) { // Do some work conditionVariable.notify_one(); } // Thread 3 while(1) { // Do some work conditionVariable.notify_one(); } What I would like to achieve is that thread 1 is guaranteed to be waiting on the condition variable when thread 2 or Thread 3

问题 Write a class named RaceHorse that extends Thread. Each RaceHorse has a name and run() method that displays the name 5000 times. Write a Java application that instantiates 2 RaceHorse objects. The last RaceHorse to finish is the loser. This is the question. I have written the code for the two classes two run the thread Here are the codes: RaceHorse class RaceHorse extends Thread { public String name; public RaceHorse(String name) { this.name = name; } public void run() { for(int i = 1 ; i <=

问题 The definition of a race condition: A race condition or race hazard is a flaw in a system or process whereby the output or result of the process is unexpectedly and critically dependent on the sequence or timing of other events. Consider the following pseudocode: Global variable i initialized to 6; Thread 1: acquire(lock l) increment global variable i, i.e. i++; Thread 2: acquire(lock l) double the value of global var i, i.e.: i*=2; If T1 acquires the lock l first and T2 second the value of i

问题 I'm building an SMS-enabled app that allows the user to communicate with an external contact. As incoming messages arrive, I save them to my database so that they can be displayed in the app. In testing, everything is working great for one-off messages, but I'm having trouble with messages longer than 160 characters, which are actually broken up into multiple messages. When an incoming message comes in via SMS, I'm checking whether we already got a message from that number within the past few

问题 I'm hitting a race-condition in my app, where all or all but 1 threads get stuck on syscall_thread_switch whenever I pause debugging. It reproduces much more often on the simulator, but also on the iPad Air. There is ALWAYS at least 2 threads stuck in CocoaLumberjack's queueLogMessage: -- see screenshots. I've never seen this before on 8.1 and 8.2, but i'm hitting it often on 8.3. I'm not claiming this is an 8.3 bug :) This is a level of complexity i've never had to debug before, so i'm not

问题 I thought of trying out thread sanitizer ( http://code.google.com/p/data-race-test/wiki/ThreadSanitizer#Using_ThreadSanitizer ) so I made a simple program: #include <thread> #include <atomic> #include <vector> #include <iostream> #include <algorithm> #include <mutex> using namespace std; int violated=0; mutex mtx; void violator() { lock_guard<mutex> lg(mtx); violated++; } int main() { thread t1(violator); t1.join(); thread t2(violator); t2.join(); } AFAIK program is OK since access to