semaphore | 易学教程

Wait for completion block of writeImageToSavedPhotosAlbum by semaphore

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问题 In my app I open the camera by a picker and after the photo has been taken I'd like to safe it by the following method the assets library. The method freezes after the call of the writeImageToSavedPhotosAlbum. Without the semaphores the methods work perfectly. But than I miss to receive the assetURL. + (NSURL*)safeImageToAssetsLibrary:(UIImage *)image metadata:(NSDictionary *)metadata { ALAssetsLibrary *library = [[ALAssetsLibrary alloc] init]; __block NSURL *retAssestURL = nil; dispatch

利用信号量实现线程同步

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本篇使用信号量机制实现对全局资源的正确使用，包括以下两点：各个子线程对全局资源的互斥使用主线程对子线程的同步信号量简单的说，信号量内核对象，也是多线程同步的一种机制，它可以对资源访问进行计数，包括最大资源计数和当前资源计数，是两个32位的值；另外，计数是以原子访问的方式进行，由操作系统维护；最大资源计数，表示可以控件的最大资源数量当前资源计数，表示当前可用资源的数量信号量的规则：如果当前资源计数器大于0，那么信号量处于触发状态如果当前资源计数器等于0，那么信号量处于未触发状态系统绝对不会让当前资源计数器变为负数当前资源计数器决定不会大于最大资源计数信号量机制：以一个停车场的运作为例。假设停车场只有三个车位，一开始三个车位都是空的。这时如果同时来了五辆车，看门人允许其中三辆直接进入，然后放下车拦，剩下的车则必须在入口等待，此后来的车也都不得不在入口处等待。这时，有一辆车离开停车场，看门人得知后，打开车拦，放入外面的一辆进去，如果又离开两辆，则又可以放入两辆，如此往复。在这个停车场系统中，车位是公共资源，每辆车好比一个线程，看门人起的就是信号量的作用，当当前资源计数大于0，信号量处于触发状态，线程编程可调度；相关API 创建信号量 HANDLE WINAPI CreateSemaphore( LPSECURITY_ATTRIBUTES

利用信号量实现线程同步

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Windows Using Semaphore Objects

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The following example uses a semaphore object to limit the number of threads that can perform a particular task. First, it uses the CreateSemaphore function to create the semaphore and to specify initial and maximum counts, then it uses the CreateThread function to create the threads. Before a thread attempts to perform the task, it uses the WaitForSingleObject function to determine whether the semaphore's current count permits it to do so. The wait function's time-out parameter is set to zero, so the function returns immediately if the semaphore is in the nonsignaled state.

What happens to mutex acquirement in wait()/signal() block?

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问题 So the description of the exercise: You have this restaurant in which there are N points where you can ask for a portion of fries. Each point has M portions. There is 1 frycheff. When an order-point has 2 portions, it warns the cheff that it needs a refill. The cheff delivers the portions in order of FIFO. We made this pseudo-code: init { Semafoor[] mutex; Condition[] cond_point = new Condition[N]; int[] portions = new int[N]; ArrayList<int> waitline = new ArrayList<int>(); for(int i = 0; i <

Semaphore signalling vs mutex

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问题 I was looking over some topics that describes the difference between mutex and binary semaphore. In many topics it is stated that semaphore act as a signalling mechanism i.e if a thread has locked a semaphore then another thread can unlock(release) the semaphore(acting as signal). But in case of mutex only the thread that locks the mutex can unlock it .It can't be unlocked by any other thread and if other thread tries to unlock it this will return error. I have tried writing a code that uses

Semaphores and shared memory

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问题 I have a question regarding multiprocess programming in C, I have several reader processes that will be reading from a file into a shared buffer and several writer processes reading from the buffer and into another file, what type of semaphores will we need to use for this. and how can we use shared memory with the semaphores. 回答1: If you're on linux, one easy option is to use pshared mutexes and condition variables. A recet version of glibc will be necessary. Essentially inside your shared

wait until the previous instance of process finish

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问题 I am new to the world of Linux C Programming so request your patience. Found several threads about inter-process synchronization (same process but different instance) using mutex and semaphores but not exactly matching my case. I tried to follow them and tried to create few samples but none of them is working for me. Finally posting here to get some help. I am working on to create a utility which will executed over Ethernet telnet session. As noted below in USAGE comment, 1st call will pass

Java Semaphore maximum?

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问题 Is there a way of knowing what is the MAXIMUM number of permits that a semaphore object has ever had in its' lifetime? We initialize it like this: Semaphore sem = new Semaphore(n); and at times we acquire, and at times we release what we acquired. But there are certain situations when we need to release more than we acquired in order to increase the number of permits. Is there a way to know the MAXIMUM number of permits that ever was in this semaphore? 回答1: The constructor is defined as

Semaphores for fortran

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问题 Can anyone help with writing a semaphore function in fortran? I have multiple processes running and I have to synchronize them using semaphore. Such a code could be found for C++ etc., but I could not find any such code for fortran. If I could call C/C++ function from fortran code, that would also suffice, since C++ semaphore function is already there. PS: (Additional Explanation) Here is the code which works for C++. I have some fortran applications (as part of standard benchmark) but no

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