在共享内存中初始化pthread互斥锁 [英] Initializing a pthread mutex in shared memory

问题描述

我可以使用初始化程序在静态内存中初始化互斥锁:

I can initialize a mutex in static memory with an initializer:

pthread_mutex_t mymutex = PTHREAD_MUTEX_INITIALIZER;

但是如何在共享内存中初始化一个呢?在共享内存中，我必须与初始化变量分开来分配内存?我可以做memcpy()吗?

but how do I initialize one in shared memory where I have to allocate the memory separately from initializing the variable? Can I do a memcpy()?

pthread_mutex_t mymutex = PTHREAD_MUTEX_INITIALIZER;
memcpy(&globalmutex, &mymutex, sizeof(mymutex);

我以为我记得很久以前读过，当将一个互斥量分配到保证要初始化为零的内存中时，它不需要初始化-恰恰是在这种情况下-但我找不到写下来的内容任何地方.那是真的吗?-我注意到PTHREAD_MUTEX_INITIALIZER在我的redhat系统上定义为{{0，0，0，0，0，0，{0，0}}}.

I thought I remembered reading a long time ago that when allocating a mutex into memory guaranteed to be initialized to zero's, it doesn't need initialization--intended for exactly this case--but I can't find that written down anywhere. Is that really true?--I notice that PTHREAD_MUTEX_INITIALIZER is defined as { { 0, 0, 0, 0, 0, 0, { 0, 0 } } } on my redhat system.

推荐答案

在这里您需要小心，并非所有实现都支持可以在各个进程中工作的互斥体.

You need to be careful here, not all implementations support mutexes that can work across processes.

PThreads本身通过使用进程共享属性来支持此功能，但是，如果需要的话，将不会使用默认的初始化程序.

PThreads itself supports this by use of the process shared attribute but, if you want that, you won't be using the default initialiser.

相反，在正确构造互斥量属性结构之后，您将需要使用pthread_mutex_init():

Instead you'll need to use pthread_mutex_init() after properly constructing a mutex attribute structure:

int pthread_mutex_init(
    pthread_mutex_t *mutex,
    const pthread_mutexattr_t *attr);

最好的选择是只在共享内存中分配空间，然后将其强制转换为正确的类型，然后将其传递给init函数.我认为这可能比初始化后将其复制更为安全.

Best bet would be to just allocate space in shared memory then cast it to the correct type and pass it to the init function. I think that may be safer than copying it after the initialisation.

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