静态pthread互斥锁初始化 [英] Static pthreads mutex initialization

问题描述

使用pthread，如何在C中初始化静态的互斥数组?

Using pthreads, how would one, in C, initialize a static array of mutexes?

对于单个静态互斥锁，似乎可以使用PTHREAD_MUTEX_INITIALIZER.但是它们的静态数组呢?例如，

For a single static mutex, it seems I can use PTHREAD_MUTEX_INITIALIZER. But what about an static array of them? As, in for example,

#include <pthread.h>
#define NUM_THREADS 5

/*initialize static mutex array*/
static pthread_mutex_t mutexes[NUM_THREADS] = ...?

还是必须动态分配它们?

Or must they be allocated dynamically?

推荐答案

如果您使用的是符合C99的编译器，则可以使用

If you have a C99 conforming compiler you can use P99 to do your initialization:

static pthread_mutex_t mutexes[NUM_THREADS] =
  { P99_DUPL(NUM_THREADS, PTHREAD_MUTEX_INITIALIZER) };

这只是将令牌序列PTHREAD_MUTEX_INITIALIZER,重复请求的次数.

This just repeats the token sequence PTHREAD_MUTEX_INITIALIZER, the requested number of times.

为此，您只需确保NUM_THREADS不会扩展为变量，而是扩展为预处理器可见的十进制整数常数，并且该常数不会太大.

For this to work you only have to be sure that NUM_THREADS doesn't expand to a variable but to a decimal integer constant that is visible to the preprocessor and that is not too large.

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