Question

我有以下示例代码（请参阅下面的代码），它由线程执行：

A: rd-lock
B: wr-lock (waiting)
A: rd-lock (recursive)
A: rd-unlock (recursive)
A: rd-unlock
B: wr-locked (wake after wait)
B: wr-unlock.

基本上，read-lock是递归的。它是POSIX标准所要求的（要求读锁是递归的，但没有为写锁指定）。这有效在Linux，FreeBSD，Solaris上，但它不适用于Darwin / Mac OS X.

以下示例在Linux上提供以下输出：

read locking
read locked
write locking
read locking 2
read locked 2
read unlocked 2
read unlocked
write locked
write unlocked 2

在达尔文上打印时：

read locking
read locked
write locking
read locking 2

这里的死锁（不会继续），基本上它不尊重递归读锁。

是否可以按预期工作（标记，定义，链接特殊库版本）？

示例代码

#include <pthread.h>
#include <stdio.h>

pthread_rwlock_t lock;

void *thread_r(void *p)
{
    printf("read locking\n");
    pthread_rwlock_rdlock(&lock);
    printf("read locked\n");
    usleep(500*1000);
    printf("read locking 2\n");
    pthread_rwlock_rdlock(&lock);
    printf("read locked 2\n");
    usleep(500*1000);
    pthread_rwlock_unlock(&lock);
    printf("read unlocked 2\n");
    usleep(500*1000);
    pthread_rwlock_unlock(&lock);
    printf("read unlocked\n");
}

void *thread_w(void *p)
{
    usleep(250*1000);
    printf("write locking\n");
    pthread_rwlock_wrlock(&lock);
    printf("write locked\n");
    pthread_rwlock_unlock(&lock);
    printf("write unlocked 2\n");
}

int main()
{
    pthread_t a,b;
    pthread_rwlock_init(&lock,NULL);
    pthread_create(&a,NULL,thread_r,0);
    pthread_create(&b,NULL,thread_w,0);
    pthread_join(a,NULL);
    pthread_join(b,NULL);
    return 0;
}