在一个pthread_once方法内又再次调用了这个pthread_once导致死锁。 分析下这个pthread_once的源码: 可以看到这个pthread_once_t结构体就是一个整形数字加自旋锁。 int___pthread_once (pthread_once_t*once_control,void(*init_routine) (void)){/* Fast path. See __pthread_once_slow. */intval;val = ato...
可以看到這個pthread_once_t結構體就是一個整形數字加自旋鎖。 int ___pthread_once (pthread_once_t *once_control, void (*init_routine) (void)) { /* Fast path. See __pthread_once_slow. */ int val; val = atomic_load_acquire (once_control); if (__glibc_likely ((val & __PTHREAD_O...
#1 futex_wait_simple (private=0, expected=1, futex_word=0x25a1b1c <bthread::init_sys_mutex_lock_once>) at ../sysdeps/nptl/futex-internal.h:172 #2 __pthread_once_slow (once_control=0x25a1b1c <bthread::init_sys_mutex_lock_once>, init_routine=0x12cfb30 <bthread::init_sys_mutex_loc...
pthread_once_t _pthread_tls_once; DWORD _pthread_tls; static int _pthread_once_raw(pthread_once_t *o, void (*func)(void)) { long state = *o; _ReadWriteBarrier(); while (state != 1) { if (!state) { if (!_InterlockedCompareExchange(o, 2, 0)) { /* Success */ func(); ...
Once a mutex is created, it cannot be validly copied or moved to a new location. If the mutex is copied or moved to a new location, the new object is not valid and should not be used. Any attempt to use the invalid object will produce unpredictable results....
sync.(*Once).doSlow(0x1458d54, 0x8) /usr/lib/go/src/sync/once.go:68 +0xd2 sync.(*Once).Do(...) /usr/lib/go/src/sync/once.go:59 crypto/x509.systemRootsPool(...) /usr/lib/go/src/crypto/x509/root.go:22 crypto/x509.(*Certificate).Verify(0xc0005b9600, {{0x0, 0x0}, 0x...
Instead, on first use, pthread_mutex_lock() or pthread_mutex_trylock() branches into a slow path and causes the initialization of the mutex. Because a mutex is not just a simple memory object and requires that some resources be allocated by the system, an attempt to call pthread_mutex_...
进入_dispatch_root_queues_init源码实现,发现是一个dispatch_once_f单例(这里不作说明,以后重点分析),其中传入的func是_dispatch_root_queues_init_once。 8:_dispatch_root_queues_init_once dispatch_root_queues_init_once 开始进入底层os的处理了。通过断点,bt打印堆栈信息如下: ...
Once both threads are over, main() will measure the time again and print difference between two measurements. This difference is how we measure effectiveness. The less time it took to run, the better. Lets run the code and see how it performs, first with mutex and then with spinlock. ~...
// Create a mutex for the thread in TLS_SLOT_SELF to wait on once it starts so we can keep // it from doing anything until after we notify the debugger about it start_mutex = (pthread_mutex_t *) &tls[TLS_SLOT_SELF]; pthread_mutex_init(start_mutex, NULL); ...