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use count=0 instead of 1 for recursive mutex with only one lock reference

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this simplifies the code paths slightly, but perhaps what's nicer is
that it makes recursive mutexes fully reentrant, i.e. locking and
unlocking from a signal handler works even if the interrupted code was
in the middle of locking or unlocking.
This commit is contained in:
Rich Felker 2011-10-03 00:09:08 -04:00
commit 7fe58d3511
2 changed files with 2 additions and 4 deletions
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2
src/thread/pthread_mutex_trylock.c
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@ -30,8 +30,6 @@ int pthread_mutex_trylock(pthread_mutex_t *m)
if ((own && !(own & 0x40000000)) || a_cas(&m->_m_lock, old, tid)!=old)
return EBUSY;
m->_m_count = 1;
if (m->_m_type < 4) return 0;
if (m->_m_type >= 8) {

4
src/thread/pthread_mutex_unlock.c
View file

@ -12,8 +12,8 @@ int pthread_mutex_unlock(pthread_mutex_t *m)
self = __pthread_self();
if ((m->_m_lock&0x1fffffff) != self->tid)
return EPERM;
if ((m->_m_type&3) == PTHREAD_MUTEX_RECURSIVE && --m->_m_count)
return 0;
if ((m->_m_type&3) == PTHREAD_MUTEX_RECURSIVE && m->_m_count)
return m->_m_count--, 0;
if (m->_m_type >= 4) {
self->robust_list.pending = &m->_m_next;
*(void **)m->_m_prev = m->_m_next;