读取弱指针在被解除分配时是否安全?

时间:2015-06-05 18:08:36

标签: objective-c automatic-ref-counting race-condition

从不同的线程读取非原子弱指针是否安全,而不是取消分配对象?

通常,我知道只要可以同时访问,属性就应该是原子的,其中至少有一个是写操作。但我不得不怀疑ARC的写操作(将指针设置为nil)是否有点特殊。否则,我希望看到有关此可能问题的更多警告。也许弱指针隐含原子?

1 个答案:

答案 0 :(得分:4)

很安全。访问弱指针和归零弱指针位于spinlock_lock和spinlock_unlock之间。

看一下运行时源代码 http://opensource.apple.com/source/objc4/objc4-646/runtime/NSObject.mm

访问弱指针

id
objc_loadWeakRetained(id *location)
{
    id result;

    SideTable *table;
    spinlock_t *lock;

 retry:
    result = *location;
    if (!result) return nil;

    table = SideTable::tableForPointer(result);
    lock = &table->slock;

    spinlock_lock(lock);
    if (*location != result) {
        spinlock_unlock(lock);
        goto retry;
    }

    result = weak_read_no_lock(&table->weak_table, location);

    spinlock_unlock(lock);
    return result;
}

归零弱指针

void 
objc_object::sidetable_clearDeallocating()
{
    SideTable *table = SideTable::tableForPointer(this);

    // clear any weak table items
    // clear extra retain count and deallocating bit
    // (fixme warn or abort if extra retain count == 0 ?)
    spinlock_lock(&table->slock);
    RefcountMap::iterator it = table->refcnts.find(this);
    if (it != table->refcnts.end()) {
        if (it->second & SIDE_TABLE_WEAKLY_REFERENCED) {
            weak_clear_no_lock(&table->weak_table, (id)this);
        }
        table->refcnts.erase(it);
    }
    spinlock_unlock(&table->slock);
}

对象解除分配流程 https://stackoverflow.com/a/14854977/629118