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请问swoole的table强制解锁中，把lock = 1是正确的吗

### 问题描述
你好，我在看swoole_table代码的时候发现加锁的代码中遇到进程异常退出或者长时间持有锁的情况下，会强制解锁。

```cpp
void TableRow::lock() {
    sw_atomic_t *lock = &lock_;
    uint32_t i, n;
    long t = 0;

while (1) {
		// 省略
        /**
         * The process occupied by the resource no longer exists,
         * indicating that OOM occurred during the locking process,
         * forced to unlock
         */
        if (kill(lock_pid, 0) < 0 && errno == ESRCH) {
            *lock = 1;
            swWarn("lock process[%d] not exists, force unlock", lock_pid);
            goto _success;
        }
        /**
         * Mark time
         */
        if (t == 0) {
            t = swoole::time<std::chrono::milliseconds>(true);
        }
        /**
         * The deadlock time exceeds 2 seconds (SW_TABLE_FORCE_UNLOCK_TIME),
         * indicating that the lock process has OOM,
         * and the PID has been reused, forcing the unlock
         */
        else if ((swoole::time<std::chrono::milliseconds>(true) - t) > SW_TABLE_FORCE_UNLOCK_TIME) {
            *lock = 1;
            swWarn("timeout, force unlock");
            goto _success;
        }
        sw_yield();
    }
}
```

这里进程退出或者持有时间超时会强制解锁，这里是把lock设置为1。

但是正常解锁逻辑的解锁操作是：
```cpp
    void unlock() {
        sw_spinlock_release(&lock_);
    }

#define sw_spinlock_release(lock) *(lock) = 0
```

这里是直接把lock = 0表示解锁。
请问这两者冲突吗，谢谢。

发布于4年前 · 7 次浏览 · 来自提问