c# - msft为什么将Interlocked.Increment(ref uniqueId)与零进行比较？

Question

我正在查看.NET 4.0的System.Threading.Tasks.TaskScheduler.Id实现，请参见以下代码:

[__DynamicallyInvokable]
public int Id
{
    [__DynamicallyInvokable]
    get
    {
        if (this.m_taskSchedulerId == 0)
        {
            int num = 0;
            do
            {
                num = Interlocked.Increment(ref s_taskSchedulerIdCounter);
            }
            while (num == 0);
            Interlocked.CompareExchange(ref this.m_taskSchedulerId, num, 0);
        }
        return this.m_taskSchedulerId;
    }
}

为什么互锁后msft比较num == 0？ Interlocked.Increment()的实现说它返回递增后的值(递增后)，因此检查零似乎是不必要的(除非计数器回绕，但是如果发生这种情况，您还有更大的问题，在这里也无法解决。

如果要这样做，我只会做:

public int Id
        {
            get
            {
                if(m_taskSchedulerId==0)
                {
                    var result = Interlocked.Increment(ref s_taskSchedulerIdCounter);
                    Interlocked.CompareExchange(ref m_taskSchedulerId, result, 0);
                }
                return m_taskSchedulerId;
            }
        }

Answer 1

but if that happens you have bigger problems

不，这就是他们这样做的确切原因。从Reference Source:

public Int32 Id
{
    get
    {
        if (m_taskSchedulerId == 0)
        {
            int newId = 0;

            // We need to repeat if Interlocked.Increment wraps around and returns 0.
            // Otherwise next time this scheduler's Id is queried it will get a new value
            do
            {
                newId = Interlocked.Increment(ref s_taskSchedulerIdCounter);
            } while (newId == 0);

            Interlocked.CompareExchange(ref m_taskSchedulerId, newId, 0);
        }

        return m_taskSchedulerId;
    }
}