DiscoveryMessageSequence.Inequality 运算符
定义
重要
一些信息与预发行产品相关,相应产品在发行之前可能会进行重大修改。 对于此处提供的信息,Microsoft 不作任何明示或暗示的担保。
确定 DiscoveryMessageSequence 的两个指定实例是否不相等。
public:
static bool operator !=(System::ServiceModel::Discovery::DiscoveryMessageSequence ^ messageSequence1, System::ServiceModel::Discovery::DiscoveryMessageSequence ^ messageSequence2);
public static bool operator != (System.ServiceModel.Discovery.DiscoveryMessageSequence messageSequence1, System.ServiceModel.Discovery.DiscoveryMessageSequence messageSequence2);
static member op_Inequality : System.ServiceModel.Discovery.DiscoveryMessageSequence * System.ServiceModel.Discovery.DiscoveryMessageSequence -> bool
Public Shared Operator != (messageSequence1 As DiscoveryMessageSequence, messageSequence2 As DiscoveryMessageSequence) As Boolean
参数
- messageSequence1
- DiscoveryMessageSequence
发现消息序列。
- messageSequence2
- DiscoveryMessageSequence
发现消息序列。
返回
如果两个发现消息序列实例不相等,则为 true
;否则为 false
。