public class InetAddresTest {
public static void main(String ... agrs) {
try {
InetAddress inet = InetAddress.getByName("1.2");
System.out.println("Good ip address");
} catch (UnknownHostException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
顺便说一下,InetAddress产生的ip地址回复为“1.0.0.2”.我无法从InetAddress的javadoc中找到合理的答案.你能解释一下这种行为吗？
@H_502_6@解决方法
从
Javadoc(在
Javadoc for InetAddress中以“IP地址的文本表示”链接)：
When a two part address is supplied,the last part is interpreted as a 24-bit quantity and placed in the right most three bytes of the network address. This makes the two part address format convenient for specifying Class A network addresses as net.host.
编辑添加：如果24bit部分让您感到困惑：
24位中的2将看起来像：00000000 00000000 00000010
然后将其映射到IPv4地址中右边的3个八位字节：.0.0.2
另外一点：正如CoolBeans在对你的问题的评论中提到的那样,来自Apache commons的InetAddressValidator就可以解决问题了.话虽这么说,如果你只想验证IP地址而不是外部依赖,你可以使用Regular Expressions to check IP addresses as well
@H_502_6@ @H_502_6@ 原文链接：https://www.f2er.com/java/239741.html