Tipstir is asking for the subnet mask and that is required to properly answer your question.IP 127 . 1 . 1 . 8 and IP 127 . 10 . 1 . 8
Pinging 127.10.1.8 with 32 bytes of data:
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
Ping statistics for 127.10.1.8:
Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),
Approximate round trip times in milli-seconds:
Minimum = 0ms, Maximum = 0ms, Average = 0ms
C:\Users\Jeff>ping 127.3.2.1
Pinging 127.3.2.1 with 32 bytes of data:
Reply from 127.3.2.1: bytes=32 time<1ms TTL=128
Reply from 127.3.2.1: bytes=32 time<1ms TTL=128
Reply from 127.3.2.1: bytes=32 time<1ms TTL=128
Reply from 127.3.2.1: bytes=32 time<1ms TTL=128
Hold on before everyone hijacks this thread.
I was told the way to tell if two IPs are on the same network is that the range
- for class A (0.0.0.0-127.255.255.255) only needs the first octet to be the same. For example 126.xxx.xxx.xxx is on the same network as 125.yyy.yyy.yyy.
- For class B the first two octets must be the same if it's on the same network (eg 172.16.x.x)
- for class C the first 3 octets must be on the same network. For example 192.168.2.1 and 192.168.3.2 are on different networks but 192.168.2.1 and 192.168.2.2 are on the same network. Is this true?
NOT with a mask of 255.255.255.0 - the mask would need to be 255.255.254.0192.168.0.xxx and 192.168.1.xxx can be on the same network too..
NOT with a mask of 255.255.255.0 - the mask would need to be 255.255.254.0