Linux - NetworkingThis forum is for any issue related to networks or networking.
Routing, network cards, OSI, etc. Anything is fair game.
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A more explicit answer is that an ip address is composed of 32 bits. Of those 32 bits, you have the network address and the host address. The net mask defines the break point between the two.
i.e. /24 or 255.255.255.0 are two ways of saying that the network address starts 24 bits from the left... i.e. 192.168.0.?
That means the standard netmasks are
/8=255.0.0.0 example 192.?.?.? (Class A)
/16=255.255.0.0 example 192.168.?.? (Class B)
/24=255.255.255.0 example 192.168.0.? (Class C)
Now... if you want to get into splitting network classes, you'll have to do some math and that requires you understand why 255 is 8 bits. We can go there if you like but I won't bother unless you ask.
Correct. And I notice you listed the range of ip's from 64-70, rather than 64-71; Judging by your immediate grasp of addressing you neglected to include 71 because it is the broadcast address. Good stuff.
Distribution: OpenBSD 4.6, OS X 10.6.2, CentOS 4 & 5
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You also need to make sure that a subnet starts on an allowed boundary. You can't just start a block of contigous addresses anywhere. Any network that you're subnetting has to break evenly into subnets. Using /29 subnets (say we are going to subnet 192.168.254.0 into /29 subnets) they would be 192.168.254.0/29, 192.168.254.8/29, 192.168.254.16/29, etc... so yes, with 192.168.254.64 you could use up to .70, using .71 for broadcast.
Also remember the basic rule that one less bit in the mask means twice as big an address range, and one more bit means half as big as previous.
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