Suppose I have the IP stored in a String:
String ip = "192.168.2.1"
and I want to get the byte array with the four ints. How can I do it? Thanks!
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3 Answers
Something like this:
InetAddress ip = InetAddress.getByName("192.168.2.1");
byte[] bytes = ip.getAddress();
for (byte b : bytes) {
System.out.println(b & 0xFF);
}
3 Comments
Inv3r53
oh and btw the masking with 0xFF is for values over 127
c.sankhala
in this case, developer have to handle UnknownHostException
Kango_V
From the javadoc: "If a literal IP address is supplied, only the validity of the address format is checked." Doh, of course you still have to catch it :(
Each number is a byte, so in your case the appropriate byte[] would be { 192, 168, 2, 1 }.
To be more specific, if you have the string, you first have to split it by the "."s and then parse a byte from each resulting string.
6 Comments
eternay
A byte has a maximum value of 127. How can you put 192 in this array?
Michael Campbell
If only there were an unsigned byte type in java.
RB_
There is a library from google - Guava. Add it and use like this: import com.google.common.primitives.UnsignedBytes
Sam Ginrich
@eternay it is (-64) & 0xFF == 192, -64 == (byte)192
Greelings
The most efficient way. It needs only 4 bytes.
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This works well for me (Kotlin):
open fun reachable(host: String?): Boolean { // 'host' is string, e.g., "177.111.155.11"
return try {
val ipAddress = InetAddress.getByName(host) // get IP address
ipAddress.isReachable(2000) // Is it reachable? T or F
} catch (e: IOException) {
CoroutineScope(Main).launch {
myNote("MyWX LAN Reachable Error", e.message!!) // Display error
}
false
}
}
