Java: convert int to InetAddress

user_99876 0
Viewed 37758

I have an int which contains an IP address in network byte order, which I would like to convert to an InetAddress object. I see that there is an InetAddress constructor that takes a byte[], is it necessary to convert the int to a byte[] first, or is there another way?

inetaddress java
10 Answers

This should work:

int ipAddress = ....
byte[] bytes = BigInteger.valueOf(ipAddress).toByteArray();
InetAddress address = InetAddress.getByAddress(bytes);

You might have to swap the order of the byte array, I can't figure out if the array will be generated in the correct order.

user_212340

Not enough reputation to comment on skaffman's answer so I'll add this as a separate answer.

The solution skaffman proposes is correct with one exception. BigInteger.toByteArray() returns a byte array which could have a leading sign bit.

byte[] bytes = bigInteger.toByteArray();

byte[] inetAddressBytes;

// Should be 4 (IPv4) or 16 (IPv6) bytes long
if (bytes.length == 5 || bytes.length == 17) {
    // Remove byte with most significant bit.
    inetAddressBytes = ArrayUtils.remove(bytes, 0);
} else {
    inetAddressBytes = bytes;
}

InetAddress address = InetAddress.getByAddress(inetAddressBytes);

PS above code uses ArrayUtils from Apache Commons Lang.

user_2360920

Since comments cannot be formatted, let me post the code derived from the comment by @Mr.KevinThomas:

if (ByteOrder.nativeOrder().equals(ByteOrder.LITTLE_ENDIAN)) {
    ipAddress = Integer.reverseBytes(ipAddress);
}
sReturn = String.format(Locale.US, "%d.%d.%d.%d", (ipAddress >> 24 & 0xff), (ipAddress >> 16 & 0xff), (ipAddress >> 8 & 0xff), (ipAddress & 0xff));

It has been tested on Android.

user_3554560

This may work try


public static String intToIp(int i) {
        return ((i >> 24 ) & 0xFF) + "." +
               ((i >> 16 ) & 0xFF) + "." +
               ((i >>  8 ) & 0xFF) + "." +
               ( i        & 0xFF);
    }
user_1754220
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