给java中的十六进制整数带来很大的价值？

Question

I am actually reading int values from the java socket bufferedread and later converting into hex representation. I am using a method as below.

StringBuilder sb = new StringBuilder();
sb.append(Integer.toHexString(nextChar));

if (sb.length() < 2) {
    sb.insert(0, '0'); //pad with leading zero if needed
}

String hexChar = sb.toString();
System.out.println("\n\n hex value is "+hexChar +"   "+"Int value is:"+nextChar);

The codes are working fine. Only I got one or two data which in normal case after conversion into hex is either BE or A9 but I am getting ffdd and the integer value is 65533. Could it be my conversion method is wrong or is it input value is it self is having that value?

Answer 1

It's not really clear what you're trying to do, but as defined in the documentation the return from read is: The character read, as an integer in the range 0 to 65535 (0x00-0xffff), or -1 if the end of the stream has been reached.

This is of course because you are using a text reader.

You might want to read this answer .

OK, Got it.

You thought you are reading bytes, but since you are using BufferedReader you actually read chars ( The char data type is a single 16-bit Unicode character. It has a minimum value of '\' (or 0) and a maximum value of '\￿' (or 65,535 inclusive). )

I believe that this is what you wanted to achieve. Please note this is not the best way, its just a sample!

InputStream is = sock.getInputStream()

BufferedInputStream bis = new BufferedInputStream(is);

int nextByte;
StringBuilder sb = new StringBuilder();

while( (nextByte = bis.read()) != -1 ) {
    sb.append(String.format("%02X ", nextByte));
}