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I know that unsigned byte is missing in Java Then how can I initialize the byte array using integer from 0 to 255 (in hex)?
final byte assoc_resp_msg_int[] = new byte[] {
0xe3, 0x00, //APDU CHOICE Type(AareApdu)
0x00, 0x2c, //CHOICE.length = 44
0x00, 0x00, //result=accept
0x50, 0x79, //data-proto-id = 20601
0x00, 0x26, //data-proto-info length = 38
0x80, 0x00, 0x00, 0x00, //protocolVersion
0x80, 0x00, //encoding rules = MDER
0x80, 0x00, 0x00, 0x00, //nomenclatureVersion
0x00, 0x00, 0x00, 0x00, //functionalUnits, normal Association
0x80, 0x00, 0x00, 0x00, //systemType = sys-type-manager
0x00, 0x08, //system-id length = 8 and value (manufacturer- and device- specific)
0x88, 0x77, 0x66, 0x55, 0x44, 0x33, 0x22, 0x11,
0x00, 0x00, //Manager's response to config-id is always 0
0x00, 0x00, //Manager's response to data-req-mode-flags is always 0
0x00, 0x00, //data-req-init-agent-count and data-req-init-manager-count are always 0
0x00, 0x00, 0x00, 0x00, //optionList.count = 0 | optionList.length = 0
};
286
To convert byte array to a hex value, we loop through each byte in the array and use String 's format() . We use %02X to print two places ( 02 ) of Hexadecimal ( X ) value and store it in the string st . This is a relatively slower process for large byte array conversion.
Byte to Hexadecimal. The bytes are 8 bit signed integers in Java. Therefore, we need to convert each 4-bit segment to hex separately and concatenate them. Consequently, we'll get two hexadecimal characters after conversion.
The java. util. Random. nextBytes() method generates random bytes and provides it to the user defined byte array.
You can simply iterate the byte array and print the byte using System. out. println() method.
You have to store 0x80 in byte like this :
final byte assoc_resp_msg_int[] = new byte[] {
(byte)0xe3, 0x00, //APDU CHOICE Type(AareApdu)
0x00, 0x2c, //CHOICE.length = 44
0x00, 0x00, //result=accept
0x50, 0x79, //data-proto-id = 20601
0x00, 0x26,
(byte)0x80,
...
}
System.out.println(assoc_resp_msg_int[10]&0xFF);
//128
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