Wrong result using arraycopy

Question

I'm trying to copy two Byte arrays into another. The problem is that the first element of the result Byte is strange, I get 0xFFFFFF80 instead of 0x80. the code I'm using is:

    this.IC_SUBMIT_APDU = new byte[13];
    byte[] prefix = {
        (byte) 0x80,
        (byte) 0x20,
        (byte) 0x07,
        (byte) 0x00,
        (byte) 0x08
    };
    System.arraycopy(prefix , 0, this.IC_SUBMIT_APDU, 0, prefix.length);

    for(int i=0; i<this.IC_SUBMIT_APDU.length ; i++)
        System.out.println("" + Integer.toHexString(this.IC_SUBMIT_APDU[i]));

when I give it this argument:

{
    (byte) 0x41,
    (byte) 0x43,
    (byte) 0x4F,
    (byte) 0x53,
    (byte) 0x54,
    (byte) 0x45,
    (byte) 0x53,
    (byte) 0x54
}

it produces the following result:

WHy do I get that 0xFFFFFF80 ? am I not supposed to get 0x80 ??

You must read it from right to left two numbers at a time, so you have 80 FF FF FF. As said mattm in his answer, the FFFFFF part comes from somewhere else or is filler. — Turtle
– Turtle, Commented May 15, 2017 at 14:06

mattm · Accepted Answer · 2017-05-15 14:04:31Z

3

0xFFFFFF80 is more than one byte. The 0xFFFFFF portion is probably from a different print statement that you have not shown.

answered May 15, 2017 at 14:04

mattm

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1 Comment

noureddine-as Over a year ago

I have no print statement before this one.

Mena · Accepted Answer · 2017-05-15 14:05:55Z

2

Nothing to do with System.arrayCopy.

In fact, System.out.println(Integer.toHexString((byte) 0x80)); will also print ffffff80.

You have an arithmetic overflow when casting hex 0x80 as byte, since its decimal value is 128, i.e. 1 more than byte's capacity (Byte.MAX_VALUE, or 127) and overflows to -128.

Note that in turn, Integer.toHexString returns the representation of an unsigned integer, hence the additional overflow.

answered May 15, 2017 at 14:05

Mena

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Comments

ΦXocę 웃 Пepeúpa ツ · Accepted Answer · 2017-05-15 14:05:54Z

0

your are copying correctly the only 5 elements present in ic_submit_apdu_prefix ...

hint: 0x80 with 8 bits is -128 (2 complement)

0xffffff80 = -128

0x20 = 32

7

0

8

0

the rest is zero. (default initial primitive value)

answered May 15, 2017 at 14:05

ΦXocę 웃 Пepeúpa ツ

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