How to build an int value with bytes in Java?

Question

My question is about Java int data structure.

If I have a buffer of 4 bytes

byte[] buffer = {(byte)A, (byte)B, (byte)C, (byte)D};

How to build a positive and a negative int values using this array and logic operators (&, |, <<, >>)?

Say int = 4 and int = 130; and -4 and -130.

Does it need to be logic operators, or rather bitwise operators? — akuzminykh, Feb 21 '20 at 18:36
When dealing with byte data, you should use hexadecimal notation instead of decimal. With that in mind, show four byte values for your array and what the resulting `int` value should be **in hex**. — Code-Apprentice, Feb 21 '20 at 18:37
You could use a `BitSet` https://stackoverflow.com/questions/2473597/bitset-to-and-from-integer-long — Andreas, Feb 21 '20 at 18:38
Let's assume you have `byte[] buffer = { (byte) 1, (byte) -2, (byte) 3, (byte) -4 };`. What is the expected value out pf this array? — Arvind Kumar Avinash, Feb 21 '20 at 19:31
@Rubik endianness is not property of programming language per se, rather cpu, or the way integer was originally packed into byte array, in java you can pack them both ways — Iłya Bursov, Feb 21 '20 at 21:39
Déjà vu: [Negative Int value from Big Endian to Small](https://stackoverflow.com/q/60329073/85421) — user85421, Feb 22 '20 at 20:40

Iłya Bursov · Accepted Answer · 2020-02-21T21:25:14.457

2

to construct 32 bit value from four 8 bit values you need to know Endianness

if we assume that high bits are stored into first byte, then you can do:

int v = 0;
for (int i=0; i<4; i++)
    v = (v<<8) | (buffer[i]&0xff);

if high bits are stored in last byte, then you need to reverse loop

int v = 0;
for (int i=3; i>=0; i--)
    v = (v<<8) | (buffer[i]&0xff);

edited Feb 21 '20 at 21:25

answered Feb 21 '20 at 19:38

Iłya Bursov

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Assume, you've received the buffer from Windows [130, 0, 0, 0].. as int = 130. now, please, give me the code that will rebuild the same value from that buffer into new int and with the same value = 130. – Rubik Feb 21 '20 at 21:23
@Rubik just reverse the loop – Iłya Bursov Feb 21 '20 at 21:26
Pls, do it with byte[] buffer = [(byte)130, (byte)0, (byte)0, (byte)0]; and get the result = 130 in the value – Rubik Feb 21 '20 at 21:35
@Rubik just did it and 2nd loop prints 130 – Iłya Bursov Feb 21 '20 at 21:38
Gotya!! Thank you!! Tell me why do you add & 0xff ? – Rubik Feb 21 '20 at 21:42
@Rubik java performs byte to int conversion preserving sign, so byte value `0x80` will be casted to int as `0xff ff ff 80`, to trim these sign bit extension we need to & with 0xff – Iłya Bursov Feb 21 '20 at 21:43
What if it is negative value in buffer? – Rubik Feb 21 '20 at 21:44
@Rubik negative values have sign bit set, which is extended during cast and this is why we're doing &0xff – Iłya Bursov Feb 21 '20 at 21:45
PLEASE, ADD YOUR LAST PHRASE TO YOUR ANSWER – Rubik Feb 21 '20 at 21:54
I tried it with negative value and it worked. Then Why in Windows they add 255, so that -130 is translated into { 126, 255, 255, 255 ) but not in {130, 0, 0, 1 } where 1 is negative sign – Rubik Feb 21 '20 at 22:02
1

@Rubik this happens because of sign extension and the way negative numbers are represented – Iłya Bursov Feb 21 '20 at 22:04
thank you, my question was about 0xff, i tried all and the value was wrong, and it made me stuck. – Rubik Feb 23 '20 at 04:25

How to build an int value with bytes in Java?

1 Answers1