Putting my 32 bits into a 4 bytearray

Question

I am putting different binary numbers into a byte array. One of the numbers are: 10001101010010010000000000001000 this number is giving me a NumberFormatException on the line where I try to parse it, obviously because it's too big. See code below where string is the binary number.

int number = Integer.parseInt(string, 2);
ByteBuffer bytes = ByteBuffer.allocate(4).putInt(number);
byte[] byteInstruction = bytes.array();

What I want is to put the numbers in my byte array but as they are 32-bit numbers I don't want to take up more than 4 bytes in my array. When I use long to parse it works but then I take up 8 spaces in my byte array.

long number = Long.parseLong(string, 2);
ByteBuffer bytes = ByteBuffer.allocate(8).putLong(number);
byte[] byteInstruction = bytes.array();

If I print out the array later I get this:

[0, 0, 0, 0, -115, 73, 0, 8]

where we can see that there are 4 spots free. How can I solve this? How did I mess up? All help is appreciated.

The point is that the number does not fit into an `int` because in Java an `int` is signed. — Willem Van Onsem, Jan 03 '16 at 17:24
@cricket_007 But what can I do? As I want to store my number in only 4 spaces in my byte array and not 8. — Fjodor, Jan 03 '16 at 17:31
I don't think that number is too big for an int. It's just a large negative number if it is signed — OneCricketeer, Jan 03 '16 at 17:32
@MattTimmermans Wouldn't number get a false value due to casting? — Fjodor, Jan 03 '16 at 17:39
only if the string was longer than 32 digits. If that's going to be a problem, then you should check for that before parsing it — Matt Timmermans, Jan 03 '16 at 17:40
@MattTimmermans Won't it fail because 10001101010010010000000000001000 = 2,370,371,592 and an int has max value of 2,147,483,647? — Fjodor, Jan 03 '16 at 17:45
Nope. When you cast a long to an int, it keeps the lower 32 bits and throws the upper 32 bits away. This will result in a negative int if bit 31 (the sign bit) is set. See https://en.wikipedia.org/wiki/Two%27s_complement — Matt Timmermans, Jan 03 '16 at 21:50

score 1 · Accepted Answer · edited May 23 '17 at 12:23

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Your input string "10001101010010010000000000001000" represents value that is too big for signed Integer. Integer.MAX_VALUE = 2147483647 and the input string you've passed has a value of 2370371592.

The Integer.parseInt does not interpret the leading 1 at position 32 from right as sign. If you would like parse a negative value it would have to be preceded with - sign.

See this answer for more through explanation.

If you expect the input "10001101010010010000000000001000" to in fact mean "-0001101010010010000000000001000" just replace the first character with +/- depending on the value of 32 bit from right.

Alternatively if you would like to treat the binary string in Two's complement compatible way you can use approach from this answer:

int msb = Integer.parseInt("1000110101001001", 2);
int lsb = Integer.parseInt("0000000000001000", 2);

int result = msb<<16 | lsb;

edited May 23 '17 at 12:23

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answered Jan 03 '16 at 17:44

miensol

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For example, it turns 10000000000000000000000000000000 into -0 – Matt Timmermans Jan 03 '16 at 21:53
Integer.MIN_VALUE. See the two's compliment link in my comment on the question – Matt Timmermans Jan 03 '16 at 21:56
@MattTimmermans see my edit. You could also add the two's complement requirement to the question so that it would be easier to find – miensol Jan 03 '16 at 22:08

Putting my 32 bits into a 4 bytearray

1 Answers1