Convert Signed Byte Array to Int array

Question

I have an integer array

int res[] = {176, 192, 312, 1028, 1064, 1016};

I get the signed byte array of the corresponding int array like this

int signed_byte_array[] = {-80, 0, -64, 0, 56, 1, 4, 4, 40, 4, -8, 3};

Each index in the int array is represented by two indexes in the byte array, means each value in int array is represented as 2 bytes.

I don't have access to the int array and I want to convert this signed byte array exactly to the int array

How can I do this?

Thanks

possible duplicate of stackoverflow.com/questions/11437203/byte-array-to-int-array — Marco Forberg
– Marco Forberg, Commented Jun 21, 2013 at 10:05
I have the signed byte array and i want to convert into int array, just consider the above byte array as input and convert it to int array so tht it gives me the above integer array — Wearybands
– Wearybands, Commented Jun 21, 2013 at 10:05
"each value in int array is represented as 2 bytes" - what would you want to do with an integer value more than 2^16? — Jon Skeet
– Jon Skeet, Commented Jun 21, 2013 at 10:05

Sergey Kalinichenko · Accepted Answer · 2013-06-21 10:18:26Z

1

It appears that the bytes represent unsigned 16-bit integers, with the most significant byte coming second, and the bits above 8-th truncated. You can do the conversion like this:

int[] signed_byte_array = {-80, 0, -64, 0, 56, 1, 4, 4, 40, 4, -8, 3};
int[] int_array = new int[signed_byte_array.length / 2];
for (int i = 0 ; i != int_array.length ; i++) {
    int_array[i] = (signed_byte_array[2*i+1] & 0xFF) << 8
                 | (signed_byte_array[2*i+0] & 0xFF);
}

When I added printing of int_array[i] in the loop I got these values:

176 192 312 1028 1064 1016

answered Jun 21, 2013 at 10:18

Sergey Kalinichenko

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Joop Eggen · Accepted Answer · 2013-06-21 12:09:58Z

0

private static int[] toB(int[] res) {
    int[] bytes = new int[res.length * 2];
    for (int i = 0; i < res.length; ++i) {
        int value = res[i];
        int lo = value & 0xFF;
        int hi = (value >> 8) & 0xFF;
        bytes[2 * i] = (int)(byte) lo;
        bytes[2 * i + 1] = (int)(byte) hi;
    }
    return bytes;
}

private static int[] toI(int[] bytes) {
    int[] res = new int[bytes.length / 2];
    for (int i = 0; i < res.length; ++i) {
        int lo = (int) bytes[2 * i] & 0xFF;
        int hi = (int) bytes[2 * i + 1] & 0xFF;
        res[i] = lo | (hi << 8);
    }
    return res;
}

public static void main(String[] args) {
    int[] res = {176, 192, 312, 1028, 1064, 1016};
    int[] signed_byte_array = {-80, 0, -64, 0, 56, 1, 4, 4, 40, 4, -8, 3};
    System.out.println("b: " + Arrays.toString(toB(res)));
    System.out.println("i: " + Arrays.toString(toI(signed_byte_array)));
}

gives

b: [-80, 0, -64, 0, 56, 1, 4, 4, 40, 4, -8, 3]
i: [176, 192, 312, 1028, 1064, 1016]

It is evidently a short[] to byte[] conversion. As you have an int[] I would say, there exists no utility conversion function.

edited Jun 21, 2013 at 12:09

answered Jun 21, 2013 at 10:13

Joop Eggen

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3 Comments

Wearybands Over a year ago

Joop Eggen How can I take lo and hi bytes as one integer value and save it as an array index

Joop Eggen Over a year ago

Sorry did a save on the first half. Now complete. Otherwise hard to test automatically.

Joop Eggen Over a year ago

@UmairIqbal made it executable. Some casts are too much as I changed byte to the used int.

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