int[] to byte[]
Author
In Java, if i have an integer array and a method takes a byte array, is there any way I can convert the integer array to the byte array, short of going through and shifting bits around. I tried casting, but that did not seem to work . . .
Thanks
i dont imaging there is one... if an integer, 4 bytes, uses each of those bytes, there is no way to convert it to a char (1 byte), without loosing much data (3 bytes worth).
why would you want to do this?
why would you want to do this?
Author
DatagramPackets take byte[] as data. I have data that is in int[] . . . So I figured there would be a function somewhere that coverted the first integer in the int[] to take up the first 4 spaces of the byte array and so on. In C this would not be such a problem i suppose . . .
size_t arraySize = 1024;int* myIntArray = new int[arraySize];char* myCharBuffer = (char*)myIntArray;size_t numberOfBytesInBuffer = arraySize * sizeof(int); 
Author
i guess its obvious to me now that there is no easy way . . . thanks for all of your help
maybe make a func like this
public byte[] getBytesFromIntArray(int[] data)
{
byte newData = new byte[4 * data.length];
for (int i = 0;i < data.length; i++)
for (int j = 0; j < 4; j++)
newData = (data<i> & <br> (Math.pow(2,(j+1) * 8)) - (Math.pow(2, j * 8)));<br> return newData;<br>}<br><br><br> </i>
public byte[] getBytesFromIntArray(int[] data)
{
byte newData = new byte[4 * data.length];
for (int i = 0;i < data.length; i++)
for (int j = 0; j < 4; j++)
newData = (data<i> & <br> (Math.pow(2,(j+1) * 8)) - (Math.pow(2, j * 8)));<br> return newData;<br>}<br><br><br> </i>
No Java guru, but I have an idea.
To get least significant byte, take the modulo (assuming there''s a modulo operator/function in Java) of the integer and 256.
I C, that is
Then, to get next byte, divide by 256, and take the modulo again. And for the third and fourth byte, divide by 256*256 and 256*256*256 repectively, and take the modulo of it.
To get least significant byte, take the modulo (assuming there''s a modulo operator/function in Java) of the integer and 256.
I C, that is
int var;char b0 = var % 256; Then, to get next byte, divide by 256, and take the modulo again. And for the third and fourth byte, divide by 256*256 and 256*256*256 repectively, and take the modulo of it.
char b1 = (var / 256) % 256;char b2 = (var / 65536) % 256;char b3 = (var / 16777216) % 256;
There is an easy way, I do this in VB all the time.
Firstly, it makes things a lot simpler if the integer and byte are unsigned. Then:
b3 b2 b1 b0
(bx = byte x)
Mask out b2, b1, b0. Bitshift right 24. This is Byte[0]
Mask out b3, b1, b0. Bitshift right 16. This is Byte[1]
Mask out b3, b2, b0. Bitshift right 8. This is Byte[2]
Mask out b3, b2, b1. This is Byte[3]
Trying is the first step towards failure.
Firstly, it makes things a lot simpler if the integer and byte are unsigned. Then:
b3 b2 b1 b0
(bx = byte x)
Mask out b2, b1, b0. Bitshift right 24. This is Byte[0]
Mask out b3, b1, b0. Bitshift right 16. This is Byte[1]
Mask out b3, b2, b0. Bitshift right 8. This is Byte[2]
Mask out b3, b2, b1. This is Byte[3]
Trying is the first step towards failure.
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