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# convert int(short or long) to byte array

Posted on 2002-03-04
79,027 Views
I intend to convert a integer (or short or long) that is greater than a byte can hold (eg, 1111) to a byte array, since casting is simply cannot be used here, any other way can do it without lose precise? I have tried to use Integer wrapper class and use its toBinaryString() to convert the int to a string representation, eg 5->"101", then I really cannot find a way of converting "101" to a byte array with correct binary representation. Please help
0
Question by:gordon_guan
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LVL 9

Expert Comment

ID: 6839117
try this:
"101".getBytes();
0

LVL 9

Expert Comment

ID: 6839172
String str = Integer.toBinaryString(12345);
byte[] byte_array = str.getBytes();
0

LVL 7

Accepted Solution

Igor Bazarny earned 20 total points
ID: 6839424
Hi,

Try this:

int value = 0xcafebabe;
byte[] bytes = new byte[4];

for( int i=0; i<bytes.length; ++i ){
int offset = (bytes.length-i-1)*8;
bytes[i] = (value & (0xff << offset)) >>> offset;
}

for( int i=0; i<bytes.length; ++i ){
System.out.print(Integer.toHexString(bytes[i])+" ");
}
System.out.println();

Regards,
Igor Bazarny,
Brainbench MVP for Java 1
www.brainbench.com
0

Author Comment

ID: 6841193
hello vinci, I would say your answer is not the thing that I want and in my case, it is wrong. I would rather prefer bazany's answer since what he/she does is more or less the same as what I did. So please change the answer back. Thanks.
0

Author Comment

ID: 6841194
hello vinci, I would say your answer is not the thing that I want and in my case, it is wrong. I would rather prefer bazany's answer since what he/she does is more or less the same as what I did. So please change the answer back. Thanks.
0

LVL 4

Expert Comment

ID: 6842116
Here's another way:

private byte[] getBytes(int n) {
BigInteger big = new BigInteger(Integer.toString(n));
return big.toByteArray();
}
0

LVL 2

Expert Comment

ID: 9337726
Assuming v is the value of the integer you are trying to convert to bytes, try this:

byte b[] = new byte[4];
int i, shift;

for(i = 0, shift = 24; i < 4; i++, shift -= 8)
b[i] = (byte)(0xFF & (v >> shift));
0

Expert Comment

ID: 13517210
This is a more readable (and faster) solution than unidyne's.

I've also included the code to recreate the int from the byte array.  This is a fragment of a class I wrote called ByteArrayUtils; if anyone can give me a faster solution, I'd be very grateful!

Enjoy!

public static final int   NUMBER_OF_BITS_IN_A_BYTE = 8;
public static final short MASK_TO_BYTE             = 0xFF;
public static final int   SIZE_OF_AN_INT_IN_BYTES  = 4;

public static void writeInt( byte[] p_dest,  // not allocated and returned so array can be reused
int    p_toWrite )
{
assert( p_dest.length >= SIZE_OF_AN_INT_IN_BYTES )
: "Programming error: p_dest is too short to hold an int";

// unrolled loop of 4 iterations
p_dest[0] = p_toWrite & MASK_TO_BYTE );
p_toWrite >>= NUMBER_OF_BITS_IN_A_BYTE;

p_dest[1] = p_toWrite & MASK_TO_BYTE );
p_toWrite >>= NUMBER_OF_BITS_IN_A_BYTE;

p_dest[2] = p_toWrite & MASK_TO_BYTE );
p_toWrite >>= NUMBER_OF_BITS_IN_A_BYTE;

p_dest[3] = p_toWrite & MASK_TO_BYTE );
}

public static int readInt( byte[] p_src ) // must be of size 4
{
assert( p_src.length >= SIZE_OF_AN_INT_IN_BYTES )
: "Programming error: p_src is too short to hold an int";

// unrolled loop of 4 iterations
int result = (   p_src[ 0 ] & MASK_TO_BYTE );
result    |= ( ( p_src[ 1 ] & MASK_TO_BYTE ) << 8 );
result    |= ( ( p_src[ 2 ] & MASK_TO_BYTE ) << 16 );
result    |= ( ( p_src[ 3 ] & MASK_TO_BYTE ) << 24 );

return result;
}
0

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