Converting BitSet to Byte[]
I have a BitSet, which needs to be converted to a Byte. However, by using BitSet.toByteArray(), I don't get the correct output. I have tried converting the Byte to its binary form in order to check whether the Bitset and the binary form of the Byte are similiar.
public static void generate() {
BitSet temp1 = new BitSet(64);
for (int i = 0; i < 64; i++) {
if(i % 8 != 0 && i < 23) {
temp1.set(i, true);
}
}
StringBuilder s = new StringBuilder();
for (int i = 0; i < 64; i++) {
s.append(temp1.get(i) == true ? 1 : 0);
}
System.out.println(s);
byte tempByteKey1 = temp1.toByteArray();
for (byte b : tempByteKey1) {
System.out.print(Integer.toBinaryString(b & 255 | 256).substring(1));
}
}
Output:
Bitset: 0111111101111111011111100000000000000000000000000000000000000000
Converted Byte: 1111111011111110011111100000000000000000000000000000000000000000
They are both 64 bits, but the first 0 in the BitSet is placed somewhere else after the conversion. Why is this happening, and how can I fix it?
java arrays bitset
asked Nov 20 '18 at 22:17
user8231110user8231110
586
|
show 1 more comment
I have a BitSet, which needs to be converted to a Byte. However, by using BitSet.toByteArray(), I don't get the correct output. I have tried converting the Byte to its binary form in order to check whether the Bitset and the binary form of the Byte are similiar.
public static void generate() {
BitSet temp1 = new BitSet(64);
for (int i = 0; i < 64; i++) {
if(i % 8 != 0 && i < 23) {
temp1.set(i, true);
}
}
StringBuilder s = new StringBuilder();
for (int i = 0; i < 64; i++) {
s.append(temp1.get(i) == true ? 1 : 0);
}
System.out.println(s);
byte tempByteKey1 = temp1.toByteArray();
for (byte b : tempByteKey1) {
System.out.print(Integer.toBinaryString(b & 255 | 256).substring(1));
}
}
Output:
Bitset: 0111111101111111011111100000000000000000000000000000000000000000
Converted Byte: 1111111011111110011111100000000000000000000000000000000000000000
They are both 64 bits, but the first 0 in the BitSet is placed somewhere else after the conversion. Why is this happening, and how can I fix it?
java arrays bitset
asked Nov 20 '18 at 22:17
user8231110user8231110
586
1
the bit order is being printed in reverse bytoBinaryStringas you are printing the BitSet. You start with the less significant bit,toBinaryStringstarts with the most significant one.(see how the zero is in first position on first output, and at 8th position on second output)
– Carlos Heuberger
Nov 20 '18 at 22:28
But they both have six trues as low order and seven trues as high order
– user8231110
Nov 20 '18 at 22:33
1
what are you counting??? BitSet:01111111 01111111 01111110 0...; toBinaryString:11111110 11111110 01111110 0...=> each corresponding 8 bits are reversed. Or, using chars for bits, BitSet:hgfedcba ponmlkji ..., toBinaryString:abcdefgh ijklmnop ...
– Carlos Heuberger
Nov 20 '18 at 22:36
That makes sense, but is it possible to store it as it originally was i.e. without the reversing?
– user8231110
Nov 20 '18 at 22:40
??? it is stored as originally , just the way you represent it differs as howtoBinaryStringdoes
– Carlos Heuberger
Nov 20 '18 at 22:40
|
show 1 more comment
I have a BitSet, which needs to be converted to a Byte. However, by using BitSet.toByteArray(), I don't get the correct output. I have tried converting the Byte to its binary form in order to check whether the Bitset and the binary form of the Byte are similiar.
public static void generate() {
BitSet temp1 = new BitSet(64);
for (int i = 0; i < 64; i++) {
if(i % 8 != 0 && i < 23) {
temp1.set(i, true);
}
}
StringBuilder s = new StringBuilder();
for (int i = 0; i < 64; i++) {
s.append(temp1.get(i) == true ? 1 : 0);
}
System.out.println(s);
byte tempByteKey1 = temp1.toByteArray();
for (byte b : tempByteKey1) {
System.out.print(Integer.toBinaryString(b & 255 | 256).substring(1));
}
}
Output:
Bitset: 0111111101111111011111100000000000000000000000000000000000000000
Converted Byte: 1111111011111110011111100000000000000000000000000000000000000000
They are both 64 bits, but the first 0 in the BitSet is placed somewhere else after the conversion. Why is this happening, and how can I fix it?
java arrays bitset
asked Nov 20 '18 at 22:17
user8231110user8231110
586
I have a BitSet, which needs to be converted to a Byte. However, by using BitSet.toByteArray(), I don't get the correct output. I have tried converting the Byte to its binary form in order to check whether the Bitset and the binary form of the Byte are similiar.
public static void generate() {
BitSet temp1 = new BitSet(64);
for (int i = 0; i < 64; i++) {
if(i % 8 != 0 && i < 23) {
temp1.set(i, true);
}
}
StringBuilder s = new StringBuilder();
for (int i = 0; i < 64; i++) {
s.append(temp1.get(i) == true ? 1 : 0);
}
System.out.println(s);
byte tempByteKey1 = temp1.toByteArray();
for (byte b : tempByteKey1) {
System.out.print(Integer.toBinaryString(b & 255 | 256).substring(1));
}
}
Output:
Bitset: 0111111101111111011111100000000000000000000000000000000000000000
Converted Byte: 1111111011111110011111100000000000000000000000000000000000000000
They are both 64 bits, but the first 0 in the BitSet is placed somewhere else after the conversion. Why is this happening, and how can I fix it?
java arrays bitset
java arrays bitset
asked Nov 20 '18 at 22:17
user8231110user8231110
586
asked Nov 20 '18 at 22:17
user8231110user8231110
586
asked Nov 20 '18 at 22:17
user8231110user8231110
586
asked Nov 20 '18 at 22:17
user8231110user8231110
586
asked Nov 20 '18 at 22:17
user8231110user8231110
586
586
1
the bit order is being printed in reverse bytoBinaryStringas you are printing the BitSet. You start with the less significant bit,toBinaryStringstarts with the most significant one.(see how the zero is in first position on first output, and at 8th position on second output)
– Carlos Heuberger
Nov 20 '18 at 22:28
But they both have six trues as low order and seven trues as high order
– user8231110
Nov 20 '18 at 22:33
1
what are you counting??? BitSet:01111111 01111111 01111110 0...; toBinaryString:11111110 11111110 01111110 0...=> each corresponding 8 bits are reversed. Or, using chars for bits, BitSet:hgfedcba ponmlkji ..., toBinaryString:abcdefgh ijklmnop ...
– Carlos Heuberger
Nov 20 '18 at 22:36
That makes sense, but is it possible to store it as it originally was i.e. without the reversing?
– user8231110
Nov 20 '18 at 22:40
??? it is stored as originally , just the way you represent it differs as howtoBinaryStringdoes
– Carlos Heuberger
Nov 20 '18 at 22:40
|
show 1 more comment
1
the bit order is being printed in reverse bytoBinaryStringas you are printing the BitSet. You start with the less significant bit,toBinaryStringstarts with the most significant one.(see how the zero is in first position on first output, and at 8th position on second output)
– Carlos Heuberger
Nov 20 '18 at 22:28
But they both have six trues as low order and seven trues as high order
– user8231110
Nov 20 '18 at 22:33
1
what are you counting??? BitSet:01111111 01111111 01111110 0...; toBinaryString:11111110 11111110 01111110 0...=> each corresponding 8 bits are reversed. Or, using chars for bits, BitSet:hgfedcba ponmlkji ..., toBinaryString:abcdefgh ijklmnop ...
– Carlos Heuberger
Nov 20 '18 at 22:36
That makes sense, but is it possible to store it as it originally was i.e. without the reversing?
– user8231110
Nov 20 '18 at 22:40
??? it is stored as originally , just the way you represent it differs as howtoBinaryStringdoes
– Carlos Heuberger
Nov 20 '18 at 22:40
1
1
the bit order is being printed in reverse by
toBinaryString as you are printing the BitSet. You start with the less significant bit, toBinaryString starts with the most significant one.(see how the zero is in first position on first output, and at 8th position on second output)– Carlos Heuberger
Nov 20 '18 at 22:28
the bit order is being printed in reverse by
toBinaryString as you are printing the BitSet. You start with the less significant bit, toBinaryString starts with the most significant one.(see how the zero is in first position on first output, and at 8th position on second output)– Carlos Heuberger
Nov 20 '18 at 22:28
But they both have six trues as low order and seven trues as high order
– user8231110
Nov 20 '18 at 22:33
But they both have six trues as low order and seven trues as high order
– user8231110
Nov 20 '18 at 22:33
1
1
what are you counting??? BitSet:
01111111 01111111 01111110 0...; toBinaryString: 11111110 11111110 01111110 0... => each corresponding 8 bits are reversed. Or, using chars for bits, BitSet: hgfedcba ponmlkji ... , toBinaryString: abcdefgh ijklmnop ...– Carlos Heuberger
Nov 20 '18 at 22:36
what are you counting??? BitSet:
01111111 01111111 01111110 0...; toBinaryString: 11111110 11111110 01111110 0... => each corresponding 8 bits are reversed. Or, using chars for bits, BitSet: hgfedcba ponmlkji ... , toBinaryString: abcdefgh ijklmnop ...– Carlos Heuberger
Nov 20 '18 at 22:36
That makes sense, but is it possible to store it as it originally was i.e. without the reversing?
– user8231110
Nov 20 '18 at 22:40
That makes sense, but is it possible to store it as it originally was i.e. without the reversing?
– user8231110
Nov 20 '18 at 22:40
??? it is stored as originally , just the way you represent it differs as how
toBinaryString does– Carlos Heuberger
Nov 20 '18 at 22:40
??? it is stored as originally , just the way you represent it differs as how
toBinaryString does– Carlos Heuberger
Nov 20 '18 at 22:40
|
show 1 more comment
1 Answer
1
active
oldest
votes
From BitSet#toByteArray() javadoc:
Returns a new byte array containing all the bits in this bit set.
More precisely, if..
byte bytes = s.toByteArray();
then
bytes.length == (s.length()+7)/8
and
s.get(n) == ((bytes[n/8] & (1<<(n%8))) != 0)
for all
n < 8 * bytes.length.
@return a byte array containing a little-endian representation of all the bits in this bit set
@since 1.7
Attention: toByteArray() doesn't even claim to know size(), it is only "reliable" regarding length()!
..So I would propose as implementation (alternative for your toBinaryString()) a method like:
static String toBinaryString(byte barr, int size) {
StringBuilder sb = new StringBuilder();
int i = 0;
for (; i < 8 * barr.length; i++) {
sb.append(((barr[i / 8] & (1 << (i % 8))) != 0) ? '1' : '0');
}
for (; i < size; i++) {
sb.append('0');
}
return sb.toString();
}
..to call it like:
System.out.println(toBinaryString(bitSet.toByteArray(), 64);
run:
0111111101111111011111100000000000000000000000000000000000000000
0111111101111111011111100000000000000000000000000000000000000000
BUILD SUCCESSFUL (total time: 0 seconds)
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
add a comment |
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From BitSet#toByteArray() javadoc:
Returns a new byte array containing all the bits in this bit set.
More precisely, if..
byte bytes = s.toByteArray();
then
bytes.length == (s.length()+7)/8
and
s.get(n) == ((bytes[n/8] & (1<<(n%8))) != 0)
for all
n < 8 * bytes.length.
@return a byte array containing a little-endian representation of all the bits in this bit set
@since 1.7
Attention: toByteArray() doesn't even claim to know size(), it is only "reliable" regarding length()!
..So I would propose as implementation (alternative for your toBinaryString()) a method like:
static String toBinaryString(byte barr, int size) {
StringBuilder sb = new StringBuilder();
int i = 0;
for (; i < 8 * barr.length; i++) {
sb.append(((barr[i / 8] & (1 << (i % 8))) != 0) ? '1' : '0');
}
for (; i < size; i++) {
sb.append('0');
}
return sb.toString();
}
..to call it like:
System.out.println(toBinaryString(bitSet.toByteArray(), 64);
run:
0111111101111111011111100000000000000000000000000000000000000000
0111111101111111011111100000000000000000000000000000000000000000
BUILD SUCCESSFUL (total time: 0 seconds)
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
add a comment |
From BitSet#toByteArray() javadoc:
Returns a new byte array containing all the bits in this bit set.
More precisely, if..
byte bytes = s.toByteArray();
then
bytes.length == (s.length()+7)/8
and
s.get(n) == ((bytes[n/8] & (1<<(n%8))) != 0)
for all
n < 8 * bytes.length.
@return a byte array containing a little-endian representation of all the bits in this bit set
@since 1.7
Attention: toByteArray() doesn't even claim to know size(), it is only "reliable" regarding length()!
..So I would propose as implementation (alternative for your toBinaryString()) a method like:
static String toBinaryString(byte barr, int size) {
StringBuilder sb = new StringBuilder();
int i = 0;
for (; i < 8 * barr.length; i++) {
sb.append(((barr[i / 8] & (1 << (i % 8))) != 0) ? '1' : '0');
}
for (; i < size; i++) {
sb.append('0');
}
return sb.toString();
}
..to call it like:
System.out.println(toBinaryString(bitSet.toByteArray(), 64);
run:
0111111101111111011111100000000000000000000000000000000000000000
0111111101111111011111100000000000000000000000000000000000000000
BUILD SUCCESSFUL (total time: 0 seconds)
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
add a comment |
From BitSet#toByteArray() javadoc:
Returns a new byte array containing all the bits in this bit set.
More precisely, if..
byte bytes = s.toByteArray();
then
bytes.length == (s.length()+7)/8
and
s.get(n) == ((bytes[n/8] & (1<<(n%8))) != 0)
for all
n < 8 * bytes.length.
@return a byte array containing a little-endian representation of all the bits in this bit set
@since 1.7
Attention: toByteArray() doesn't even claim to know size(), it is only "reliable" regarding length()!
..So I would propose as implementation (alternative for your toBinaryString()) a method like:
static String toBinaryString(byte barr, int size) {
StringBuilder sb = new StringBuilder();
int i = 0;
for (; i < 8 * barr.length; i++) {
sb.append(((barr[i / 8] & (1 << (i % 8))) != 0) ? '1' : '0');
}
for (; i < size; i++) {
sb.append('0');
}
return sb.toString();
}
..to call it like:
System.out.println(toBinaryString(bitSet.toByteArray(), 64);
run:
0111111101111111011111100000000000000000000000000000000000000000
0111111101111111011111100000000000000000000000000000000000000000
BUILD SUCCESSFUL (total time: 0 seconds)
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
From BitSet#toByteArray() javadoc:
Returns a new byte array containing all the bits in this bit set.
More precisely, if..
byte bytes = s.toByteArray();
then
bytes.length == (s.length()+7)/8
and
s.get(n) == ((bytes[n/8] & (1<<(n%8))) != 0)
for all
n < 8 * bytes.length.
@return a byte array containing a little-endian representation of all the bits in this bit set
@since 1.7
Attention: toByteArray() doesn't even claim to know size(), it is only "reliable" regarding length()!
..So I would propose as implementation (alternative for your toBinaryString()) a method like:
static String toBinaryString(byte barr, int size) {
StringBuilder sb = new StringBuilder();
int i = 0;
for (; i < 8 * barr.length; i++) {
sb.append(((barr[i / 8] & (1 << (i % 8))) != 0) ? '1' : '0');
}
for (; i < size; i++) {
sb.append('0');
}
return sb.toString();
}
..to call it like:
System.out.println(toBinaryString(bitSet.toByteArray(), 64);
run:
0111111101111111011111100000000000000000000000000000000000000000
0111111101111111011111100000000000000000000000000000000000000000
BUILD SUCCESSFUL (total time: 0 seconds)
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
edited Feb 10 at 8:03
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
answered Feb 6 at 12:28
xerx593xerx593
3,56921534
3,56921534
add a comment |
add a comment |
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1
the bit order is being printed in reverse by
toBinaryStringas you are printing the BitSet. You start with the less significant bit,toBinaryStringstarts with the most significant one.(see how the zero is in first position on first output, and at 8th position on second output)– Carlos Heuberger
Nov 20 '18 at 22:28
But they both have six trues as low order and seven trues as high order
– user8231110
Nov 20 '18 at 22:33
1
what are you counting??? BitSet:
01111111 01111111 01111110 0...; toBinaryString:11111110 11111110 01111110 0...=> each corresponding 8 bits are reversed. Or, using chars for bits, BitSet:hgfedcba ponmlkji ..., toBinaryString:abcdefgh ijklmnop ...– Carlos Heuberger
Nov 20 '18 at 22:36
That makes sense, but is it possible to store it as it originally was i.e. without the reversing?
– user8231110
Nov 20 '18 at 22:40
??? it is stored as originally , just the way you represent it differs as how
toBinaryStringdoes– Carlos Heuberger
Nov 20 '18 at 22:40