How to set 5 bits to value 3 at offset 387 bit in byte data sequence? - dart

Question

I need set some bits in ByteData at position counted in bits.
How I can do this?
Eg.
var byteData = new ByteData(1024);
var bitData = new BitData(byteData);
// Offset in bits: 387
// Number of bits: 5
// Value: 3
bitData.setBits(387, 5, 3);

Answer 1

Yes it is quite complicated. I dont know dart, but these are the general steps you need to take. I will label each variable as a letter and also use a more complicated example to show you what happens when the bits overflow.
1. Construct the BitData object with a ByteData object (A)
2. Call setBits(offset (B), bits (C), value (D));
I will use example values of:
A: 11111111 11111111 11111111 11111111
B: 7
C: 10
D: 00000000 11111111
3. Rather than using an integer with a fixed length of bits, you could
use another ByteData object (D) containing your bits you want to write.
Also create a mask (E) containing the significant bits.
e.g.
A: 11111111 11111111 11111111 11111111
D: 00000000 11111111
E: 00000011 11111111 (2^C - 1)
4. As an extra bonus step, we can make sure the insignificant
bits are really zero by ANDing with the bitmask.
D = D & E
D 00000000 11111111
E 00000011 11111111
5. Make sure D and E contain at least one full zero byte since we want
to shift them.
D 00000000 00000000 11111111
E 00000000 00000011 11111111
6. Work out these two integer values:
F = The extra bit offset for the start byte: B mod 8 (e.g. 7)
G = The insignificant bits: size(D) - C (e.g. 14)
7. H = G-F which should not be negative here. (e.g. 14-7 = 7)
8. Shift both D and E left by H bits.
D 00000000 01111111 10000000
E 00000001 11111111 10000000
9. Work out first byte number (J) floor(B / 8) e.g. 0
10. Read the value of A at this index out and let this be K
K = 11111111 11111111 11111111
11. AND the current (K) with NOT E to set zeros for the new bits.
Then you can OR the new bits over the top.
L = (K & !E) | D
K & !E = 11111110 00000000 01111111
L = 11111110 01111111 11111111
12. Write L to the same place you read it from.

Answer 2

There is no BitData class, so you'll have to do some of the bit-pushing yourself.
Find the corresponding byte offset, read in some bytes, mask out the existing bits and set the new ones at the correct bit offset, then write it back.
The real complexity comes when you need to store more bits than you can read/write in a single operation.
For endianness, if you are treating the memory as a sequence of bits with arbitrary width, I'd go for little-endian. Endianness only really makes sense for full-sized (2^n-bit, n > 3) integers. A 5 bit integer as the one you are storing can't have any endianness, and a 37 bit integer also won't have any natural way of expressing an endianness.
You can try something like this code (which can definitely be optimized more):
import "dart:typed_data";
void setBitData(ByteBuffer buffer, int offset, int length, int value) {
assert(value < (1 << length));
assert(offset + length < buffer.lengthInBytes * 8);
int byteOffset = offset >> 3;
int bitOffset = offset & 7;
if (length + bitOffset <= 32) {
ByteData data = new ByteData.view(buffer);
// Can update it one read/modify/write operation.
int mask = ((1 << length) - 1) << bitOffset;
int bits = data.getUint32(byteOffset, Endianness.LITTLE_ENDIAN);
bits = (bits & ~mask) | (value << bitOffset);
data.setUint32(byteOffset, bits, Endianness.LITTLE_ENDIAN);
return;
}
// Split the value into chunks of no more than 32 bits, aligned.
do {
int bits = (length > 32 ? 32 : length) - bitOffset;
setBitData(buffer, offset, bits, value & ((1 << bits) - 1));
offset += bits;
length -= bits;
value >>= bits;
bitOffset = 0;
} while (length > 0);
}
Example use:
main() {
var b = new Uint8List(32);
setBitData(b.buffer, 3, 8, 255);
print(b.map((v)=>v.toRadixString(16)));
setBitData(b.buffer, 13, 6*4, 0xffffff);
print(b.map((v)=>v.toRadixString(16)));
setBitData(b.buffer, 47, 21*4, 0xaaaaaaaaaaaaaaaaaaaaa);
print(b.map((v)=>v.toRadixString(16)));
}