c# - bit parsing 8 byte array

Question

Bytes: 240 255 255 9 0 224 9 0

f0 ff ff 09 00 E0 09 00

Little endian unsigned int 64 translation:

00 09 E0 00 09 ff ff f0

int value1 = 0-19 bits  
int value2 = 20-39 bits 
int value3 = 40-59 bits
int value4 = 60-62 bits 
bool value5 = 63 bit

value1 = (uint)(byteArray[0] | byteArray[1] << 8 | (byteArray[2] << 16)) & 0x14;

Am I doing this correctly? I keep getting value of 0 but should be 158.

What is the intent of the & 0x14? That might be your problem. — Tim S.
– Tim S., Commented Jul 22, 2013 at 17:10
Also: why are you trying to store 64 bits in 5 ints? Why not byte[] or long/ulong? — Tim S.
– Tim S., Commented Jul 22, 2013 at 17:12
Are you trying to store RGBA - value with 20bits per channel (20-20-20-4) or something similar? 0x14 should be about 5 "F" - 0xFFFFF... or your braces are wrong as The Moof suggested. — Alexei Levenkov
– Alexei Levenkov, Commented Jul 22, 2013 at 17:24
I'm parsing a file which has byte array of size 8. I have to parse 5 variables from the 8 bytes. — Goss
– Goss, Commented Jul 22, 2013 at 17:25
((byteArray[0] | byteArray[1] << 8 | (byteArray[2] << 16))) & 0xfffff this still gives me wrong value. — Goss
– Goss, Commented Jul 22, 2013 at 17:26

The Moof · Accepted Answer · 2013-07-22 17:12:37Z

1

The last operation in your calculation is & 0x14. This will do a bitwise and against the binary value of 0001 0100. You're looking for the first 20 bits, so your mask should be 0xfffff.

answered Jul 22, 2013 at 17:12

The Moof

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5 Comments

Goss Over a year ago

yeah I was trying to mask off the 20 bits because it was currently 24 bits long. So i assume 0x14 being 20 in decimal, it would mask it. (uint)(byteArray[0] | byteArray[1] << 8 | (byteArray[2] << 16)) & 0xfffff; still gives me 2304.

The Moof Over a year ago

Based on your rules, the first number would be 0x00900, which is 2304.

The Moof Over a year ago

Keep in mind that bit values are read right-to-left. Based on your expected answer, it leads me to believe that the rules are incorrect. They should be Val1 = bits 44-63, Val2 = bits 24-43, etc. Using these rules with the Little Endian bytes, you will get the expected values.

Goss Over a year ago

The specification sheet for this particular file(which I can't specify what it is), says that's how the bits are stored and what order.

The Moof Over a year ago

The bit order isn't the problem. As I said, numbering your bits backwards.

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