C#: Converting 24-bit integer into a three byte array?

Question

So basically here is what I have. The user gives an integer and I'm converting it into 3 bytes.

int number = 167123;

byte[0] = (int)number / 65536;
byte[1] = (int)number / 256;
byte[2] = (int)number;

stream.Position = 0x503;   
stream.WriteByte((byte)byte[2]);
stream.WriteByte((byte)byte[1]);
stream.WriteByte((byte)byte[0]);

(Note: I'm cycling through the byte array backwards on purpose at the end.)

When I check the value later it works as intended. Now I'm looking hard at the code and trying the calculation by hand and I'm not getting the right answer. What am I doing wrong? How is this working? And what is Visual C# writing into the third byte when it casts 167123 as a "byte"?

Answer 1

The reason this works is because the assignment of the int value to the byte truncates the value. This may be why your math isn't working out - you're not truncating.

Essentially what you're doing by dividing is bitshifting. Your code is the same as this:

byte[0] = (int)number >> 16;
byte[1] = (int)number >> 8;
byte[2] = (int)number;

To make your manual math work, do the math, then convert it to binary, and chop off anything above the last 8 digits. That's the number you're assigning to the byte array.

One example:

byte[1] = (int)number / 256;

This is 167123 / 256 = 652. In binary, this is 001010001100. Now, truncate everything above the size of a byte (8 bits), and you have 10001100, which is 140 in decimal. This is what is assigned to this byte array index.

Answer 2

Try to use this method instead: BitConvet.GetBytes(int). More about this problem is also available in this question.