How do I add 3 bytes and return an integer number?

Question

Lets assume that I have a hex string of

00 00 04 01 11 00 08 00 06 C2 C1 BC

With this the 7th, 8th, and 9th octet are a number I need to generate. The hex is

00 06 C2

This number turns out to be 1730. With the following, how can I simplify this?

byte b1 = 0x00;
byte b2 = 0x06;
byte b3 = 0xC2;

Console.WriteLine(Convert.ToInt32((Convert.ToString(b1, 16)) + (Convert.ToString(b2, 16)) + (Convert.ToString(b3, 16)), 16));

I know there has to be a simpler way. I tried Console.WriteLine((b1 + b2 + b3).ToString()); but it doesn't work.

score 8 · Accepted Answer · edited May 16 '13 at 06:46

8

Try:

int result = b3 | (b2 << 8) | (b1 << 16);

Assuming that b1, b2 and b3 are the byte values that you need to convert.

The << operator shifts its operand left by the specified number of bits.

edited May 16 '13 at 06:46

Marc Gravell

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answered May 16 '13 at 06:36

Matthew Watson

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This worked perfectly! I will mark this as the answer as soon as it lets me! – Gabriel Graves May 16 '13 at 06:40
3

As a small note: when dealing with `byte` arithmetic, `|` would be preferable to `+`; this may be important when "negative" numbers are involved (where the MSB is set) – Marc Gravell May 16 '13 at 06:41
(the OP put the `+` => `|` through as a suggested edit... feel free to change it back, but I maintain `|` improves the answer) – Marc Gravell May 16 '13 at 06:47
I was just changing the + to a | but it required me to change 6 characters to make an edit so I added a comment at the end. – Gabriel Graves May 16 '13 at 06:55
@MarcGravell True dat - I wasn't too concerned since it was only 3 bytes (and so `+` would always work), but it's best to do it the right way. – Matthew Watson May 16 '13 at 07:44
@MarcGravell Actually in this case it is always safe to use `+` because the bits in the bytes don't overlap - but it's better to use `|` because that will always work even in cases when the bits do overlap. Best practice and all that. – Matthew Watson May 16 '13 at 08:26

score 2 · Answer 2 · answered May 16 '13 at 06:43

2

You can use the BitConverter class to convert an array of bytes to an int.

// Add the bytes to an array, starting with b3, then b2, b1 and 0 to
// make it 4 bytes in total.
byte[] b = new byte[] { 0xC2, 0x06, 0x00, 0x00 };
int i = BitConverter.ToInt32(b, 0);

i will now have the value 1730.

answered May 16 '13 at 06:43

John Willemse

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That's a pretty hard way of doing something fairly simple, though - and it opens all sorts of questions about "endianness" etc. I'll work, but to be done properly you'd need to check `BitConverter.IsLittleEndian`, etc – Marc Gravell May 16 '13 at 06:45
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Yes, I realize that, but didn't want to make it more complicated than it already was. I prefer the bitshifting way, but wanted to propose an alternative to the other answers. – John Willemse May 16 '13 at 06:46

Shreyos Adikari · Answer 3 · 2013-05-16T07:20:30.863

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You can try this:

    byte b1 = (byte)0x00;
    byte b2 = (byte)0x06;
    byte b3 = (byte)0xC2;
    int i = ((b1 & 0xFF) << 16) | ((b2 & 0xFF) << 8) | (b3 & 0xFF);

EDIT:

    byte b1 = (byte)0x00;
    byte b2 = (byte)0x06;
    byte b3 = (byte)0xC2;
    int i = (b1 << 16) | (b2 << 8) | b3;

edited May 16 '13 at 07:20

answered May 16 '13 at 06:41

Shreyos Adikari

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I saw that sneaky edit ;) Note: there's no benefit doing an `& 0xFF` if you know the source is `byte` - there are no other bits that can be set – Marc Gravell May 16 '13 at 06:43

How do I add 3 bytes and return an integer number?

3 Answers3