Question

I’m trying to convert 3 bytes to signed integer (Big-endian) in C#.

I’ve tried to use BitConverter.ToInt32 method, but my problem is what value should have the lats byte.

Can anybody suggest me how can I do it in different way?

I also need to convert 5 (or 6 or 7) bytes to signed long, is there any general rule how to do it?

Thanks in advance for any help.

Answer 1

As a last resort you could always shift+add yourself:

byte b1, b2, b3;

int r = b1 << 16 | b2 << 8 | b3;

Just swap b1/b2/b3 until you have the desired result.

On second thought, this will never produce negative values.
What result do you want when the msb >= 0x80 ?

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Part 2, brute force sign extension:

    private static int Bytes2Int(byte b1, byte b2, byte b3)
    {
        int r = 0;
        byte b0 = 0xff;

        if ((b1 & 0x80) != 0) r |= b0 << 24;
        r |= b1 << 16;
        r |= b2 << 8;
        r |= b3;
        return r;
    }

I’ve tested this with:

      byte[] bytes = BitConverter.GetBytes(p);
      int r = Bytes2Int(bytes[2], bytes[1], bytes[0]);
      Console.WriteLine("{0} == {1}", p, r);

for several p.