set some consecutive bits in a byte

I need an efficient method with the following signature:

public byte SetBits(byte oldValue, byte newValue, int startBit, int bitCount)

Which returns oldValue, only that starting from its startbit bit 开发者_运维问答up to its startbit + bitcount bit (zero-based), it's replaced with the first bitcount bits of newValue

For example, if:

• oldValue = 11101101
• newValue = 10000011
• startBit = 1
• bitCount = 2

Then the result would be: 11101111 (the segment 10 in oldValue is replaced with the corresponding 11 segment in newValue)

Here you go... Bitshift both directions to get the mask... then use it to generate the new byte

public static byte SetBits(byte oldValue, byte newValue, int startBit, int bitCount)
{
    if (startBit < 0 || startBit > 7 || bitCount < 0 || bitCount > 7 
                     || startBit + bitCount > 8)
        throw new OverflowException();

    int mask = (255 >> 8 - bitCount) << startBit;
    return Convert.ToByte((oldValue & (~mask)) | ((newValue << startBit) & mask));
}

startBit--; //account for 0 indexing
byte flag = 1 << startBit;
for (int i = startBit; i < bitCount; i++, flag <<= 1)
{
    byte mask = newValue & flag;
    if (mask != 0)
        oldValue |= mask;
    else
        oldValue &= ~(flag);
}
return oldValue;

Some brain compiled code here but it should be along the lines that you want if I read the question correctly.

If I understood your question, I think this is what you are after:

byte mask = 0xFF;
for (int i = startPos-1; i < numBits; i++)
{
     if ((newValue & (1 << i)) == 1)
     {
          mask = (byte)(mask | (1 << i));
     }
     else
     {
          mask = (byte)(mask &~(1<<i));
     }
 }
 return (byte)(oldValue & mask);

This code is based on some neat tricks from Low Level Bit Hacks You Absolutely Must Know

I know setting a bit in a byte initialized to 0xFF is really a no-op, but I felt the code should be left in as it can help show off what is really going on. I encourage users of the code to optimize it as needed.