byte[] sample = new byte[]{10,20,30};
- 该值为6位,从第三位(从右到左)开始
new byte [] {10,20,30}看起来像“00001010 00010100 00011110”(应按字节顺序排列) 所以我需要“00001010 00010100 * 000111 * 10” - 我的价值是7
基于帮助的解决方案(通过 Yaur 回答1),只是改变位方向
public static bool GetValue(byte[] data, int position)
{
var bytePos = data.Length - 1 - position / 8;//right -> left
//var bytePos = position / 8;//left -> right
var bitPos = position % 8;
return ((data[bytePos] & (1 << bitPos)) != 0);//right -> left
//return ((data[bytePos] & (1 << (7 - bitPos))) != 0); //left -> right
}
public static long GetValue(byte[] data, int position, int length)
{
if (length > 62)
{
throw new ArgumentException("not going to work properly with 63 bits if the first bit is 1");
}
long retv = 0;
for (int i = position + length - 1; i > position - 1; i--)
//for(int i = position;i<position+length;i++)//left -> right
{
if (GetValue(data, i)) retv |= 1;
retv = retv << 1;
}
retv = retv >> 1;
return retv;
}
答案 0 :(得分:3)
这适用于大多数输入:
public bool GetValue(byte[] data, int position)
{
var bytePos = position / 8;
var bitPos = position % 8;
return ((data[bytePos] & (1 << bitPos))!=0)
// depending on the order in which you expect the bits you might need this instead
//return ((data[bytePos] & (1 << (7-bitPos)))!=0)
}
public long GetValue(byte[] data, int position, int length)
{
if(length > 62)
{
throw new ArgumentException("not going to work properly with 63 bits if the first bit is 1");
}
long retv=0;
for(int i = position;i<position+length;i++)
{
if(GetValue(data,i)
{
retv |=1;
}
retv = retv << 1;
}
retv = retv >> 1;
}