3

4ビット値を元の32ビット整数の特定の位置に置き換える方法に固執しています。すべての助けは大歓迎です！

/**
 * Set a 4-bit nibble in an int.
 * 
 * Ints are made of eight bytes, numbered like so:
 *   7777 6666 5555 4444 3333 2222 1111 0000
 *
 * For a graphical representation of this:
 *   1 1 1 1 1 1                 
 *   5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 
 *  +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
 *  |Nibble3|Nibble2|Nibble1|Nibble0|
 *  +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
 * 
 * Examples:
 *     setNibble(0xAAA5, 0x1, 0); // => 0xAAA1
 *     setNibble(0x56B2, 0xF, 3); // => 0xF6B2
 * 
 * @param num The int that will be modified.
 * @param nibble The nibble to insert into the integer.
 * @param which Selects which nibble to modify - 0 for least-significant nibble.
 *            
 * @return The modified int.
 */
public static int setNibble(int num, int nibble, int which)
{
           int shifted = (nibble << (which<<2));
       int temp = (num & shifted);
           //this is the part I am stuck on, how can  I replace the original
           // which location with the nibble that I want? Thank you!
       return temp;
}
4

2 に答える 2

2 
public static int setNibble(int num, int nibble, int which) {
    return num & ~(0xF << (which * 4)) | (nibble << (which * 4));
}

ここ：

  • & ~(0xF << (which * 4))ニブルの元の値をマスクします。
  • は| (nibble << (which * 4))それを新しい値に設定します。
于 2013-01-21T21:05:45.530 に答える
0 
nibble &= 0xF; // Make sure
which &= 0x3;
int shifts = 4 * which; // 4 bits
num &= ~(0xF << shifts);
num |= nibble << shifts;
return num;
于 2013-01-21T21:06:02.940 に答える