将 32 位整数的半字节设置为某个值 [重复]

Question

我被困在如何将 4 位值替换为原始 32 位整数的某个位置。非常感谢所有帮助！

/**
 * 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;
}

Answer 1

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

这里：

• & ~(0xF << (which * 4)) 掩盖了半字节的原始值；
• | (nibble << (which * 4)) 将其设置为新值。

Answer 2

nibble &= 0xF; // Make sure
which &= 0x3;
int shifts = 4 * which; // 4 bits
num &= ~(0xF << shifts);
num |= nibble << shifts;
return num;