计算机系统导论——小班作业(1)——位运算

(x & 0xFF) | (y & ~0xFF)

(x | ~0xFF) & (y | 0xFF)

A: !(~x) 或 !(x+1)
B: !x
C: !((x & 0xFF) – 0xFF) 
   或 !(~x << ((sizeof(int) – 1) << 3))
D: !(x >> ((sizeof(int) – 1) << 3)) 
   或 !(x & (0xFF << ((sizeof(int) – 1) << 3)))

/* 注意k=0的情况，不能左移w-k位*/

unsigned srl(unsigned x, int k){
    /* Perform shift arithmetically */
    unsigned xsra = (int) x >> k;
    int w = sizeof(int) << 3;
    int temp = (2 << (w - k - 1)) - 1;//最低的w-k位是1，高位是0
    return xsra & temp;//高位变0
}

int sra(unsigned x, int k){
    /* Perform shift logically */
    int xsrl = (unsigned) x >> k;
    int w = sizeof(int) << 3;
    int sign = xsrl & (1 << (w - k - 1));//仅保留右移后的符号位
    int temp = ~sign + 1;//符号位为0，则temp所有位为0；符号位为1，则temp的高k+1位为1，其余位为0
    return xsrl | temp;//高位全部变符号位
}

int sra(unsigned x, int k){
    /* Perform shift logically */
    int xsrl = (unsigned) x >> k;
    int w = sizeof(int) << 3;
    ( x & (1<<(w-1)) ) && ( xsrl |= (~0<<(w-k)) );//A && B, 若A为真则执行B,若A为假则跳过B
    return xsrl;
}