# 2022q1 Homework3 (quiz3)
contributed by < `a12345645` >

## 測驗1

#define GENMASK(h, l) \
    (((~0UL) >> (LEFT)) & ((~0UL) >> (l) << (RIGHT)))

`LEFT` = `63 - h`
`RIGHT` = `l`

- `~0UL` 為所有位元皆為 `1`
- `(~0UL) >> (63 - h)` 為最低位元到第 `h` 位元皆為 `1`
- `(~0UL) >> (l) << (RIGHT))` 為第 `l` 到最高位元都為 `1`
- 最後在把兩個做 AND 運算得到 mask

## 測驗 2

struct foo;
  
struct fd {
    struct foo *foo;
    unsigned int flags;
};

enum {
    FOO_DEFAULT = 0,
    FOO_ACTION,                           
    FOO_UNLOCK,
} FOO_FLAGS;

static inline struct fd to_fd(unsigned long v)
{
    return (struct fd){EXP1, v & 3};
}

要將 `v` 對 4 個位元組進行向下對齊
代表要清除 `v` 的最後兩個位元
所以是 `v & ~3`
`EXP1` = `v & ~3`

## 測驗 3

#include <stdint.h>
uint8_t rev8(uint8_t x)
{
    x = (x >> 4) | (x << 4);               
    x = ((x & 0xCC) >> 2) | (EXP2);
    x = ((x & 0xAA) >> 1) | (EXP3);
    return x;
}

`EXP2` = `0x33`
`EXP3` = `0x55`

舉例原本為 01101010
1. 第一行 `x = (x >> 4) | (x << 4)`
把前 4 個 bits 與後 4 個 bits對調

0110 | 1010

1010 | 0110

2. 第二行 `x = ((x & 0xCC) >> 2) | (0x33)`
對調後的分成兩個兩個一組在對調

10 | 10 | 01 | 10

10 | 10 | 10 | 01

3. 最後再兩兩對調

1 | 0 | 1 | 0 | 1 | 0 | 0 | 1

0 | 1 | 0 | 1 | 0 | 1 | 1 | 0