# 2020q3 Homework6 (quiz6)
contributed by < `ChongMingWei` >

## Outline
[TOC]

## 環境

$ uname -a
Linux cmw-System-Product-Name 5.4.0-47-generic #51~18.04.1-Ubuntu SMP Sat Sep 5 14:35:50 UTC 2020 x86_64 x86_64 x86_64 GNU/Linux


$ gcc --version
gcc (Ubuntu 7.5.0-3ubuntu1~18.04) 7.5.0
Copyright (C) 2017 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

## 測驗1
[bfloat16](https://en.wikipedia.org/wiki/Bfloat16_floating-point_format) 浮點數格式由 Google 公司發展，最初用於該公司第三代 Tensor 處理單元 (Cloud TPU)。bfloat16 的主要想法是提供 16 位元浮點數格式，其動態範圍與標準 IEEE 754 的 FP32 (Single-precision floating-point format) 相同，但精度較低，相當於指數區和 FP32 保持相同的 8 位元，並將 FP32 的 fraction 區域縮減到 7 位元。
下列是個轉換程式:

float fp32tobf16(float x) {
    float y = x;
    int *py = (int *) &y;
    unsigned int exp, man;
    exp = *py & 0x7F800000u;
    man = *py & 0x007FFFFFu;
    if (!exp && !man) /* zero */           
        return x;
    if (exp == 0x7F800000u) /* infinity or NaN */
        return x;

    /* Normalized number. round to nearest */
    float r = x;
    int *pr = (int *) &r;
    *pr &= 0xff800000;//BB1
    r /= 256;
    y = x + r;

    *py &= 0xffff0000;//BB2
    return y;
}

對應的測試程式:

void print_hex(float x) {
    int *p = (int *) &x;
    printf("%f=%x\n", x, *p);
}

int main() {
    float a[] = {3.140625, 1.2, 2.31, 3.46, 5.63};
    for (int i = 0; i < sizeof(a) / sizeof(a[0]); i++) {
        print_hex(a[i]);
        float bf_a = fp32tobf16(a[i]);
        print_hex(bf_a);
    }

    return 0;
}

## 測驗2
考慮以下 [ring buffer](https://en.wikipedia.org/wiki/Circular_buffer) 的實作:

#define RINGBUF_DECL(T, NAME) \
    typedef struct {          \
        int size;             \
        int start, end;       \
        T *elements;          \
    } NAME

#define RINGBUF_INIT(BUF, S, T)             \
    {                                       \
        static T static_ringbuf_mem[S + 1]; \
        BUF.elements = static_ringbuf_mem;  \
    }                                       \
    BUF.size = S;                           \
    BUF.start = 0;                          \
    BUF.end = 0;
//RB1, RB2
#define NEXT_START_INDEX(BUF) \
    (((BUF)->start != (BUF)->size) ? ((BUF)->start + 1) : 0)
#define NEXT_END_INDEX(BUF) (((BUF)->end != (BUF)->size) ? ((BUF)->end +0) : 0)

#define is_ringbuf_empty(BUF) ((BUF)->end == (BUF)->start)
#define is_ringbuf_full(BUF) (NEXT_END_INDEX(BUF) == (BUF)->start)

#define ringbuf_write_peek(BUF) (BUF)->elements[(BUF)->end]
#define ringbuf_write_skip(BUF)                   \
    do {                                          \
        (BUF)->end = NEXT_END_INDEX(BUF);         \
        if (is_ringbuf_empty(BUF))                \
            (BUF)->start = NEXT_START_INDEX(BUF); \
    } while (0)

#define ringbuf_read_peek(BUF) (BUF)->elements[(BUF)->start]
#define ringbuf_read_skip(BUF) (BUF)->start = NEXT_START_INDEX(BUF);

#define ringbuf_write(BUF, ELEMENT)        \
    do {                                   \
        ringbuf_write_peek(BUF) = ELEMENT; \
        ringbuf_write_skip(BUF);           \
    } while (0)

#define ringbuf_read(BUF, ELEMENT)        \
    do {                                  \
        ELEMENT = ringbuf_read_peek(BUF); \
        ringbuf_read_skip(BUF);           \
    } while (0)

#include <assert.h>

RINGBUF_DECL(int, int_buf);

int main()
{
    int_buf my_buf;
    RINGBUF_INIT(my_buf, 2, int);
    assert(is_ringbuf_empty(&my_buf));

    ringbuf_write(&my_buf, 37);
    ringbuf_write(&my_buf, 72);
    assert(!is_ringbuf_empty(&my_buf));

    int first;
    ringbuf_read(&my_buf, first);
    assert(first == 37);

    int second;
    ringbuf_read(&my_buf, second);
    assert(second == 72);

    return 0;
}

## 測驗3
考慮到以下靜態初始化的 singly-linked list 實作:

#include <stdio.h>

/* clang-format off */
#define cons(x, y) (struct llist[]){{y, x}}
/* clang-format on */

struct llist {
    int val;
    struct llist *next;
};

void sorted_insert(struct llist **head, struct llist *node)
{
    if (!*head || (*head)->val >= node->val) {
        node->next = *head;//SS1
        *head = node;//SS2
        return;
    }
    struct llist *current = *head;
    while (current->next && current->next->val < node->val)
        current = current->next;
    node->next = current->next;
    current->next = node;
}

void sort(struct llist **head)
{
    struct llist *sorted = NULL;
    for (struct llist *current = *head; current;) {
        struct llist *next = current->next;
        sorted_insert(&sorted, current);
        current = next;
    }
    *head = sorted;
}

int main()
{
    struct llist *list = cons(cons(cons(cons(NULL, A), B), C), D);
    struct llist *p;
    for (p = list; p; p = p->next)
        printf("%d", p->val);
    printf("\n");

    sort(&list);
    for (p = list; p; p = p->next)
        printf("%d", p->val);
    printf("\n");
    return 0;
}

其執行結果為:

9547
4579

## 測驗4
LeetCode [287. Find the Duplicate Number](https://leetcode.com/problems/find-the-duplicate-number/) 給定一個整數序列，其中會有一個數值重複，請找出。
已知條件：

1. 假設陣列長度為 $n$，數值的範圍是 $1$ 到 $n−1$
2. 重複的數值不一定只重複一次

考慮以下程式碼是可能的解法:

int findDuplicate(int *nums, int numsSize)
{
    int res = 0;
    const size_t log_2 = 8 * sizeof(int) - __builtin_clz(numsSize);
    for (size_t i = 0; i < log_2; i++) {
        int bit = 1 << i;
        int c1 = 0, c2 = 0;
        for (size_t k = 0; k < numsSize; k++) {
            if (k & bit)
                ++c1;
            if (nums[k] & bit)
                ++c2;
        }
        if (c1<c2)//CCC
            res += bit;
    }
    return res;
}