2022q1 第 8 週測驗題

tags: `linux2022`

測驗 `1`

SIMD within a register (SWAR) 是軟體最佳化技巧之一，以下展示 SWAR 運用於 64 位元微處理器架構，原本判斷 2 個 32 位元寬度的整數是否都是奇數 (odd)，可能會這樣撰寫:

#include <stdint.h>
bool both_odd(uint32_t x, uint32_t y) {
    return (x & 1) && (y & 1);
}

但我們可先組合 (compound) 2 個 32 位元寬度的整數為 1 個 64 位元整數，再運用特製的 bitmask，從而減少運算量:

static uint64_t SWAR_ODD_MASK = (1L << 32) + 1;
bool both_odd_swar(uint64_t xy) {
    return (xy & SWAR_ODD_MASK) == SWAR_ODD_MASK;
}

測試程式:

static inline uint64_t bit_compound(uint32_t x, uint32_t y) {
    return ((0L + x) << 32) | ((y + 0L) & (-1L >> 32));
}

int main()
{
    int x = 12345678;
    int y = 9012345;
    uint64_t xy = bit_compound(x, y);
    printf("%d, %d\n", both_odd(x, y), both_odd_swar(xy));
}

延伸閱讀: SIMD and SWAR Techniques

在 Linux 核心原始程式碼中，lib/string.c 具備 memchr 的實作:

/**
 * memchr - Find a character in an area of memory.
 * @s: The memory area
 * @c: The byte to search for
 * @n: The size of the area.
 *
 * returns the address of the first occurrence of @c, or %NULL
 * if @c is not found
 */
void *memchr(const void *s, int c, size_t n)
{
    const unsigned char *p = s;
    while (n-- != 0) {
        if ((unsigned char)c == *p++) {
            return (void *)(p - 1);
        }
    }
    return NULL;
}

測試程式:

int main()
{   
    const char str[] = "http://wiki.csie.ncku.edu.tw";
    const char ch = '.';
    
    char *ret = memchr_opt(str, ch, strlen(str));
    printf("String after |%c| is - |%s|\n", ch, ret);
    return 0;
}

預期執行結果:

String after |.| is - |.csie.ncku.edu.tw|

利用上述 SIMD within a register (SWAR) 的技巧，我們可改寫為以下 memchr_opt 函式:

#include <stddef.h>
#include <stdint.h>
#include <limits.h>
#include <string.h>

/* Nonzero if either X or Y is not aligned on a "long" boundary */
#define UNALIGNED(X) ((long) X & (sizeof(long) - 1))

/* How many bytes are loaded each iteration of the word copy loop */
#define LBLOCKSIZE (sizeof(long))

/* Threshhold for punting to the bytewise iterator */
#define TOO_SMALL(LEN) ((LEN) < LBLOCKSIZE)

#if LONG_MAX == 2147483647L
#define DETECT_NULL(X) (((X) -0x01010101) & ~(X) & 0x80808080)
#else
#if LONG_MAX == 9223372036854775807L
/* Nonzero if X (a long int) contains a NULL byte. */
#define DETECT_NULL(X) (((X) -0x0101010101010101) & ~(X) & 0x8080808080808080)
#else
#error long int is not a 32bit or 64bit type.
#endif
#endif

/* @return nonzero if (long)X contains the byte used to fill MASK. */
#define DETECT_CHAR(X, MASK) (DETECT_NULL(X ^ MASK))

void *memchr_opt(const void *src_void, int c, size_t length)
{
    const unsigned char *src = (const unsigned char *) src_void;
    unsigned char d = c;

    while (UNALIGNED(src)) {
        if (!length--)
            return NULL;
        if (*src == d)
            return (void *) src;
        src++;
    }

    if (!TOO_SMALL(length)) {
        /* If we get this far, we know that length is large and
         * src is word-aligned.
         */

        /* The fast code reads the source one word at a time and only performs
         * the bytewise search on word-sized segments if they contain the search
         * character, which is detected by XORing the word-sized segment with a
         * word-sized block of the search character and then detecting for the
         * presence of NULL in the result.
         */
        unsigned long *asrc = (unsigned long *) src;
        unsigned long mask = d << 8 | d;
        mask = mask << 16 | mask;
        for (unsigned int i = 32; i < LBLOCKSIZE * 8; i <<= 1)
            mask = (mask << i) | mask;

        while (length >= LBLOCKSIZE) {
            /* XXXXX: Your implementation should appear here */
        }

        /* If there are fewer than LBLOCKSIZE characters left, then we resort to
         * the bytewise loop.
         */
        src = (unsigned char *) asrc;
    }

    while (length--) {
        if (*src == d)
            return (void *) src;
        src++;
    }

    return NULL;
}

請補完程式碼，使上述 memchr_opt 的實作符合 memchr 行為，作答規範:

列出 memchr_opt 函式完整程式碼，儘量撰寫程式註解
XXXX 所在的 scope 應該利用 DETECT_CHAR 巨集
儘量以最精簡的程式碼撰寫

延伸問題:

解釋上述程式碼運作原理
比較 Linux 核心原本 (與平台無關) 的實作和 memchr_opt，設計實驗來觀察隨著字串長度和特定 pattern 變化的效能影響
在 Linux 核心原始程式碼找出 x86_64 或 arm64 對應的最佳化實作，跟上述程式碼比較，並嘗試舉出其中的策略和分析

2022q1 第 8 週測驗題

tags: linux2022

測驗 1

tags: `linux2022`

測驗 `1`