---
tags: linux2022
---
# 2022-04-04 (quiz8)
contributed by < `sternacht` >

## 測驗 1

### 作答

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) {
            /* detect if target char in this const length 
             * string or not. Break when target is found,
             * else detect target in next string.
             */
            if (DETECT_CHAR(*asrc, mask))
                break;
            asrc ++;
            length -= LBLOCKSIZE;
        }    

        /* 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;
}

## 測驗 2

### 作答

static inline ringidx_t cond_reload(ringidx_t idx, const ringidx_t *loc)
{
    ringidx_t fresh = __atomic_load_n(loc, __ATOMIC_RELAXED);
    if (before(idx, fresh)) { /* fresh is after idx, use this instead */
        idx = fresh;
    } else { /* Continue with next slot */
	    idx++;      // DDD
    }
    return idx;
}

static inline ringidx_t find_tail(lfring_t *lfr, ringidx_t head, ringidx_t tail)
{
    if (lfr->flags & LFRING_FLAG_SP) /* single-producer enqueue */
        return __atomic_load_n(&lfr->tail, __ATOMIC_ACQUIRE);

    /* Multi-producer enqueue.
     * Scan ring for new elements that have been written but not released.
     */
    ringidx_t mask = lfr->mask;
    ringidx_t size = mask + 1;     // KKK
    while (before(tail, head + size) &&
           __atomic_load_n(&lfr->ring[tail & mask].idx, __ATOMIC_ACQUIRE) ==
               tail)        // TTT
        tail++;
    tail = cond_update(&lfr->tail, tail);
    return tail;
}

uint32_t lfring_dequeue(lfring_t *lfr,
                        void **restrict elems,
                        uint32_t n_elems,
                        uint32_t *index)
{
    ringidx_t mask = lfr->mask;
    intptr_t actual;
    ringidx_t head = __atomic_load_n(&lfr->head, __ATOMIC_RELAXED);
    ringidx_t tail = __atomic_load_n(&lfr->tail, __ATOMIC_ACQUIRE);
    do { /* skipped */
    } while (!__atomic_compare_exchange_n(
        &lfr->head, &head, /* Updated on failure */
        /* XXXXX */ HHH,
        /* weak */ false, __ATOMIC_RELAXED, __ATOMIC_RELAXED));
    *index = (uint32_t) head;
    return (uint32_t) actual;
}

`cond_reload` 被用在當目前記錄的 `tail` 已經不再是整個 ring 的 tail 時，將變數 `tail` 更新，方式是透過 `before` 來比較，若 `tail` < `fresh`，則更新 `tail`。

需要 `find_tail` 的時機是多個 producer ，已經寫入 ring 但尚未更新到 `lfr->tail`，這時候就需要用 `find_tail` 來做更新，使 consumer 可以得到正確的資訊。
## 測驗 3