Function dictRehash

app/redis-6.2.6/src/dict.c:211–254

Performs N steps of incremental rehashing. Returns 1 if there are still * keys to move from the old to the new hash table, otherwise 0 is returned. * * Note that a rehashing step consists in moving a bucket (that may have more * than one key as we use chaining) from the old to the new hash table, however * since part of the hash table may be composed of empty spaces, it is not * guaranteed t

 * will visit at max N*10 empty buckets in total, otherwise the amount of
 * work it does would be unbound and the function may block for a long time. */
int dictRehash(dict *d, int n) {
    int empty_visits = n*10; /* Max number of empty buckets to visit. */
    if (!dictIsRehashing(d)) return 0;

    while(n-- && d->ht[0].used != 0) {
        dictEntry *de, *nextde;

        /* Note that rehashidx can't overflow as we are sure there are more
         * elements because ht[0].used != 0 */
        assert(d->ht[0].size > (unsigned long)d->rehashidx);
        while(d->ht[0].table[d->rehashidx] == NULL) {
            d->rehashidx++;
            if (--empty_visits == 0) return 1;
        }
        de = d->ht[0].table[d->rehashidx];
        /* Move all the keys in this bucket from the old to the new hash HT */
        while(de) {
            uint64_t h;

            nextde = de->next;
            /* Get the index in the new hash table */
            h = dictHashKey(d, de->key) & d->ht[1].sizemask;
            de->next = d->ht[1].table[h];
            d->ht[1].table[h] = de;
            d->ht[0].used--;
            d->ht[1].used++;
            de = nextde;
        }
        d->ht[0].table[d->rehashidx] = NULL;
        d->rehashidx++;
    }

    /* Check if we already rehashed the whole table... */
    if (d->ht[0].used == 0) {
        zfree(d->ht[0].table);
        d->ht[0] = d->ht[1];
        _dictReset(&d->ht[1]);
        d->rehashidx = -1;
        return 0;
    }

    /* More to rehash... */
    return 1;
}

long long timeInMilliseconds(void) {
    struct timeval tv;