* bit(i) can alter output bit(j).
   */
  static void checkNoFunnels(HashFunction function) {
    Random rand = new Random(0);
    int keyBits = 32;
    int hashBits = function.bits();

    // output loop tests input bit
    for (int i = 0; i < keyBits; i++) {
      int same = 0x0; // bitset for output bits with same values
      int diff = 0x0; // bitset for output bits with different values
      int count = 0;

			// ~1.5-3x faster than Rabin-Karp since n is small.
			c0 := b[last]
			c1 := b[last-1] // There are at least 2 chars to match
		loop:
			for ; i < len(s); i++ {
				if s[i] == c0 && s[i-1] == c1 {
					for k := 2; k < n; k++ {
						if s[i-k] != b[last-k] {
							continue loop
						}
					}
					return i - last
				}
			}
		}
		return -1
	}
}

        Beau--ootiful Soo--oop!
        Beau--ootiful Soo--oop!
    Soo--oop of the e--e--evening,
        Beautiful, beautiful Soup!

    `Beautiful Soup!  Who cares for fish,
    Game, or any other dish?
    Who would not give all else for two
    Pennyworth only of beautiful Soup?
    Pennyworth only of beautiful Soup?
        Beau--ootiful Soo--oop!
        Beau--ootiful Soo--oop!

	}

	for i := range dataDown {
		writers[i] = nil
	}
	for i := data; i < data+parityDown; i++ {
		writers[i] = nil
	}

	b.SetBytes(size)
	b.ReportAllocs()
	for b.Loop() {
		bitrotReaders := make([]io.ReaderAt, len(disks))
		for index, disk := range disks {
			if writers[index] == nil {
				continue
			}
			tillOffset := erasure.ShardFileOffset(0, size, size)

   * iterable}. The iterator's {@code hasNext()} method returns {@code true} until {@code iterable}
   * is empty.
   *
   * <p><b>Warning:</b> Typical uses of the resulting iterator may produce an infinite loop. You
   * should use an explicit {@code break} or be certain that you will eventually remove all the
   * elements.
   *
   * <p>To cycle over the iterable {@code n} times, use the following: {@code

		p.errorf("expected identifier, found %q", p.lex.Text())
		return "", "", nil, false // Might as well stop now.
	}
	word, cond = p.lex.Text(), ""
	operands = scratch[:0]
	// Zero or more comma-separated operands, one per loop.
	nesting := 0
	colon := -1
	for tok != '\n' && tok != ';' {
		// Process one operand.
		var items []lex.Token
		if cap(operands) > len(operands) {
			// Reuse scratch items slice.

 * same efficient signaling as we had to hand-code in the {@code ReentrantLock} version above.
 * Finally, the programmer no longer has to hand-code the wait loop, and therefore doesn't have to
 * remember to use {@code while} instead of {@code if}.
 *
 * {@snippet :
 * public class SafeBox<V> {
 *   private V value;
 *   private final Monitor monitor = new Monitor();

/// note | Technical Details

It used <a href="https://github.com/MagicStack/uvloop" class="external-link" target="_blank">`uvloop`</a> instead of the default Python `asyncio` loop. That's what made it so fast.

It clearly inspired Uvicorn and Starlette, that are currently faster than Sanic in open benchmarks.

///

/// check | Inspired **FastAPI** to

   * fluent iterable. The iterator's {@code hasNext()} method returns {@code true} until this fluent
   * iterable is empty.
   *
   * <p><b>Warning:</b> Typical uses of the resulting iterator may produce an infinite loop. You
   * should use an explicit {@code break} or be certain that you will eventually remove all the
   * elements.
   *
   * <p><b>{@code Stream} equivalent:</b> if the source iterable has only a single element {@code

   * fluent iterable. The iterator's {@code hasNext()} method returns {@code true} until this fluent
   * iterable is empty.
   *
   * <p><b>Warning:</b> Typical uses of the resulting iterator may produce an infinite loop. You
   * should use an explicit {@code break} or be certain that you will eventually remove all the
   * elements.
   *
   * <p><b>{@code Stream} equivalent:</b> if the source iterable has only a single element {@code