// Tolerance is not set, defaults to half of the locker clients.
	tolerance := len(restClnts) / 2

	// Quorum is effectively = total clients subtracted with tolerance limit
	quorum := len(restClnts) - tolerance
	if !isReadLock {
		// In situations for write locks, as a special case
		// to avoid split brains we make sure to acquire
		// quorum + 1 when tolerance is exactly half of the
		// total locker clients.

szemben
szerint
szinte
talán
tehát
teljes
tovább
továbbá
több
úgy
ugyanis
új
újabb
újra
után
utána
utolsó
vagy
vagyis
valaki
valami
valamint
való
vagyok
van
vannak
volt
voltam
voltak
voltunk
vissza
vele
viszont

Alo
lume

MinIO protects data against hardware failures and silent data corruption using erasure code and checksums. With the highest level of redundancy, you may lose up to half (N/2) of the total drives and still be able to recover the data.

## What is Erasure Code?

  private final long k1;

  /**
   * @param c the number of compression rounds (must be positive)
   * @param d the number of finalization rounds (must be positive)
   * @param k0 the first half of the key
   * @param k1 the second half of the key
   */
  SipHashFunction(int c, int d, long k0, long k1) {
    checkArgument(
        c > 0, "The number of SipRound iterations (c=%s) during Compression must be positive.", c);

{!../../docs_src/python_types/tutorial001.py!}
```

### Düzenle

Bu çok basit bir program.

Ama şimdi sıfırdan yazdığınızı hayal edin.

Bir noktada fonksiyonun tanımına başlayacaktınız, parametreleri hazır hale getirdiniz...

Ama sonra "ilk harfi büyük harfe dönüştüren yöntemi" çağırmanız gerekir.

 `upper` mıydı ? Yoksa  `uppercase`' mi? `first_uppercase`? `capitalize`?

g","flen","forshaga","färgelanda","gagnef","gislaved","gnesta","gnosjö","gotland","grum","grästorp","gullspång","gällivare","gävle","göteborg","götene","habo","hagfor","hallsberg","hallstahammar","halmstad","hammarö","haninge","haparanda","heby","hedemora","helsingborg","herrljunga","hjo","hofor","huddinge","hudiksvall","hultsfred","hylte","håbo","hällefor","härjedalen","härnösand","härryda","hässleholm","höganäs","högsby","hörby","höör","jokkmokk","järfälla","jönköping","kalix","kalmar","karlsb...

   *   <li>awaits executor service termination for half of the specified timeout.
   *   <li>if the timeout expires, it calls {@link ExecutorService#shutdownNow()}, cancelling
   *       pending tasks and interrupting running tasks.
   *   <li>awaits executor service termination for the other half of the specified timeout.
   * </ol>
   *

      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;
    return oldSlowFactorial(n1, mid).multiply(oldSlowFactorial(mid, n2));
  }

  @Benchmark
  int slowFactorial(int reps) {
    int tmp = 0;

      return entry.value;
    }
    entry = cacheEntry2;
    if (entry != null && entry.key == key) {
      // Promote second cache entry to first so the access pattern
      // [K1, K2, K1, K3, K1, K4...] still hits the cache half the time.
      addToCache(entry);
      return entry.value;
    }
    return null;
  }

  @Override
  void clearCache() {
    super.clearCache();
    cacheEntry1 = null;
    cacheEntry2 = null;
  }