Source:
// Calculate hash value 32
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
Explain: key.hashCode () to get 32-bit hash value shaping
h >>> 16 represents a 16-bit logical right shift, left patch 16 0, corresponds to the high 16-bit 16-bit moves to the low
^ Denotes exclusive OR operation, the same is 0, 1 is different, the probability of 0 and 1 are 1/2
It means that the 32-bit hash of high worth, returns treated 32-bit hash values are different and the low 16-bit or 16-bit operations
// The length of the hash calculation of array subscript n
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
Explain: n is the length of the array, n being a power of 2, in the form of binary 10000 ...
N-1 is the form of 1111, with any number of equal phase are less than n-1
1 // move left four get 16
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
Explanation: The left four binary 1 is 10000, the decimal turn, is 2 ^ 4 = 16 power
// size of the array 16 as an initial
newCap = DEFAULT_INITIAL_CAPACITY;
Question: Why is the array of length 2 ^ n-th power
Answer: Since the 2 ^ n-1 selected array subscript, binary form 2 ^ n-1 to n-1 from the composition, when the phase with the hash value, the number will give less 2 ^ n-1 and The array index is calculated the same low probability
