在hash中查找key的时候,经常会发现用&取代%,先看两段代码吧,
/**
* Returns index for hash code h.
*/
static int indexFor(int h, int length) {
return h & (length-1);
}
n.size = realsize;
n.sizemask = realsize-1;
//此处略去xxx行
while(de) {
unsigned int 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;
}
大家可以看到a%b取模的形式都被替换成了a&(b-1) ,当hashtable的长度是2的幂的情况下(疏忽,一开始没写),这两者是等价的,那为什么要用后者呢?
另一方面,为什么hashtable的长度最好要是2的n次方呢,这个不在本次讨论范围之列,原因简单说一下就是1、分布更均匀 2、碰撞几率更小 详情自己思考,JDK中的HashMap就会在初始化时,保证这一点:
1. public HashMap(int initialCapacity, float loadFactor) {
2. if (initialCapacity < 0)
3. throw new IllegalArgumentException("Illegal initial capacity: " +
4. initialCapacity);
5. if (initialCapacity > MAXIMUM_CAPACITY)
6. initialCapacity = MAXIMUM_CAPACITY;
7. if (loadFactor <= 0 || Float.isNaN(loadFactor))
8. throw new IllegalArgumentException("Illegal load factor: " +
9. loadFactor);
10.
11. // Find a power of 2 >= initialCapacity
12. int capacity = 1;
13. while (capacity < initialCapacity)
14. capacity <<= 1;
15.
16. this.loadFactor = loadFactor;
17. threshold = (int)(capacity * loadFactor);
18. table = new Entry[capacity];
19. init();
20. }
redis中也有类似的保证:
1. /* Our hash table capability is a power of two */
2. static unsigned long _dictNextPower(unsigned long size)
3. {
4. unsigned long i = DICT_HT_INITIAL_SIZE;
5.
6. if (size >= LONG_MAX) return LONG_MAX;
7. while(1) {
8. if (i >= size)
9. return i;
10. i *= 2;
11. }
12. }
言归正传,大家都知道位运算的效率最高,这也是&取代%的原因,来看个程序:
1. int main(int argc, char* argv[])
2. {
3. int a = 0x111;
4. int b = 0x222;
5. int c = 0;
6. int d = 0;
7.
8. c = a & (b-1);
9. d = a % b;
10.
2. return 0;
3. }
看反汇编的结果:
反汇编代码
1. 13: c = a & (b-1);
2. 00401044 mov eax,dword ptr [ebp-8]
3. 00401047 sub eax,1
4. 0040104A mov ecx,dword ptr [ebp-4]
5. 0040104D and ecx,eax
6. 0040104F mov dword ptr [ebp-0Ch],ecx
7. 14: d = a % b;
8. 00401052 mov eax,dword ptr [ebp-4]
9. 00401055 cdq
10. 00401056 idiv eax,dword ptr [ebp-8]
11. 00401059 mov dword ptr [ebp-10h],edx
可以看到,&操作用了:3mov+1and+1sub %操作用了:2mov+1cdp+1idiv
我们可以查阅Coding_ASM_-_Intel_Instruction_Set_Codes_and_Cycles资料,发现前者只需5个CPU周期,而后者至少需要26个CPU周期(注意,是最少!!!) 效率显而易见。所以以后自己在写的时候,也可以使用前者的写法。