Redis从入门到放弃系列(四) Set
本文例子基于:5.0.4 Set是Redis中一种比较常见的数据结构,当存储的member为十进制64位有符号整数范围内的整数的字符串的时候其实现为intset,其他为hashtable
首先让我们来看一下该如何在redis里面使用Set类型
1//设置key的集合中的值为member 2sadd key member [member ...] 3
代码示例:
1> sadd books java python c 2(integer) 3 3//当我们重复添加相同的数据的时候,redis返回为0 4> sadd books java python c 5(integer) 0 6---------------------------------- 7//返回books集合的所有元素 8>smembers books 91) "c" 102) "python" 113) "java" 12---------------------------------- 13//判断某个元素是否在集合里面 14>sismember books c 15(integer) 1 16>sismember books 99 17(integer) 0 18---------------------------------- 19//两个集合的交集 20> sadd new_books java c++ R 21(integer) 3 22> SINTER books new_books 231) "java" 24---------------------------------- 25//两个集合的并集 26> SUNION books new_books 271) "java" 282) "python" 293) "c" 304) "c++" 315) "R" 32//两个集合的差集 33> SMEMBERS books 341) "c" 352) "python" 363) "java" 37> SMEMBERS new_books 381) "R" 392) "c++" 403) "java" 41> SDIFF books new_books 421) "python" 432) "c" 44> SDIFF new_books books 451) "R" 462) "c++" 47
至此,redis set的用法先告一段落.
源码解析
本文开头的时候讲set实现分为intset跟hashtable,hashtable这块讲解的话可以去回头看一下Redis从入门到放弃系列(二) Hash 本节重点来讲一下intset~当存储元素为整数的时候,redis为了节省空间,采用了intset这种数据结构来做存储,我们知道set结构存储字符串的时候都是无序的,可当采用intset来存储的整数的时候, set是有序的,内部采用了二分法方便快速查询 让我们先来看一下intset内部结构
1typedef struct intset { 2 uint32_t encoding; 3 uint32_t length; 4 int8_t contents[]; 5} intset; 6
我们发现intset里面其实是由一个变量类型,一个长度表示的,也就是说要计算当前intset占据的字节:encoding * length; redis在使用intset的时候,首先会判断当前插入的value的大小,然后返回不同字节的类型
1/* Note that these encodings are ordered, so: 2 * INTSET_ENC_INT16 < INTSET_ENC_INT32 < INTSET_ENC_INT64. */ 3#define INTSET_ENC_INT16 (sizeof(int16_t)) 4#define INTSET_ENC_INT32 (sizeof(int32_t)) 5#define INTSET_ENC_INT64 (sizeof(int64_t)) 6 7/* Return the required encoding for the provided value. */ 8static uint8_t _intsetValueEncoding(int64_t v) { 9 if (v < INT32_MIN || v > INT32_MAX) 10 return INTSET_ENC_INT64; 11 else if (v < INT16_MIN || v > INT16_MAX) 12 return INTSET_ENC_INT32; 13 else 14 return INTSET_ENC_INT16; 15} 16
当每次插入的value的值大于当前类型的话,redis会将intset升级为更大的编码
1/* Upgrades the intset to a larger encoding and inserts the given integer. */ 2static intset *intsetUpgradeAndAdd(intset *is, int64_t value) { 3 uint8_t curenc = intrev32ifbe(is->encoding); 4 uint8_t newenc = _intsetValueEncoding(value); 5 int length = intrev32ifbe(is->length); 6 int prepend = value < 0 ? 1 : 0; 7 8 /* First set new encoding and resize */ 9 is->encoding = intrev32ifbe(newenc); 10 is = intsetResize(is,intrev32ifbe(is->length)+1); 11 12 /* Upgrade back-to-front so we don't overwrite values. 13 * Note that the "prepend" variable is used to make sure we have an empty 14 * space at either the beginning or the end of the intset. */ 15 while(length--) 16 _intsetSet(is,length+prepend,_intsetGetEncoded(is,length,curenc)); 17 18 /* Set the value at the beginning or the end. */ 19 if (prepend) 20 _intsetSet(is,0,value); 21 else 22 _intsetSet(is,intrev32ifbe(is->length),value); 23 is->length = intrev32ifbe(intrev32ifbe(is->length)+1); 24 return is; 25} 26
前面我们说过,intset是一个有序的,然后查找的时候采用了二分法来查找元素,那么其内部是如何实现的呢?
1/* Search for the position of "value". Return 1 when the value was found and 2 * sets "pos" to the position of the value within the intset. Return 0 when 3 * the value is not present in the intset and sets "pos" to the position 4 * where "value" can be inserted. */ 5static uint8_t intsetSearch(intset *is, int64_t value, uint32_t *pos) { 6 int min = 0, max = intrev32ifbe(is->length)-1, mid = -1; 7 int64_t cur = -1; 8 9 /* The value can never be found when the set is empty */ 10 if (intrev32ifbe(is->length) == 0) { 11 if (pos) *pos = 0; 12 return 0; 13 } else { 14 /* Check for the case where we know we cannot find the value, 15 * but do know the insert position. */ 16 if (value > _intsetGet(is,max)) { 17 if (pos) *pos = intrev32ifbe(is->length); 18 return 0; 19 } else if (value < _intsetGet(is,0)) { 20 if (pos) *pos = 0; 21 return 0; 22 } 23 } 24 25 while(max >= min) { 26 mid = ((unsigned int)min + (unsigned int)max) >> 1; 27 cur = _intsetGet(is,mid); 28 if (value > cur) { 29 min = mid+1; 30 } else if (value < cur) { 31 max = mid-1; 32 } else { 33 break; 34 } 35 } 36 37 if (value == cur) { 38 if (pos) *pos = mid; 39 return 1; 40 } else { 41 if (pos) *pos = min; 42 return 0; 43 } 44} 45/* Insert an integer in the intset */ 46intset *intsetAdd(intset *is, int64_t value, uint8_t *success) { 47 uint8_t valenc = _intsetValueEncoding(value); 48 uint32_t pos; 49 if (success) *success = 1; 50 51 /* Upgrade encoding if necessary. If we need to upgrade, we know that 52 * this value should be either appended (if > 0) or prepended (if < 0), 53 * because it lies outside the range of existing values. */ 54 if (valenc > intrev32ifbe(is->encoding)) { 55 /* This always succeeds, so we don't need to curry *success. */ 56 return intsetUpgradeAndAdd(is,value); 57 } else { 58 /* Abort if the value is already present in the set. 59 * This call will populate "pos" with the right position to insert 60 * the value when it cannot be found. */ 61 if (intsetSearch(is,value,&pos)) { 62 if (success) *success = 0; 63 return is; 64 } 65 66 is = intsetResize(is,intrev32ifbe(is->length)+1); 67 if (pos < intrev32ifbe(is->length)) intsetMoveTail(is,pos,pos+1); 68 } 69 70 _intsetSet(is,pos,value); 71 is->length = intrev32ifbe(intrev32ifbe(is->length)+1); 72 return is; 73} 74
重点在while操作那一段~我们可以看到,其查找采用了二分法,那么如何让其有序呢?
1if (intsetSearch(is,value,&pos)) { 2 if (success) *success = 0; 3 return is; 4} 5
看到这一段了吧?判断查找的时候,将pos的位置查找了出来,给下面_intsetSet操作做前奏~
应用场景
1.去重~ 2.查看两个人的共同爱好
写在最后
祝大家520快乐~
