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sqlite variable-length integers 实现方式

发布时间:2020-12-12 19:32:47 所属栏目:百科 来源:网络整理
导读:网站地址:http://sqlite.org/src4/doc/trunk/www/varint.wiki 这种存储方式的好处在于, 1 可以用比较少的位数来存储比较小的数字(在实际应用中,存储的比较多),而用比较多的位数来存储比较大的数字(在实际应用中,存储的比较少)。 2 长度可以在第一个

网站地址:http://sqlite.org/src4/doc/trunk/www/varint.wiki


这种存储方式的好处在于,

1 可以用比较少的位数来存储比较小的数字(在实际应用中,存储的比较多),而用比较多的位数来存储比较大的数字(在实际应用中,存储的比较少)。

2 长度可以在第一个字节就可以知道

3 编码之后的字节串的排序没改变。一个已经排序好的数组,编码之后,还是排好序的


编码:

  • If V<=240 then output a single by A0 equal to V.
  • If V<=2287 then output A0 as (V-240)/256 + 241 and A1 as (V-240)%256.
  • If V<=67823 then output A0 as 249,A1 as (V-2288)/256,and A2 as (V-2288)%256.
  • If V<=16777215 then output A0 as 250 and A1 through A3 as a big-endian 3-byte integer.
  • If V<=4294967295 then output A0 as 251 and A1..A4 as a big-ending 4-byte integer.
  • If V<=1099511627775 then output A0 as 252 and A1..A5 as a big-ending 5-byte integer.
  • If V<=281474976710655 then output A0 as 253 and A1..A6 as a big-ending 6-byte integer.
  • If V<=72057594037927935 then output A0 as 254 and A1..A7 as a big-ending 7-byte integer.
  • Otherwise then output A0 as 255 and A1..A8 as a big-ending 8-byte integer.

解码

  • If A0 is between 0 and 240 inclusive,then the result is the value of A0.
  • If A0 is between 241 and 248 inclusive,then the result is 240+256*(A0-241)+A1.
  • If A0 is 249 then the result is 2288+256*A1+A2.
  • If A0 is 250 then the result is A1..A3 as a 3-byte big-ending integer.
  • If A0 is 251 then the result is A1..A4 as a 4-byte big-ending integer.
  • If A0 is 252 then the result is A1..A5 as a 5-byte big-ending integer.
  • If A0 is 253 then the result is A1..A6 as a 6-byte big-ending integer.
  • If A0 is 254 then the result is A1..A7 as a 7-byte big-ending integer.
  • If A0 is 255 then the result is A1..A8 as a 8-byte big-ending integer.

位数 最大数字 1 240 2 2287 3 67823 4 224-1 5 232-1 6 240-1 7 248-1 8 256-1 9 264-1

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