emmansun
commented
11 months ago 对比一下STD编码的base64解码：
// The input consists of six character sets in the Base64 alphabet, which we
// need to map back to the 6-bit values they represent. There are three ranges,
// two singles, and then there's the rest.
//
// # From To Add Characters
// 1 [43] [62] +19 +
// 2 [47] [63] +16 /
// 3 [48..57] [52..61] +4 0..9
// 4 [65..90] [0..25] -65 A..Z
// 5 [97..122] [26..51] -71 a..z
// (6) Everything else => invalid input
//
// We will use lookup tables for character validation and offset computation.
// Remember that 0x2X and 0x0X are the same index for _mm_shuffle_epi8, this
// allows to mask with 0x2F instead of 0x0F and thus save one constant
// declaration (register and/or memory access).
//
// For offsets:
// Perfect hash for lut = ((src >> 4) & 0x2F) + ((src == 0x2F) ? 0xFF : 0x00)
// 0000 = garbage
// 0001 = /
// 0010 = +
// 0011 = 0-9
// 0100 = A-Z
// 0101 = A-Z
// 0110 = a-z
// 0111 = a-z
// 1000 >= garbage
//
// For validation, here's the table.
// A character is valid if and only if the AND of the 2 lookups equals 0:
//
// hi \ lo 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
// LUT 0x15 0x11 0x11 0x11 0x11 0x11 0x11 0x11 0x11 0x11 0x13 0x1A 0x1B 0x1B 0x1B 0x1A
//
// 0000 0x10 char NUL SOH STX ETX EOT ENQ ACK BEL BS HT LF VT FF CR SO SI
// andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
//
// 0001 0x10 char DLE DC1 DC2 DC3 DC4 NAK SYN ETB CAN EM SUB ESC FS GS RS US
// andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
//
// 0010 0x01 char ! " # $ % & ' ( ) * + , - . /
// andlut 0x01 0x01 0x01 0x01 0x01 0x01 0x01 0x01 0x01 0x01 0x01 0x00 0x01 0x01 0x01 0x00
//
// 0011 0x02 char 0 1 2 3 4 5 6 7 8 9 : ; < = > ?
// andlut 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x02 0x02 0x02 0x02 0x02 0x02
//
// 0100 0x04 char @ A B C D E F G H I J K L M N O
// andlut 0x04 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
//
// 0101 0x08 char P Q R S T U V W X Y Z [ \ ] ^ _
// andlut 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x08 0x08 0x08 0x08 0x08
//
// 0110 0x04 char ` a b c d e f g h i j k l m n o
// andlut 0x04 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
// 0111 0x08 char p q r s t u v w x y z { | } ~
// andlut 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x08 0x08 0x08 0x08 0x08
//
// 1000 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
// 1001 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
// 1010 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
// 1011 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
// 1100 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
// 1101 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
// 1110 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
// 1111 0x10 andlut 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10 0x10
STD编码的base64解码应该没有问题，目前是URL编码的base64验证问题没有解决，产生不了验证表
主要是0101行和0111行最后一个字节不同，最后通过调节两个高四位搞定。