我有一个特定的要求,即将字节流转换为每个字符恰好为6位的字符编码。
Here's an example:
Input: 0x50 0x11 0xa0
Character Table:
010100 T
000001 A
000110 F
100000 SPACE
Output: "TAF "
Logically I can understand how this works:
Taking 0x50 0x11 0xa0 and showing as binary:
01010000 00010001 10100000
Which is "TAF ".
以编程方式执行此操作的最佳方法是什么(伪代码或c ++)。谢谢!
答案 0 :(得分:6)
好吧,每3个字节,你最终得到4个字符。所以首先,如果输入不是三个字节的倍数,你需要弄清楚要做什么。 (它是否有某种填充,如base64?)
然后我可能依次取每个3个字节。在C#中,它足够接近C的伪代码:)
for (int i = 0; i < array.Length; i += 3)
{
// Top 6 bits of byte i
int value1 = array[i] >> 2;
// Bottom 2 bits of byte i, top 4 bits of byte i+1
int value2 = ((array[i] & 0x3) << 4) | (array[i + 1] >> 4);
// Bottom 4 bits of byte i+1, top 2 bits of byte i+2
int value3 = ((array[i + 1] & 0xf) << 2) | (array[i + 2] >> 6);
// Bottom 6 bits of byte i+2
int value4 = array[i + 2] & 0x3f;
// Now use value1...value4, e.g. putting them into a char array.
// You'll need to decode from the 6-bit number (0-63) to the character.
}
答案 1 :(得分:3)
以防万一有人感兴趣 - 另一种变体在流出现时立即从流中提取6位数字。也就是说,即使当前读取的字节少于3个字节,也可以获得结果。对于未填充的流非常有用。
代码将累加器a
的状态保存在变量n
中,该变量存储前一次读取中累加器中剩余的位数。
int n = 0;
unsigned char a = 0;
unsigned char b = 0;
while (read_byte(&byte)) {
// save (6-n) most significant bits of input byte to proper position
// in accumulator
a |= (b >> (n + 2)) & (077 >> n);
store_6bit(a);
a = 0;
// save remaining least significant bits of input byte to proper
// position in accumulator
a |= (b << (4 - n)) & ((077 << (4 - n)) & 077);
if (n == 4) {
store_6bit(a);
a = 0;
}
n = (n + 2) % 6;
}