[英]Reversing 7-bit integer encoding
浏览现有代码库相对较大的代码时,发现以下功能:
int write_actual_size(unsigned int actual_size, int &out_size)
{
unsigned char second;
unsigned char third;
unsigned char fourth;
int result;
int usedBytes;
*(unsigned char *)out_size = actual_size | 0x80;
if ( actual_size < 0x80 ) {
*(unsigned char *)out_size = ((unsigned char)actual_size | 0x80) & 0x7F;
result = 1;
} else {
second = (actual_size >> 7) | 0x80;
*(unsigned char *)(out_size + 1) = second;
if (actual_size < 0x4000) {
*(unsigned char *)(out_size + 1) = second & 0x7F;
usedBytes = 2;
} else {
third = (actual_size >> 14) | 0x80;
*(unsigned char *)(out_size + 2) = third;
if (actual_size < 0x200000) {
*(unsigned char *)(out_size + 2) = third & 0x7F;
usedBytes = 3;
}
else {
fourth = (actual_size >> 21) | 0x80;
*(unsigned char *)(out_size + 3) = fourth;
if (actual_size < 0x10000000) {
*(unsigned char *)(out_size + 3) = fourth & 0x7F;
usedBytes = 4;
}
}
}
result = usedBytes;
}
return result;
}
根据原始输入大小,这会将一个普通的无符号整数编码为一个或多个字节。
据我了解,最左边的位用于确定是否有“后续”字节。 我认为这样做的原因是为了节省带宽(即使每个数据包最多3个字节)。 这些合理的假设吗?
我想制作一个read_actual_size版本...我可以将每个字节线性“右移7”直到遇到“ 0”吗?
请不要太苛刻,我对C还是很陌生。
通用VLQ解码器如下所示:
int decode_vlq(unsigned char *input)
{
int result = 0;
do
{
result = (result << 7) | (*input & 0x7F);
}
while (*input++ & 0x80);
return result;
}
我乐于接受建议,因为我的C很生锈,我是手工编写的。
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