用Zlib解压缩文本字符串
Decompress text string with zlib
我需要以zlib格式解码十六进制字符串。一个例子是：
180000001300000000aefjyoaazidfcmqgggqqaajwacg ==
其中18000000和13000000是未压缩/压缩数据的大小(在这种情况下为24和19(。
我也知道字符串的其余部分
020000000000000000030000000001000000030000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
问题在哪里？按照任何教程，例如https://panthema.net/2007/0328-zlibstring.html压缩该字符串，我得到了
x？302@？p ??
在十六进制中可以写为
783F33330324053F503F103FF5
与我预期的压缩字符串无关，所以我没有找到解压缩原始字符串的方式(这是我的最终目标(
预先感谢您的任何提示！
ps。我正在使用解压缩程序https://github.com/systemed/intersector/blob/master/helpers.cpp
看来该字符串已在base64上编码(谢谢@zdenek和 @mark-adler(，我设法用
对其进行了解码BYTE *res;
int resSize = FromBase64Simple((BYTE*)actualData.c_str(),actualData.len(),res,sizeCompressed);
您可以从https://github.com/kengonakajima/luvit-base64/blob/master/base/base64.c 读取实施。
，但这不是问题，因为我可以使用
转让结果char* resChar = new char[resSize];
for(int i = 0;i<resSize;i++)
{
int asciiCode = (BYTE)res[i];
resChar[i]=char(asciiCode);
char buffer [2];
itoa (asciiCode,buffer,16);
qDebug()<<"["<<i<<"]t"<<asciiCode<<"t"<<buffer;
}
我在十进制和十六进制中获得每个字节的结果，两者都可以。十六进制看起来像：
78 01 63 62 80 00 66 20 C5 08 C4 20 1A 04 00 9C 00 0A
但是reschar是" x？cb？"这与 @Mark-Adler所说的" X？302 @？p"无关。(显然在哪里'？'符号是不可打印的(，我真的认为这是问题所在，但是我的数据似乎与此表相符：https：//www.asciatible.com/，Mark的一个人也不是这样。Web https://conv.darkbyte.ru/返回与我的算法相同的结果
我尝试使用上面说的实现来解压缩字符串，但失败了(也尝试了https://gist.github.com/arq5x/5315739(，但其解剖值是单个字符串"
在这里，我们使用最小可重复的情况：
#include <string>
static char LookupDigits[] = {
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, //gap: ctrl chars
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, //gap: ctrl chars
0,0,0,0,0,0,0,0,0,0,0, //gap: spc,!"#$%'()*
62, // +
0, 0, 0, // gap ,-.
63, // /
52, 53, 54, 55, 56, 57, 58, 59, 60, 61, // 0-9
0, 0, 0, // gap: :;<
99, // = (end padding)
0, 0, 0, // gap: >?@
0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14,15,16,
17,18,19,20,21,22,23,24,25, // A-Z
0, 0, 0, 0, 0, 0, // gap: []^_`
26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,
43,44,45,46,47,48,49,50,51, // a-z
0, 0, 0, 0, // gap: {|}~ (and the rest...)
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
};
int FromBase64Simple(const unsigned char* pSrc, int nLenSrc, unsigned char* pDst, int nLenDst)
{
int nLenOut = 0;
for (int j = 0; j<nLenSrc; j += 4) {
if (nLenOut > nLenDst) {
return(0); // error, buffer too small
}
unsigned char s1 = LookupDigits[*pSrc++];
unsigned char s2 = LookupDigits[*pSrc++];
unsigned char s3 = LookupDigits[*pSrc++];
unsigned char s4 = LookupDigits[*pSrc++];
unsigned char d1 = ((s1 & 0x3f) << 2) | ((s2 & 0x30) >> 4);
unsigned char d2 = ((s2 & 0x0f) << 4) | ((s3 & 0x3c) >> 2);
unsigned char d3 = ((s3 & 0x03) << 6) | ((s4 & 0x3f) >> 0);
*pDst++ = d1; nLenOut++;
if (s3 == 99) break; // end padding found
*pDst++ = d2; nLenOut++;
if (s4 == 99) break; // end padding found
*pDst++ = d3; nLenOut++;
}
return(nLenOut);
}
int main()
{
std::string inputData = "eAFjYoAAZiDFCMQgGgQAAJwACg==";
//19 is hardcoded since I know its size prior to this call
unsigned char res[19];
int resSize = FromBase64Simple((unsigned char*)inputData.c_str(), inputData.size(), res, 19);
for (int i = 0; i<resSize; i++)
{
int asciiCode = res[i];
printf("[%i]t%it%xn", i, asciiCode, asciiCode);
}
printf("nnres: %s", (char*)res);
getchar();
return 0;
}
这对我来说很好。我使用了您链接的解压功能和您提供的base64函数。我删除了错误检查并重新格式化以使其短。
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <sstream>
#include <stdexcept>
#include <string>
#include <zlib.h>
#pragma comment(lib, "zdll.lib")
static char LookupDigits[] = {
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,62,0,0,0,63,52,53,54,55,56,57,58,59,60,61,
0,0,0,99,0,0,0,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,
20,21,22,23,24,25,0,0,0,0,0,0,26,27,28,29,30,31,32,33,34,35,36,
37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
};
int FromBase64Simple(const unsigned char* pSrc, int nLenSrc, unsigned char* pDst, int nLenDst)
{
int nLenOut = 0;
for (int j = 0; j<nLenSrc; j += 4)
{
if (nLenOut > nLenDst)
{
return(0); // error, buffer too small
}
unsigned char s1 = LookupDigits[*pSrc++];
unsigned char s2 = LookupDigits[*pSrc++];
unsigned char s3 = LookupDigits[*pSrc++];
unsigned char s4 = LookupDigits[*pSrc++];
unsigned char d1 = ((s1 & 0x3f) << 2) | ((s2 & 0x30) >> 4);
unsigned char d2 = ((s2 & 0x0f) << 4) | ((s3 & 0x3c) >> 2);
unsigned char d3 = ((s3 & 0x03) << 6) | ((s4 & 0x3f) >> 0);
*pDst++ = d1; nLenOut++;
if (s3 == 99) break; // end padding found
*pDst++ = d2; nLenOut++;
if (s4 == 99) break; // end padding found
*pDst++ = d3; nLenOut++;
}
return(nLenOut);
}
std::string decompress_string(const std::string& str)
{
z_stream zs; // z_stream is zlib's control structure
memset(&zs, 0, sizeof(zs));
inflateInit(&zs);
zs.next_in = (Bytef*)str.data();
zs.avail_in = str.size();
int ret;
char outbuffer[32768];
std::string outstring;
do
{
zs.next_out = reinterpret_cast<Bytef*>(outbuffer);
zs.avail_out = sizeof(outbuffer);
ret = inflate(&zs, 0);
if (outstring.size() < zs.total_out)
{
outstring.append(outbuffer, zs.total_out - outstring.size());
}
}
while (ret == Z_OK);
inflateEnd(&zs);
return outstring;
}
int main()
{
std::string inputData = "eAFjYoAAZiDFCMQgGgQAAJwACg==";
//19 is hardcoded since I know its size prior to this call
std::string res(19, ' ');
FromBase64Simple((unsigned char*)inputData.c_str(), inputData.size(), (unsigned char*)res.data(), res.size());
std::string d = decompress_string(res);
for (int c : d)
{
printf("%02x", c);
}
printf("n");
getchar();
return 0;
}
输出：020000000000000003000000010000000300000000000000
" eafjyoaazidfcmqgggqaajwacg ==" base64编码。您需要首先将其解码给二进制文件，以获取可以解压缩的东西。在十六进制中表达的二进制是：
78 01 63 62 80 00 66 20 c5 08 c4 20 1a 04 00 00 9c 00 0a
这是一个有效的zlib流，它解压缩了，以十六进制表示：
02 00 00 00 00 00 00 00 03 00 00 00 01 00 00 00 03 00 00 00 00 00 00 00
您的压缩结果" x？302@？p？"最初是在二进制中，无法打印。这些问号并不是原始的问号，而是其他不打印的字节。所以不要打印。您将印刷结果转换为六角的尝试是不正确的，因为您在十六进制中有问号(3F(。
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