使用 zlib 解压缩文本字符串

Question

我需要解码以 zlib 格式存储的十六进制字符串。一个例子是：

1800000013000000eAFjYoAAZiDFCMQgGgQAAJwACg==

其中 18000000 和 13000000 是 uncompressed/compressed 数据的大小（在本例中为 24 和 19）。

我也知道字符串的其余部分包含

020000000000000003000000010000000300000000000000

问题出在哪里？按照 https://panthema.net/2007/0328-ZLibString.html 压缩该字符串的任何教程，我得到

x?302@?P??

十六进制可以写成

783f3330324053f503f103ff5

这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法（这是我的最终目标）

提前感谢您的提示！

PS。我正在使用来自 https://github.com/systemed/intersector/blob/master/helpers.cpp

看起来字符串已经用 base64 编码了（谢谢@zdenek 和@Mark-Adler）我设法用

解码了它

BYTE *res;
int resSize = FromBase64Simple((BYTE*)actualData.c_str(),actualData.len(),res,sizeCompressed);

您可以阅读 https://github.com/kengonakajima/luvit-base64/blob/master/base64.c

中的实现

但这不是问题，因为我可以使用

转储结果

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;
}

我得到了十进制和十六进制的每个字节的结果，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 00 9c 00 0a

但是 resChar 是 "x?cb?" 与@Mark-Adler "x?302@?P??" 所说的值无关（显然 '?' 符号不是 printable 符号），我真的认为这是问题所在，但我的数据似乎与此相对应 table: https://www.asciitable.com/ and Mark's one doesn't also this web https://conv.darkbyte.ru/ returns 与我的算法

相同的结果

我尝试用上面说的实现解压字符串但失败了（也试过https://gist.github.com/arq5x/5315739）但它的解压值是单个字符串“”

这里我们使用最小的可重现案例：

#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%i\t%x\n", i, asciiCode, asciiCode);
    }
    printf("\n\nres: %s", (char*)res);

    getchar();

    return 0;
}

Answer 1

"eAFjYoAAZiDFCMQgGgQAAJwACg==" 是 Base64 编码的。您需要先将其解码为二进制文件以获得可以解压缩的内容。以十六进制表示的二进制是：

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，解压缩到这个，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??"原来是二进制的，无法打印。那些问号在原文中并不是真正的问号，而是一些其他不打印的字节。所以不要打印它。您将打印结果转换为十六进制的结果尝试不正确，因为十六进制中有问号 (3f)。

Answer 2

这对我来说很好用。我使用了您链接的解压缩功能和您提供的 base64 功能。我删除了错误检查并重新格式化了一些内容以使其更短。

#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, '[=10=]');
    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

使用 zlib 解压缩文本字符串

Decompress text string with zlib

c++

string

zlib

compression