星痕共鸣数据分析2

今天实验内容是攻击力部分

1.思路
由于昨天数据分析出了一个函数
这个函数可以把奇怪的字节变成正常的数字

int parse_varint(unsigned const char* data, int count)
{int value = 0;int shift = 0;for (int i = 0; i < count; i++) {unsigned char byte = data[i];value |= ((byte & 0x7F) << shift);shift += 7;}return value;
}

然后，我们把函数逆推，得到下面的函数
这个函数可以把正常的数字变成奇怪的字符

void encode_varint(unsigned char* output, int* count, int value)
{*count = 0;do {unsigned char byte = value & 0x7F;value >>= 7;if (value != 0) {byte |= 0x80;  // 设置最高位表示还有后续字节}output[(*count)++] = byte;} while (value != 0);
}

然后我们攻击一下场景内的怪物，造成伤害346，然后抓包
将346代入上面的函数，得到0xda, 0x02
结合抓包信息

我们可以看到伤害0xda, 0x02已经在包里面出现了（这个包的数据量比较多）
为了确保正确性，可以多次抓包验证

然后，分析字节数据发现大多数数据都是0x12+长度
可以用这个分析结构

#include <iostream>
#include <vector>
#include <iomanip>
#include <memory>
#include"XHGM.h"
// 定义段节点结构
struct Segment {int length;  // 数据段长度（不包括起始的0x12和长度字节）std::vector<unsigned char> data;  // 段数据（可能包含子段）std::vector<std::unique_ptr<Segment>> children;  // 子段列表
};// 递归解析数据为树形结构
std::unique_ptr<Segment> parseSegment(const unsigned char* data, int data_size, int& index, int depth = 0) {if (index >= data_size) return nullptr;// 创建新节点auto node = std::make_unique<Segment>();// 验证起始字节if (data[index] != 0x12) {std::cerr << "Error: Expected 0x12 at index " << index << ", found 0x"<< std::hex << std::setw(2) << std::setfill('0')<< static_cast<int>(data[index]) << std::endl;index++;  // 跳过无效字节return nullptr;}index++;  // 跳过0x12// 获取段长度if (index >= data_size) {std::cerr << "Error: Missing length byte at index " << index << std::endl;return nullptr;}node->length = static_cast<int>(data[index]);index++;  // 跳过长度字节// 提取段数据int end_index = index + node->length;if (end_index > data_size) {std::cerr << "Error: Incomplete segment at index " << index<< ", declared length: " << node->length << std::endl;return nullptr;}// 递归解析嵌套的子段while (index < end_index) {if (data[index] == 0x12) {// 递归解析子段auto child = parseSegment(data, data_size, index, depth + 1);if (child) {node->children.push_back(std::move(child));}}else {// 添加普通数据字节node->data.push_back(data[index]);index++;}}return node;
}// 打印树形结构
void printSegmentTree(const Segment* node, int depth = 0) {if (!node) return;// 缩进表示层级std::string indent(depth * 2, ' ');// 打印当前节点信息std::cout << indent.c_str() << "Segment (Length=" << node->length << "): ";// 打印原始数据if (!node->data.empty()) {std::cout << "Data: ";for (auto byte : node->data) {std::cout << "0x" << std::hex << std::setw(2) << std::setfill('0')<< static_cast<int>(byte) << " ";}/*int lens = node->data.size();char *t = new char[lens];int t1 = 0;for (auto byte : node->data) {t[t1++] = byte;}int tt = parse_varint(t, lens);std::cout << tt;delete[] t;*/}std::cout << std::endl;// 递归打印子节点for (const auto& child : node->children) {printSegmentTree(child.get(), depth + 1);}
}int main() {// 原始数据unsigned char peer0_5[] = {0x12, 0x0b, 0x08, 0x96, 0xa1, 0x3e, 0x10, 0xa9, 0x06, 0x18, 0xe4, 0xaf, 0x01,
0x12, 0x0a, 0x08, 0x97, 0xa1, 0x3e, 0x10, 0xe0, 0x04, 0x18, 0x98, 0x75,
0x12, 0x0a, 0x08, 0x98, 0xa1, 0x3e, 0x10, 0xc7, 0x06, 0x18, 0x98, 0x75,
0x12, 0x04, 0x08, 0xf7, 0xa1, 0x3e, 
0x12, 0x0b, 0x08, 0xf8, 0xa1, 0x3e, 0x10, 0xec, 0x03, 0x18, 0x80, 0xe1, 0x01, 
0x12, 0x04, 0x08, 0xf9, 0xa1, 0x3e, 
0x12, 0x0a, 0x08, 0xfa, 0xa1, 0x3e, 0x10, 0x01, 0x18, 0x8c, 0xf6, 0x01, 
0x12, 0x0a, 0x08, 0xfb, 0xa1, 0x3e, 0x10, 0x03, 0x18, 0xf8, 0xd2, 0x01,
0x12, 0x04, 0x08, 0xfc, 0xa1, 0x3e, 
0x12, 0x0b, 0x08, 0xfd, 0xa1, 0x3e, 0x10, 0x8e, 0x01, 0x18, 0x88, 0xef, 0x01, 
0x12, 0x0b, 0x08, 0xfe, 0xa1, 0x3e, 0x10, 0xeb, 0x04, 0x18, 0xfc, 0xd9, 0x01};int data_size = sizeof(peer0_5) / sizeof(peer0_5[0]);int index = 0;std::vector<std::unique_ptr<Segment>> rootSegments;// 解析所有顶级段while (index < data_size) {auto segment = parseSegment(peer0_5, data_size, index);if (segment) {rootSegments.push_back(std::move(segment));}else {// 跳过无效字节index++;}}// 打印解析结果std::cout << "Found " << rootSegments.size() << " root segments:\n";for (size_t i = 0; i < rootSegments.size(); ++i) {std::cout << "\nRoot Segment " << i + 1 << ":\n";printSegmentTree(rootSegments[i].get());}return 0;
}