需要帮助理解如何用二进制表示向量[C++]

Question

我正在尝试学习如何破解文件格式,所以我从一个简单的例子开始,我从那里开始:如何读取/写入二进制文件中的结构？

#include <fstream>
#include <iostream>
#include <vector>
#include <string.h>

using namespace std;


typedef struct student
{
    char name[10];
    double age;
    vector<int> grades;
}student_t;

void readBinaryFile(string filename)
{
    ifstream input_file(filename, ios::binary);
    student_t master[3];
    input_file.read((char*)&master, sizeof(master));

    for (size_t idx = 0; idx < 3; idx++)
    {
        // If you wanted to search for specific records, 
        // you should do it here! if (idx == 2) ...

        cout << "Record #" << idx << endl;
        //cout << "Capacity: " << master[idx].grades.capacity() << endl;
        cout << "Name: " << master[idx].name << endl;
        cout << "Age: " << master[idx].age << endl;
        cout << "Grades: " << endl;

        for (size_t i = 0; i < master[idx].grades.size(); i++)
           cout << master[idx].grades[i] << " ";
        cout << endl << endl;
    }
    input_file.close();
}

int main()
{
    student_t apprentice[3];  
    strcpy(apprentice[0].name, "john");
    apprentice[0].age = 21;
    apprentice[1].grades.push_back(1);
    apprentice[1].grades.push_back(2);
    apprentice[1].grades.push_back(3);

    strcpy(apprentice[1].name, "jerry");
    apprentice[1].age = 22;
    apprentice[0].grades.push_back(4);
    apprentice[0].grades.push_back(5);
    apprentice[0].grades.push_back(6);

    strcpy(apprentice[2].name, "jimmy");
    apprentice[2].age = 24;
    apprentice[2].grades.push_back(7);
    apprentice[2].grades.push_back(8);
    apprentice[2].grades.push_back(9);

    string filename = "students2.data";
    // Serializing struct to student.data
    ofstream output_file(filename, ios::binary);
    output_file.write((char*)&apprentice, sizeof(apprentice));
    output_file.close();

    // Reading from it
    readBinaryFile(filename);

    system("pause");
    return 0;
}

我可以编写文件并正确读取它,当我在十六进制编辑器中打开它时,我得到了这个:

6a 6f 68 6e 00 cc cc cc cc cc cc cc cc cc cc cc
00 00 00 00 00 00 35 40 40 59 84 00 80 cf 84 00
8c cf 84 00 8c cf 84 00 6a 65 72 72 79 00 cc cc
cc cc cc cc cc cc cc cc 00 00 00 00 00 00 36 40
50 85 84 00 60 d0 84 00 6c d0 84 00 6c d0 84 00
6a 69 6d 6d 79 00 cc cc cc cc cc cc cc cc cc cc
00 00 00 00 00 00 38 40 50 79 84 00 b8 cf 84 00
c4 cf 84 00 c4 cf 84 00

我可以清楚地找到名称(6a 6f 68 6e 00 cc cc cc cc cc)和年龄(00 00 00 00 00 00 35 40),但我在找到等级值时遇到了很多麻烦.我想通过制作具有不同值的第二个文件,我可以找到差异,但我找到了一些我不理解的东西.通过将第一个学生改为:

strcpy(apprentice[0].name, "john");
apprentice[0].age = 21;
apprentice[1].grades.push_back(1);
apprentice[1].grades.push_back(2);
apprentice[1].grades.push_back(3);
apprentice[1].grades.push_back(4);
apprentice[1].grades.push_back(5);
apprentice[1].grades.push_back(6);

我希望得到一个更大的文件,但它的大小不会改变:

6a 6f 68 6e 00 cc cc cc cc cc cc cc cc cc cc cc
00 00 00 00 00 00 35 40 40 59 8e 00 48 cf 8e 00
54 cf 8e 00 54 cf 8e 00 6a 65 72 72 79 00 cc cc
cc cc cc cc cc cc cc cc 00 00 00 00 00 00 36 40
50 85 8e 00 88 79 8e 00 ac 79 8e 00 ac 79 8e 00
6a 69 6d 6d 79 00 cc cc cc cc cc cc cc cc cc cc
00 00 00 00 00 00 38 40 50 79 8e 00 98 d0 8e 00
a4 d0 8e 00 a4 d0 8e 00

怎么可能？我甚至尝试使用60+元素的矢量,文件仍然保持相同的大小...任何帮助将不胜感激!

编辑:正如tux3指出的那样,我实际上并没有将我的向量保存为二进制.我应该更加关注我复制的代码,不好.

Answer 1

output_file.write((char*)&apprentice,sizeof(apprentice));

这不符合你的想法.std::vector将其数据放在免费商店中,而不是像数组一样放在对象本身中.

所以在这里你只是编写矢量的元数据(它的大小,容量,指向数据的指针......),而不是数据本身.