如何在二进制文件中读取/写入结构?

2021-12-21 00:00:00 binaryfiles struct vector c++

我遇到了一个小问题.我有一个结构体，它有一个向量.请注意，向量在每次迭代中都是动态的.现在，在特定的迭代中，如何将包含大小为 n 的向量的结构存储到二进制文件中?

I am facing a small problem. I have a struct, which has a vector. Note that the vector is dynamic per every iteration. Now, in a particular iteration, how do I store the struct which contains a vector of size n to a binary file?

另外，在检索时，假设我知道向量的大小，如何从二进制文件中检索包含所有存储元素的向量的结构变量?

Also, when retrieving, assume that I know how the size of the vector, how to I retrieve from the binary file, the struct variable containing the vector of all the stored elements?

我能够将一些内容存储到二进制文件中(因为我可以看到写入时大小增加)，但是当我尝试检索元素时，我将向量的大小设置为零.

I am able to store something to the binary file (as I can see the size increasing when writing), but when I am trying to retrieve back the elements, I am getting size of vector to be zero.

不幸的是，我必须使用标准 STL 来实现这一点，而不是使用任何第三方库.

推荐答案

你应该看看 Boost 序列化.

如果你不能使用第三方库，你必须知道C++不直接支持序列化.这意味着你必须自己做.

If you can't use 3rd party libraries, you must know that C++ doesn't support serialization directly. This means you'll have to do it yourself.

这篇文章 展示了一种将自定义对象序列化到磁盘并取回它的好方法.而本教程向您展示了如何立即开始使用fstream.

This article shows a nice way of serializing a custom object to the disk and retrieving it back. And this tutorial shows you how to get started right now with fstream.

这是我的尝试:

EDIT:由于 OP 询问如何存储/检索比记录更多的内容，因此我决定更新原始代码.

EDIT: since the OP asked how to store/retrieve more than record I decided to updated the original code.

那么，发生了什么变化?现在有一个 array student_t apprentice[3]; 来存储 3 个学生的信息.整个阵列被序列化到磁盘，然后全部加载回 RAM，在那里可以读取/搜索特定记录.请注意，这是一个非常非常小的文件(84 字节).在大文件中搜索记录时，我不建议使用这种方法.

So, what changed? Now there's an array student_t apprentice[3]; to store information of 3 students. The entire array is serialized to the disk and then it's all loaded back to the RAM where reading/searching for specific records is possible. Note that this is a very very small file (84 bytes). I do not suggest this approach when searching records on huge files.

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

using namespace std;


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

int main()
{
    student_t apprentice[3];  
    strcpy(apprentice[0].name, "john");
    apprentice[0].age = 21;
    apprentice[0].grades.push_back(1);
    apprentice[0].grades.push_back(3);
    apprentice[0].grades.push_back(5);    

    strcpy(apprentice[1].name, "jerry");
    apprentice[1].age = 22;
    apprentice[1].grades.push_back(2);
    apprentice[1].grades.push_back(4);
    apprentice[1].grades.push_back(6);

    strcpy(apprentice[2].name, "jimmy");
    apprentice[2].age = 23;
    apprentice[2].grades.push_back(8);
    apprentice[2].grades.push_back(9);
    apprentice[2].grades.push_back(10);

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

    // Reading from it
    ifstream input_file("students.data", 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 << "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;
    }

    return 0;
}

输出:

Record #0
Name: john
Age: 21
Grades: 
1 3 5 

Record #1
Name: jerry
Age: 22
Grades: 
2 4 6 

Record #2
Name: jimmy
Age: 23
Grades: 
8 9 10

二进制文件的转储:

$ hexdump -c students.data 
0000000   j   o   h   n   237   {     ?   ?   {   ? 025      
0000010   (   ?   ?     4   ?   ?     8   ?   ?     j   e   r   r
0000020   y     ?     ?   ?   |   026         @   ?   ?  
0000030   L   ?   ?     P   ?   ?     j   i   m   m   y      
0000040   ?   6   ?   027         X   ?   ?     d   ?   ?  
0000050   h   ?   ?                                                  
0000054

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