写数组的二进制文件？ [英] Write array into the binary file?

问题描述

我需要一些帮助 - 下一块code写的长双动态数组到文件

I need some help - next piece of code writes a long double dynamic array into the file

int nx = 10, ny = 10;
long double **data = new long double *[nx]; 
long double **data_read = new long double *[nx]; 
for (int i = 0; i < nx; i++) {
    data[i] = new long double [ny];
    data_read[i] = new long double [ny];
}

data[4][4] = 10.0;
printf("%LF\n", data[4][4]);

FILE *file = fopen("data", "wb");
fwrite(data, nx * ny * sizeof(data), 1, file);
fclose(file);

file = fopen("data", "rb");
fread(data, nx * ny * sizeof(data_read), 1, file );
fclose(file);

printf("%LF\n", data_read[4][4]);

但数据[4] [4]！= DATA_READ [4] [4] ，因为从文件中读取后 DATA_READ [4] [ 4] = 0.0 。

But data[4][4] != data_read[4][4], because after reading from file data_read[4][4]=0.0.

任何人知道我在做什么错了？

Anybody knows what am I doing wrong?

推荐答案

您需要编写每一行的指针数组分别。一团写不会为假的二维数组（或Nd）的指针到指针的实施工作。

You need to write each row in your pointer array individually. A mass write will not work for pointer-to-pointer implementations of a fake 2D array (or nD).

有关写作：

for (int i=0; i<nx; ++i)
    fwrite(data[i], sizeof(data[i][0]), ny, file);

有关阅读：

for (int i=0; i<nx; ++i)
    fread(data[i], sizeof(data[i][0]), ny, file);

坦率地说，你是（联合国）幸运的进程并没有彻底崩溃，因为你写了一堆的内存地址到你的硬盘文件（十六进制转储会显示你的），并有可能走关在你的指针数组分配结束的两个的操作。

这是说，我开始学习C ++标准库的IO，而不是在C ++使用世界C- code（或修复在这个问题上的标签）。

That said, I'd start learning about the standard C++ IO library rather than using C-code in a C++ world (or fix the tag on this question).

单块写/读

您询问是否可以这样做，因为读/写单块。答案是肯定的，但你必须连续分配内存。如果你仍然想一个指针到指针数组，你当然可以使用一个。虽然我推荐使用的std ::矢量＆lt;长双＆GT; 数据缓冲区，下面就演示一下我指的是：

You asked if it is possible to do this as a single block read/write. The answer is yes, but you must allocate the memory contiguously. If you still want a pointer-to-pointer array you can certainly use one. Though I recommend using std::vector<long double> for the data buffer, the following will demonstrate what I refer to:

int main()
{
    int nx = 10, ny = 10;

    long double *buff1 = new long double[nx * ny];
    long double *buff2 = new long double[nx * ny];

    long double **data = new long double *[nx];
    long double **data_read = new long double *[nx];

    for (int i = 0; i < nx; i++)
    {
        data[i] = buff1 + (i*ny);
        data_read[i] = buff2 + (i*ny);
    }

    data[4][4] = 10.0;
    printf("%LF\n", data[4][4]);

    FILE *file = fopen("data.bin", "wb");
    fwrite(buff1, sizeof(*buff1), nx * ny, file);
    fclose(file);

    file = fopen("data.bin", "rb");
    fread(buff2, sizeof(*buff2), nx * ny, file );
    fclose(file);

    printf("%LF\n", data_read[4][4]);

    // delete pointer arrays
    delete [] data;
    delete [] data_read;

    // delete buffers
    delete [] buff1;
    delete [] buff2;
}

输出

10.000000
10.000000

---

使用的std ::矢量＆lt;＆GT; 为 RAII 解决方案

---

Using a std::vector<> for an RAII Solution

所有这些分配会导致混乱，并坦言容易出现问题。考虑一下，这是不同的：

All those allocations can get messy, and frankly prone to problems. Consider how this is different:

#include <iostream>
#include <fstream>
#include <vector>

int main()
{
    int nx = 10, ny = 10;

    // buffers for allocation
    std::vector<long double> buff1(nx*ny);
    std::vector<long double> buff2(nx*ny);

    // holds pointers into original
    std::vector<long double*> data(nx);
    std::vector<long double*> data_read(nx);

    for (int i = 0; i < nx; i++)
    {
        data[i] = buff1.data() + (i*ny);
        data_read[i] = buff2.data() + (i*ny);
    }

    data[4][4] = 10.0;
    std::cout << data[4][4] << std::endl;

    std::ofstream ofp("data.bin", std::ios::out | std::ios::binary);
    ofp.write(reinterpret_cast<const char*>(buff1.data()), buff1.size() * sizeof(buff1[0]));
    ofp.close();

    std::ifstream ifp("data.bin", std::ios::in | std::ios::binary);
    ifp.read(reinterpret_cast<char*>(buff2.data()), buff2.size() * sizeof(buff2[0]));
    ifp.close();

    std::cout << data_read[4][4] << std::endl;

    return 0;
}

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