Question

我有一个大型结构的数组，我试图输出到硬盘驱动器。我似乎能够很好地写入硬盘驱动器（尽管通过查看二进制数据很难验证），但是当我尝试将其读回时，我总是会出现乱码。我有什么想法吗？

这是结构配置：

class xyz
{
public:
    double x, y, z;
};
class trianglePackage
{
public:

    int score;
    int position;

    xyz contactCoordinates;
    xyz normalVector;
    xyz locatorOffset;

};
class quadanglesOutput
{
public:

    int locator1position, locator2position, locator3position, locator4position;

    xyz centroid;

    int surfaceAreaScore;
    int centroidDifferance1Score;
    int centroidDifferance2Score;
    int minDistance1Score;
    int minDistance2Score;

    int totalLocatorScore;
    int totalHullScore;
    int totalScore;

    double surfaceArea;
    double centroidDifferance1;
    double centroidDifferance2;
    double minDistance1;
    double minDistance2;

    int hull;

    trianglePackage locator1, locator2, locator3, locator4;
};

以下是我正在使用的读/写函数：

void outputQuadangleOutput(quadanglesOutput* output, string description, param parameters)
{

    string outputName = parameters.fileName + " " + description + ".bin";
    cout << "Output " << outputName.c_str() << "...";
    ofstream output2;
    output2.open(outputName.c_str());
    output2.write(reinterpret_cast<char*>(output), streamsize(parameters.topXlist * sizeof(quadanglesOutput)));
    output2.close();
    cout << "done" << endl;

}
void readIn(quadanglesOutput* pointer, param parameters, string description)
{
    string fileName = parameters.fileName + " " + description + ".bin";
    cout << "openining " << fileName << "...";
    ifstream readFile;
    readFile.open(fileName.c_str());
    readFile.read(reinterpret_cast<char*>(pointer), (parameters.topXlist * sizeof(quadanglesOutput)));
    readFile.close();
    cout << "done" << endl;
}

通常，结构数组的长度大约为100，但通常只有大约前25个读回正确，其他所有内容都是默认的未初始化数据。

我99％确定我的代码有问题，但有可能它与四字节对齐有关吗？

感谢。

Answer 1

字节对齐可能是一个问题，使用pragma。尝试使用

包装类

#PRAGMA PACK PUSH(1)
....
#PRAGMA PACK POP

或

#PRAGMA PACK(1)
struct{
..
}

尝试这些：
强制流的二进制标志。

ios_base :: binary

readFile.open(fileName.c_str(), ios_base::binary);

尝试冲洗流。

stream.write(...)
stream.flush()

//我知道close（）应该冲洗它。

<强>更新
一切都适合我：

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

using namespace std;
#pragma pack(1)
class xyz
{
public:
    double x, y, z;
};

#pragma pack(1)
class trianglePackage
{
public:

    int score;
    int position;

    xyz contactCoordinates;
    xyz normalVector;
    xyz locatorOffset;

};

#pragma pack(1)
class quadanglesOutput
{
public:

    int locator1position, locator2position, locator3position, locator4position;

    xyz centroid;

    int surfaceAreaScore;
    int centroidDifferance1Score;
    int centroidDifferance2Score;
    int minDistance1Score;
    int minDistance2Score;

    int totalLocatorScore;
    int totalHullScore;
    int totalScore;

    double surfaceArea;
    double centroidDifferance1;
    double centroidDifferance2;
    double minDistance1;
    double minDistance2;

    int hull;

    trianglePackage locator1, locator2, locator3, locator4;
};

class param
{
public:
    string fileName;
    int topXlist;
};


void outputQuadangleOutput(quadanglesOutput* output, string description, param parameters)
{

    string outputName = parameters.fileName + " " + description + ".bin";
    cout << "Output " << outputName.c_str() << "...";
    ofstream output2;
    output2.open(outputName.c_str());
    output2.write(reinterpret_cast<char*>(output), streamsize(parameters.topXlist * sizeof(quadanglesOutput)));
    output2.close();
    cout << "done" << endl;

}
void readIn(quadanglesOutput* pointer, param parameters, string description)
{
    string fileName = parameters.fileName + " " + description + ".bin";
    cout << "openining " << fileName << "...";
    ifstream readFile;
    readFile.open(fileName.c_str());
    readFile.read(reinterpret_cast<char*>(pointer), (parameters.topXlist * sizeof(quadanglesOutput)));
    readFile.close();
    cout << "done" << endl;
}



int main(int argc, char *argv[])
{

    quadanglesOutput a = {0};
    cout<<"total score:"<<a.totalScore<<endl;
    cout<<"locator position:"<<a.totalScore<<endl;
    cout<<"locator position:"<<a.locator1.position<<endl;
    cout<<"locator position:"<<a.locator2.normalVector.y <<endl;
    cout<<"sizeof quadangsomething:"<<sizeof(quadanglesOutput)<<endl;
    a.totalScore=1;
    a.locator1.position=333445;
    a.locator2.normalVector.y = 999.3224;
    cout<<"total score:"<<a.totalScore<<endl;
    cout<<"locator position:"<<a.locator1.position<<endl;
    cout<<"locator position:"<<a.locator2.normalVector.y <<endl;
    param p = {"C:/", 1};
    outputQuadangleOutput(&a, "file1", p);

    quadanglesOutput *b = new quadanglesOutput();
    readIn(b, p, "file1");
    cout<<"new total score:"<<b->totalScore<<endl;
    cout<<"new locator position:"<<b->locator1.position<<endl;
    cout<<"new locator position:"<<b->locator2.normalVector.y <<endl;

    delete b;





    string asdf;
    cin>>asdf;

};

<强>输出：
总分：0
定位器位置：0
locator2.normalVector.y：0
sizeof quadangsomething：436
总分：1
定位器位置：333445
locator2.normalVector.y：999.322
输出C：/ file1.bin ...完成
openining C：/ file1.bin ...完成
新总分：1
新定位器位置：333445
new locator2.normalVector.y：999.322

没有编译指示它仍然是正确的，但你可以看到大小的差异：

sizeof quadangsomething：440

但是通过网络发送结构时包装它是很好的因为这里系统总是以相同的方式分配它。

将大型结构输出到HD

1 个答案: