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当我使用 ofstream 将双数组输出到 txt 文件时出现错误。这是代码:

static void OutpuResults(std::string fileName, double * yCal, int nPtCal)
{
    std::string value;
    double * yCal_local = yCal; 

    std::ofstream outfile;
    outfile.open(fileName.c_str());

    for (int i = 0; i < nPtCal; i++)
    {
        outfile<<yCal_local[i]<<std::endl;
    }

    delete[] yCal_local;
    outfile.close();    
}

错误发生在outfile<<yCal_local[i]<<std::endl;i = 0 的地方,并且yCal_local[i]是 0.000000 作为双精度数。对我毫无意义。

这是文件名的定义:

std::string fileName = "d:\\inter.txt";

这里是 yCal 的定义:

int nPtCal = 256;
double * yCal = new double[nPtCal];

我想知道我哪里做错了?任何建议都值得赞赏。

编辑:编译时没有问题,但运行时出现错误。这是错误消息:

Unhandled exception at 0x75f5812f in ppppp.exe: Microsoft C++ exception: std::bad_alloc at memory location 0x0028f754..

这是程序从错误返回时显示的代码:

void __cdecl _unlock (
        int locknum
        )
{
        /*
         * leave the critical section.
         */
        LeaveCriticalSection( _locktable[locknum].lock );
}

编辑:

这是所有代码:

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


static void show_usage(std::string name)
{
    std::cerr << "To use ZoomCal:\n " << "ZoomCal.exe inputfile -options\n"
              << "options:\n"
              << " -l: linear\n"
              << " -c: cubic"<< std::endl;
}

static void LoadData(std::string fileName, double* y)
{
    std::string value;
    double * y_local = y;
    char *cstr = new char[fileName.length() + 1];
    std::strcpy(cstr, fileName.c_str());

    std::ifstream infile;

    infile.open(cstr);

    if (!infile.is_open())
    {
        std::cerr<<"File loading failed. Double check if file exists in the same folder as ZoomCal.exe"<<std::endl;
        exit(1);
    }

    int count = 0;
    int i = 0;
    int col = 10;
    while ( infile.good() )
    {
        getline ( infile, value, ',' ); // read a string until next comma: http://www.cplusplus.com/reference/string/getline/

        if (count % col == 0)
        {
            if (count != 0)
            {
                *y_local = std::atof(value.c_str());
                y_local++;
            }
        }
        count++;
    }

    //delete [] cstr;
    //delete y_local;
    infile.close();    
}

static void OutpuResults(std::string fileName, double * yCal, int nPtCal)
{
    std::string value;
    //double * yCal_local = yCal;   

    std::ofstream outfile;
    outfile.open(fileName.c_str());

    for (int i = 0; i < nPtCal; i++)
    {
        outfile<<yCal[i]<<std::endl;
    }

//  delete[] yCal_local;
    outfile.close();    
}

double * LinInterp(double * y, int nPt, int nPtCal)
{
    double * yCal = new double[nPtCal];
    double * tmp = new double[nPtCal-5];
    int interval = 10;
    double * pPiece = new double[interval];
    double f2, f1, m, b;
    int x2, x1;

    std::memset(yCal, 0, sizeof(double)*nPtCal);
    tmp += 6;

    for (int i = 0; i < (nPt - 1); i ++)
    {
        *pPiece = *y; 

        f2 = *(y + 1);
        f1 = *y;
        x2 = (i + 1) * interval;
        x1 = i * interval;

        m = (f2 - f1)/(x2 - x1);
        b = f2 - m * x2;

        for (int k = 1; k < interval; k++)
        {
            pPiece[k] = m * ((i * interval) + k) + b;
        }

        std::memcpy(tmp + (i * interval),  pPiece, sizeof(double)*interval);

        y++;
    }

    std::memcpy(yCal + 6,  tmp, sizeof(double)*250);


    return yCal;

}

double * CubInterp(double * y, int nPt, int nPtCal)
{

    double * yCal = new double[nPtCal];
    std::memset(yCal, 0, sizeof(double)*nPtCal);

    return yCal;
}

int main(int argc, char* argv[])
{
    if (argc < 3) 
    {
         show_usage(argv[0]);   
         return 1;
    }

    std::vector <std::string> sources;
    std::string destination;

    for (int i = 1; i < argc; ++i) 
    {
        std::string arg = argv[i];
        if ((arg == "-h") || (arg == "--help")) 
        {
            show_usage(argv[0]);
            return 0;
        } 
        else if ((arg == "-d") || (arg == "--destination")) 
        {
            if (i + 1 < argc) 
            { // Make sure we aren't at the end of argv!
                destination = argv[i++]; // Increment 'i' so we don't get the argument as the next argv[i].
            } 
            else 
            { // Uh-oh, there was no argument to the destination option.
                std::cerr << "--destination option requires one argument." << std::endl;
                return 1;
            }               
        }   
        else 
        {
             sources.push_back(argv[i]);
        }                            
    }

    int nPt = 26;
    double * y = new double[nPt];

    LoadData(sources[0], y);

    int nPtCal = 256;
    double * yCal = new double[nPtCal];

    yCal = LinInterp(y, nPt, nPtCal);

    std::string fileName = "D:\\inter.txt";
    OutpuResults(fileName, yCal, nPtCal);

    getchar();
    return 1;     
}
4

1 回答 1

0

Ok finally I got where I did wrong:

tmp += 6;

My program is to interpolate a array of 26 elements to 256 elements. But the first 6 elements of the interpolated array have to be 0; So I created a new array with length being 256-6 = 250;

double nPtCal = 256;
double * tmp = new double[nPtCal-5];

But somehow, I also moved this tmp pointer by 6. I guess it would be right if I created an array with length being 256; So no surprisingly the error occurred. Because when the operation fills 256-6 = 250 elements, if the pointer already points to the 6th elements out of the total 250, then it will eventually be out of range because I will basically move the pointer 250 times to process all the 250 elements in the new array. That is why I kept getting error. I thought it was because some file IO, but now I think not. Thanks for all your help. Very appreciate it.

于 2013-03-14T14:28:43.973 回答