我如何用c++编写一个程序来解释事先不知道它正在读取的数据文本文件中包含了多少行？

Question

我已经编写了一个程序来计算给出的数据的x平方值。但是，我需要编写它，这样我的程序就可以解释不知道它正在读取的文本文件中有多少行。如果我将我的结构点定义为点thePoints1000 (即大于文件中的行数)，那么计算就不起作用，最后得到'nan‘值。这是我的节目：

#include<iostream>
#include<fstream>
#include<cmath>
using namespace std;


//define my own data structure
struct point {
    double x;
    double y;
    double s;
    double w;
    double wx;
    double wy;
    double wxy;
    double wxx;
    double wmxcy;
};


//main program
int main() {

    //array of point
    point thePoints[11];
    int i = 0;

    //open a file to read in
    ifstream myFile ("xys.data.txt");

    //check if it opened successfully
    if (myFile.is_open() ) {

        //file is open
        //read it in
        while (!myFile.eof() ) {

            //whilst we are not at the end of the file
            myFile >> thePoints[i].x >> thePoints[i].y >> thePoints[i].s;
            //increment i
            i++;
        }

        //close the file
        myFile.close();
    }

    // something went wrong
    else {
        cout << "Error opening file!\n";
        exit(1);
    }

    // data is now in an array of point structures

    //set the summation variables to zero - sets an initial value for the rest of the appropriate array to then be added onto
    double Sw = 0;
    double Swxx = 0;
    double Swxy = 0;
    double Swx = 0;
    double Swy = 0;
    double xsq = 0;


    //create an array for w
    for (int j = 0; j <= 10; j++){
        thePoints[j].w = 1/(pow((thePoints[j].s),2));

    //sum over all w i.e. create an array for Sw
        Sw += thePoints[j].w;

    //create an array for wx
        thePoints[j].wx = (thePoints[j].w) * (thePoints[j].x);

    //sum over all wx i.e. create an array for Swx
        Swx += thePoints[j].wx;

    //create an array for wy
        thePoints[j].wy = (thePoints[j].w) * (thePoints[j].y);

    //sum over all wy i.e. create an array for Swy
        Swy += thePoints[j].wy;

    //create an array for wxy
        thePoints[j].wxy = (thePoints[j].w) * (thePoints[j].x) * (thePoints[j].y);

    //sum over all wxy i.e. create an array for Swxy
        Swxy += thePoints[j].wxy;

    //create an array for wxx
        thePoints[j].wxx = (thePoints[j].w) * (thePoints[j].x) * (thePoints[j].x);

    //sum over all wxx i.e. create an array for Swxx
        Swxx += thePoints[j].wxx;
    }

    printf("%6.2f, %6.2f, %6.2f, %6.2f, %6.2f\n", Sw, Swx, Swy, Swxy, Swxx);

    //caluculate a value for D
    double D = ((Sw * Swxx) - (Swx * Swx));

    //calculate a value for m
    double m = ((Sw * Swxy) - (Swx * Swy))/D;

    //calculate a value for dm
    double dm = sqrt(Sw/D);

    //calculate a value for c
    double c = ((Swy * Swxx) - (Swx * Swxy))/D;

    //calculate a value for dc
    double dc = sqrt(Swxx/D);

    //calculate chi-squared value, xsq = Sw(((m * x) + c - y)^2)
    for (int j = 0; j < i; j++){
        thePoints[j].wmxcy = (thePoints[j].w * (pow(((m * thePoints[j].x) + c - thePoints[j].y),2)));

    //sum over all chi-squared
    xsq += thePoints[j].wmxcy;
    }

    //prints all of the results of the data
        printf("The equation of the line for the data is y = %6.2f x + %6.2f.\n", m, c);
        printf("The gradient, m, has an associated error of %6.2f.\n", dm);
        printf("The y intercept, c, has an associated error of %6.2f.\n", dc);
        printf("The data has a chi-squared value of %6.2f.\n", xsq);

    return 0;
}

我已经附加了输入文件.txt文件

如有任何意见，将不胜感激。

Answer 1

如果您不知道编译时的点数，那么应该使用动态分配的内存。您可以通过两次读取文件来计数点数，然后使用new一次性分配内存并在第二次传递时填充这些点，或者使用数据结构存储在读取文件时可以根据需要增长的点。我建议您看一下std：：载体，从STL作为起点。下面是一个使用std::vector的简单示例。

//vector of point
std::vector<point> thePoints;

//open a file to read in
ifstream myFile ("xys.data.txt");

//check if it opened successfully
if (myFile.is_open() ) {

    //file is open
    //read it in
    while (!myFile.eof() ) {
        point aPoint;

        //whilst we are not at the end of the file
        myFile >> aPoint.x >> aPoint.y >> aPoint.s;

        //add a point to the vector of points
        thePoints.push_back(aPoint);
    }

    //close the file
    myFile.close();
}

您可以通过thePoints.size()获得点数。确保更新所有for循环以删除硬编码的10。