如何在C 中编写一个程序，以说明不知道它正在阅读的数据文本文件中包含多少行

Question

我已经编写了一个程序来使用给定的数据计算卡方值。但是，我需要编写它，以便我的程序可以解释不知道它正在阅读的文本文件中有多少行。如果我将我的结构点定义为点[1000]（即大于文件中的行数），则计算不起作用，我最终会得到" NAN"值。这是我的程序：

---

#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, %6.2fn", 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;
}

我已连接输入文件.txt文件

任何意见都将不胜感激。

Answer 1

如果您不知道编译时间的点数，则应使用动态分配的内存。您可以在两次通行证中读取文件以计数点数，然后用new一次分配内存，并在第二次通过时填充点，或者您可以使用数据结构来存储可以根据需要生长的点当您阅读文件时。我建议您从STL看一个STD ::向量作为起点。这是一个使用std :: vector的简短示例。

//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();
}

您可以使用thePoints.size()获得点数。确保您更新所有循环以删除硬编码10。