Чи в квадрате — Как мне написать программу на С ++, чтобы не знать заранее, сколько строк включено в текстовый файл данных, из которого он читает?

Я написал программу для вычисления значения хи-квадрат, используя данные. Тем не менее, мне нужно написать это так, чтобы моя программа могла не знать, сколько строк в текстовом файле, из которого она читает. Если я определяю свои точки структуры как точки thePoints [1000], (то есть больше, чем количество строк в файле), тогда вычисления не работают, и я получаю значения ‘nan’.
Вот моя программа:

#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;
}

Я приложил входной файл.текстовый файл

Любые мнения будут высоко оценены.