c++ - 无法将文件流作为函数参数传递？

Question

这是我的作业代码。每当我尝试编译时，由于“ios_base.h”中的某些内容，我的读取函数都会出错，我不确定该怎么做和/或我的代码是否执行了获取文件并将其元素移动到单独的文件的预期功能名称和平均值相邻的文件。

#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>

using namespace std;

struct Student
{
    string fname;
    string lname;
    double average;
};

int read(ifstream, Student s[]);

void print(ofstream fout, Student s[], int amount);


int main()
{
    const int size = 10;
    ifstream fin;
    ofstream fout;
    string inputFile;
    string outputFile;
    Student s[size];

    cout << "Enter input filename: ";
    cin >> inputFile;
    cout << "Enter output filename: ";
    cin >> outputFile;
    cout << endl;

    fin.open(inputFile.c_str());
    fout.open(outputFile.c_str());

    read(fin , s);
    print(fout, s, read(fin, s));

}

int read(ifstream fin, Student s[])
{
    string line;
    string firstName;
    string lastName;
    double score;
    double total;
    int i=0;
    int totalStudents=0;
    Student stu;

    while(getline(fin, line)){
        istringstream sin;
        sin.str(line);

        while(sin >> firstName >> lastName){
            stu.fname = firstName;
            stu.lname = lastName;

            while(sin >> score){
            total *= score;
            i++;
            }
            stu.average = (total/i);
        }
        s[totalStudents]=stu;
        totalStudents++;
    }
    return totalStudents;
}

void print(ofstream fout, Student s[], int amount)
{
    ostringstream sout;
    for(int i = 0; i<amount; i++)
    {
        sout << left << setw(20) << s[i].lname << ", " << s[i].fname;
        fout << sout << setprecision(2) << fixed << "= " << s[i].average;
    }
}

Answer 1

流对象不可复制。他们的复制构造函数被删除。它们必须通过引用传递，而不是通过值传递:

int read(ifstream &, Student s[]);

void print(ofstream &fout, Student s[], int amount);

等...