实验6模板类和文件IO

task3-1.cpp
#include <iostream>
#include <fstream>
#include <array>
#define N 5
int main() {
    using namespace std;
    array<int, N> x{ 97, 98, 99, 100, 101 };
    ofstream out;
    out.open("data1.txt", ios::binary);
    if (!out.is_open()) {
        cout << "fail to open data1.dat\n";
        return 1;
    }
    // 把从地址&x开始连续sizeof(x)个字节的数据块以字节数据块方式写入文件data1.txt
    out.write(reinterpret_cast<char*>(&x), sizeof(x));
    out.close();
}

task3-2.cpp
#include <iostream>
#include <fstream>
#include <array>
#define N 5
int main() {
    using namespace std;
    array<char, N> x;
    ifstream in;
    in.open("data1.txt", ios::binary);
    if (!in.is_open()) {
        cout << "fail to open data1.dat\n";
        return 1;
    }
    // 从文件流对象in关联的文件data1.txt中读取sizeof(x)字节数据写入&x开始的地址单元
    in.read(reinterpret_cast<char*>(&x), sizeof(x));
    in.close();
    for (auto i = 0; i < N; ++i)
        cout << x[i] << ", ";
    cout << "\b\b \n";
}

产生该结果的原因是因为写文件的时候写入的是int类型，int类型占四个字节，读文件的时候采用的是一次读一个char，即一个字节，所以造成该结果

Vector.hpp
#pragma once
#include<iostream>
using namespace std;
template<typename T>
class Vector {
    template<typename T>
    friend void output(Vector<T> &v);
public:
     Vector(int num);
     Vector(int num, T value);
     Vector(const Vector<T>& v);
     ~Vector();
     int get_size();
     T& at(int i);
     T& operator[](int i);
private:
     T* array;
     int size;
};
template<typename T>
void output(Vector<T> &v) {
    for (auto i = 0; i < v.get_size(); i++) {
        cout << v.array[i]<<",";
    }
    cout << endl;
}
template<typename T>
Vector<T>::Vector(int num) {
    this->size = num;
    array = new T[size];
}
template<typename T>
Vector<T>::Vector(int num, T value) {
    this->size = num;
    array = new T[size];
    for (int i = 0; i < size; i++) {
        array[i] = value;
    }
}
template<typename T>
Vector<T>::Vector(const Vector<T> &v) {
    this->size = v.size;
    this->array = new T[size];
    for (int i = 0; i < size; i++) {
        this->array[i] = v.array[i];
    }
}
template<typename T>
Vector<T>::~Vector() {
    if (this->array != NULL) {
        delete[] array;
    }
    array = NULL;
}
template<typename T>
int Vector<T>::get_size() {
    return size;
}
template<typename T>
T& Vector<T>::at(int i) {
    return array[i];
}
template<typename T>
T& Vector<T>::operator[](int i) {
    return array[i];
 }

#include <iostream>
#include "Vector.hpp"
void test() {
    using namespace std;
    int n;
    cin >> n;
    Vector<double> x1(n);
    for (auto i = 0; i < n; ++i)
        x1.at(i) = i * 1.1;
    output(x1);
    Vector<int> x2(n, 66);
    Vector<int> x3(x2);
    output(x2);
    output(x3);
    x2.at(0) = 66;
    output(x2);
    x3[0] = 888;
    output(x3);
}
int main() {
    test();
}

task5.cpp
#include<iostream>
#include<fstream>
#include<iomanip>
#include<iomanip>
using namespace std;
void output(ostream& out) {
    ifstream ifs;
    ifs.open("cipher_key.txt");
    int num[27][27];
    for (int i = 0; i < 27; i++) {
        for (int j = 0; j < 27; j++) {
            if (j == 0 && i == 0) {
                out<<"  ";
                continue;
            }
            if (j == 0 && i != 0) {
                ifs >> num[i][j];
                out << setw(2) << num[i][j];
                continue;
            }
            ifs >> num[i][j];
            out << setw(2) << (char)num[i][j];
        }
        out << endl;
    }
    ifs.close();
}
int main() {
     ofstream ofs;
     ofs.open("cipher_key.txt", ios::out);
     int a[27][27];
     int cnt = 1;
     int num = 97;
     int num2 = 66;
     for (int j = 1; j < 27; j++) {
         a[0][j] = num++;
         a[j][0] = cnt++;
    }
     for (int i = 1; i < 27; i++) {
         for (int j = 1; j < 27; j++) {
             if (a[i][j - 1]== 90)a[i][j] = 65;
             else if (j == 1) {
                 if (num2 == 91)a[i][j] = 65;
                 else a[i][j] = num2++;
             }
             else a[i][j] = a[i][j - 1] + 1;
         }
     }
     for (int i = 0; i < 27; i++) {
         for (int j = 0; j < 27; j++) {
             if (i == 0 && j == 0)ofs << "    ";
             else
             ofs << a[i][j] << " ";
         }
         ofs << endl;
     }
     ofs.close();
     output(cout);
}