实验6 模板类和文件I/O

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.dat", ios::binary);
    if(!out.is_open()) {
        cout << "fail to open data1.dat\n";
        return 1;
    }

    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<int, N> x;

    ifstream in;
    in.open("data1.dat", ios::binary);
    if(!in.is_open()) {
        cout << "fail to open data1.dat\n";
        return 1;
    }

    in.read(reinterpret_cast<char *>(&x), sizeof(x));
    in.close();

    for(auto i = 0; i < N; ++i)
        cout << x[i] << ", ";
    cout << "\b\b \n";
}

 

问题：存储时用int类型，占四个字节，读取时按char一个字节读，97的四个字节赋值给char变量，隐式转换为a，后面的三个字节为空，接着赋值97，转换为b

task4.hpp

#pragma once
#include <iostream>
using namespace std;

template <typename T>
class Vector
{
public:
    Vector(int n)
    {
        this->size = n;
        v = new T[n];
    }
    Vector(int n, T t)
    {
        this->size = n;
        v = new T[n];
        for (int i = 0; i < n; i++)
        {
            this->v[i] = t;
        }
    }
    T &at(int n)
    {
        return this->v[n];
    }
    T &operator[](int n)
    {
        return this->v[n];
    }
    int getsize()
    {
        return this->size;
    }

private:
    int size;
    T *v;
};
template <typename T>
void output(Vector<T> &v)
{
    for (int i = 0; i < v.getsize(); i++)
        cout << v[i] << ' ';
    puts("");
}

task5.cpp

#include <iostream>
#include <fstream>
#include <iomanip>
using namespace std;
void output(std::ostream &out)
{
    char cipher[100][100];
    cipher[0][0] = ' ';
    for (int i = 1; i <= 26; i++)
        cipher[0][i] = 'a' + i - 1;
    for (int i = 1; i <= 26; i++)
    {
        cipher[i][0] = '0' + i;
        for (int j = 1; j <= 26 - i; j++)
        {
            cipher[i][j] = 'A' + i + j - 1;
        }
        for (int k = 26 - i + 1, m = 0; k <= 26; k++, m++)
        {
            cipher[i][k] = 'A' + m;
        }
    }
    for (int i = 0; i <= 26; i++)
    {
        if (i != 0)
            cout << setw(2) << cipher[i][0] - '0';
        else
            cout << setw(2) << cipher[i][0];
        for (int j = 1; j <= 26; j++)
            cout << setw(2) << cipher[i][j];
        puts("");
    }
    for (int i = 0; i <= 26; i++)
    {
        if (i != 0)
            out << setw(2) << cipher[i][0] - '0';
        else
            out << setw(2) << cipher[i][0];
        for (int j = 1; j <= 26; j++)
            out << setw(2) << cipher[i][j];
        out << '\n';
    }
}
int main()
{
    ofstream out;
    out.open("cipher_key.txt", ios::out);
    if (!out.is_open())
    {
        cout << "fail!!" << endl;
        return 1;
    }
    output(out);
    return 0;
}

cipher_key.txt