实验6

task3_1

#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

#include <iostream>
#include <fstream>
#include <array>
#define N 5

int main() {
    using namespace std;
    array<char, 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";
}

运行截图

int类型占四个字节，但97等只保存在一个地址空间里，其余三个字节保存0，char类型占一个字节，读入char类型数组时，第一次读到97，转化为a,接下来读到数据为空，直到第五个读到98，变为b,结束

vector.hpp

#pragma once
#include<iostream>
using namespace std;
template<class T>
class Vector {
public:
	Vector() {

	}
	Vector(int n) {
		p = new T[n];
		size = n;
	}
	Vector(int n,T value) {
		p = new T[n];
		for (int i = 0; i < n; i++) {
			p[i] = value;
		}
		size = n;
	}
	
	Vector(const Vector<T> &vec) {
		p = new T[vec.get_size()];
		for (int i = 0; i < vec.get_size(); i++) {
			p[i] = vec.p[i];
		}
		size = vec.size;
	}
	~Vector(){
		delete p;
	}

	int get_size() const;
	T& at(int index);
	T& operator[](int index);
	template<typename T>
	friend void output(Vector<T> vec);

private:
	T* p;
	int size;
};
template<class T>
int Vector<T>::get_size() const{
	return size;
}
template<class T>
T& Vector<T>::at(int index) {
	return p[index];
}
template<class T>
T& Vector<T>::operator[](int index) {
	return p[index];
}
template<class T>
void output(Vector<T> vec) {
	for (int i = 0; i < vec.get_size(); i++) {
		cout << vec.p[i] << " ";
	}
	cout << endl;
}

task4.cpp

#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 * 0.7;

    output(x1);

    Vector<int> x2(n, 42);
    Vector<int> x3(x2);

    output(x2);
    output(x3);

    x2.at(0) = 77;
    output(x2);

    x3[0] = 999;
    output(x3);
}

int main() {
    test();
}

task5

#include<iostream>
using namespace std;
#include<fstream>

void output(ofstream& out);

int main() {
	ofstream out;
	out.open("cipher_key.txt", ios::out);
	output(out);
}

void output(ofstream& out) {
	cout << " ";
	out << " ";

	for (int i = 0; i < 26; i++) {
		char c = i + 'a';
		cout <<" " << c;
		out <<" "<< c;
	}
	cout << endl;
	out << endl;
	for (int i = 1; i<=26; i++) {
		cout << i;
		out << i;
		for (int j = i; j < i+26; j++) {
			char c = 'A' + j%26;
			cout << " " << c;
			out << " " << c;
		}
		cout << endl;
		out << endl;
	}
}