实验6

实验任务3：

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<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";
}

 

 当把

array<int, N> x; ----> 修改成: array<char, N> x;

 

     原因： 当array数组储存字符型时，读入文件内容”a b c d e f“将把空格也读成字符。

 实验任务4：

Vector.hpp:

#include <iostream>
using namespace std;
template<typename T>
class Vector {
public:
	Vector(int n) :size{ n } {
		p = new T[n];

	}
	Vector(T n, T value) :size{ n } {
		p = new T[n];
		
		for (auto i = 0; i < size; i++) {
			p[i] = value;
		}
	}
	Vector(Vector& vi) :size{ vi.size } {
		cout << "copy constructor called." << endl;
		p = new T[size];
		for (auto i = 0; i < size; ++i)
		{
			p[i] = vi.p[i];
		}

	}
	~Vector() = default;

	T get_size() { return size; }
	
	T& at(int index) { return p[index]; }
	
	friend void output(Vector& x)
	{
		int i;
		for (i = 0; i < x.size - 1; i++)
		{
			cout << x.p[i] << ", ";
		}
		cout << x.p[i] << endl;
	}
	T & operator [](int n) { return p[n]; }
private:
	T size;
	T* p;
};

　　

 

task:

#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();
}

　　

 

 

 

 实验任务5：

 

#include <iostream>
#include<fstream>
#include<string>
using namespace std;
void output() {
    string a[27][27] = { " " };
    int i, j, n = 0;
    for (i = 1; i < 27; i++) {
        a[i][0] = to_string(i);
        a[0][i] = 97 + n;
        n++;
    }
    for (i = 1; i < 27; i++) {
        n = i;
        for (j = 1; j < 27; j++) {
            if (65 + n > 90)
            {
                a[i][j] = 65 + n - 26;
            }
            else {
                a[i][j] = 65 + n;
            }
            n++;
        }

    }


    ofstream out;
    out.open("cipher_ke.txt");
    if (!out.is_open()) {
        cout << "fail to open cipher_key.txt to write" << endl;
    }

    for (i = 0; i < 27; i++) {

        for (j = 0; j < 27; j++) {
            cout << a[i][j] << " ";
            out << a[i][j] << " ";
        }
        cout << endl;
        out << endl;
    }



}

int main() {
    
    output();

}