WELCOME

任何一个伟大的目标,都有一个微不足道的开始。

实验四

task 5

vectorInt.hpp

#pragma once
#include <iostream>
using namespace std;
class vectorInt {
public:
    vectorInt (int n) {
        cout << "constructor 2 called." << endl;
        this->capacity = n;
        this->size = n;
        this->arr = new int[n];
    }
    vectorInt (int n ,int x) {
        this->capacity = n;
        this->size = n;
        this->arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = x;
        }
    }
    vectorInt(vectorInt & x) {
        cout << "constructor 2 called." << endl;
        cout << "copy constructor called." << endl;
        this->size = x.size;
        this->arr = new int[x.size];
        for(int i = 0; i < x.size; i++)
        {
            this->arr[i] = x.arr[i];
        }
    }

    int& at(int idx) {
        return this->arr[idx];
    }
    // int operator[] (int idx){
    //     return this->arr[idx];
    // }
    ~vectorInt() {
        if(arr != NULL)
        {
            delete[] arr;
            this->arr = NULL;
            cout << "destructor called" << endl;
        }
    }
public:
    const int INCREACE_LEN = 64;
    int *arr;
    int size;
    int capacity;
};
void output(vectorInt &x){
    for (int i = 0; i < x.size; i++) {
        cout << x.at(i) << ' ';
    }
    puts("");
}

task5.cpp

 

#include <iostream>
#include "vectorInt.hpp"

void test() {
    using namespace std;

    int n;
    cin >> n;
    
    vectorInt x1(n);
    for(auto i = 0; i < n; ++i)
        x1.at(i) = i * i;
    output(x1);

    vectorInt x2(n, 42);
    vectorInt x3(x2);

    output(x2);
    output(x3);

    x2.at(0) = 77;

    output(x2);
    output(x3);
}

int main() {
    test();
}

 

 

 task6

Matrix.hpp

#pragma once
#include <iostream>
using namespace std;
using std::cout;
using std::endl;

class Matrix {
public:
    Matrix(int n);                     // 构造函数,构造一个n*n的矩阵
    Matrix(int n, int m);              // 构造函数,构造一个n*m的矩阵
    Matrix(const Matrix &X);           // 复制构造函数,使用已有的矩阵X构造
    ~Matrix();                         //析构函数

    void set(const double *pvalue);     // 用pvalue指向的连续内存块数据按行为矩阵赋值
    void set(int i, int j, int value); //设置矩阵第i行第j列元素值为value
    double &at(int i, int j);          //返回矩阵第i行第j列元素的引用
    double at(int i, int j) const;     // 返回矩阵第i行第j列元素的值
    int get_lines() const;             //返回矩阵行数
    int get_cols() const;              //返回矩列数
    void print() const;                // 按行打印输出矩阵

private:
    int lines; // 矩阵行数
    int cols;  // 矩阵列数
    double *p; // 指向存放矩阵数据的内存块的首地址
};

// 类Matrix的实现:待补足
Matrix::Matrix(int n) :lines(n), cols(n) 
{
    p = new double[n * n];
}
Matrix::Matrix(int n, int m) : lines(n), cols(m)
{
    p = new double[n * m];
}
Matrix::Matrix(const Matrix& X) : lines(X.lines), cols(X.cols)
{
    p = new double[lines * cols];
    for (int i = 1; i <= lines * cols; i++)
        p[i] = X.p[i];
}
Matrix::~Matrix()
{
    delete[]p;
}
void Matrix::set(const double* pvalue)
{
    for (int i = 0; i < lines * cols; i++)
        p[i] = pvalue[i];
}
void Matrix::set(int i, int j, int value)
{
    p[i * cols + j] = value;
}
double& Matrix::at(int i, int j)
{
    return p[i * cols + j];
}
double Matrix::at(int i, int j) const
{
    return p[i * cols + j];
}
int Matrix::get_lines() const
{
    return lines;
}
int Matrix::get_cols() const
{
    return cols;
}

void Matrix::print() const
{
    for (int i = 0; i < lines; i++)
    {
        for (int j = 0; j < cols; j++)
            cout << p[i * cols + j] << ' ';
        cout << endl;
    }
}

task6.cpp

#include <iostream>
#include "matrix.hpp"

void test() {
    using namespace std;

    double x[] = {1, 2, 3, 4, 5, 6};

    Matrix m1(3, 2);    // 创建一个3×2的矩阵
    m1.set(x);          // 用一维数组x的值按行为矩阵m1赋值
    m1.print();         // 打印矩阵m1的值
    cout << "the first line is: " << endl;
    cout << m1.at(0, 0) << " " << m1.at(0, 1) << endl;  // 输出矩阵m1第1行两个元素的值
    cout << endl;

    Matrix m2(2, 3);
    m2.set(x);
    m2.print();
    cout << "the first line is: " << endl;
    cout << m2.at(0, 0) << " " << m2.at(0, 1) << " " << m2.at(0, 2) << endl;
    cout << endl;

    Matrix m3(m2);      // 用矩阵m2构造新的矩阵m3
    m3.set(0, 0, 999);  // 将矩阵m3第0行第0列元素值设为999
    m3.print();
}

int main() {
    test();
}

 

posted @ 2022-11-07 23:47  周杰棍的双截伦  阅读(25)  评论(0)    收藏  举报