首页 > 编程语言 >【暑假例题】20230727 矩阵基本运算(C++)

【暑假例题】20230727 矩阵基本运算(C++)

时间：2023-07-31 22:55:40浏览次数：47  
标签：Matrix int 矩阵 C++ ++ vector 例题 20230727 size

题目

请使用C++实现矩阵的各种运算

  1. 矩阵创建
  2. 矩阵相加
  3. 矩阵相减
  4. 矩阵相乘
  5. 数字乘矩阵
  6. 矩阵上叠加
  7. 矩阵左右叠加
  8. 矩阵转置
  9. 矩阵旋转
  10. 矩阵求逆
  11. 矩阵输出

题目分析

矩阵创建

这里只需注意由于我们需要通过不同的函数对数组进行操作，所以我们需要将数组存储在容器或者使用指针防止数据丢失

const double epsilon = 1e-12;//小于该数判断为0

vector<vector<double>> v;//矩阵

vector<vector<double>> Create_Matrix(int h, int l)//创建 h行l列的矩阵，并将初始各值设定为0
{
	for (int i = 0; i < h; i++)
	{
		vector<double>v1(l, 0);
		v.push_back(v1);
	}
	return v;
}

矩阵相加、减、乘

需创建一个新的得数矩阵，并且注意两个矩阵对应位置相加减、乘即可

矩阵相加

vector<vector<double>> Matrix_add(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵相加
{
	int h = A.size();
	int l = A[0].size();

	vector<vector<double>> C;
	C = Create_Matrix(h, l);//创建矩阵C

	for (int i = 0; i < h; i++)
	{
		for (int j = 0; j < l; j++)
		{
			C[i][j] = A[i][j] + B[i][j];
			if (abs(C[i][j]) < epsilon)
			{
				C[i][j] = 0;
			}
		}
	}
	return C;
}

矩阵相减

vector<vector<double>> Matrix_sub(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵相减
{
	int h = A.size();
	int l = A[0].size();

	vector<vector<double>> C;
	C = Create_Matrix(h, l);//创建矩阵C

	for (int i = 0; i < h; i++)
	{
		for (int j = 0; j < l; j++)
		{
			C[i][j] = A[i][j] - B[i][j];
			if (abs(C[i][j]) < epsilon)
			{
				C[i][j] = 0;
			}
		}
	}
	return C;
}

矩阵相乘

vector<vector<double>> Matrix_mul(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵相乘
{
	int A_h = A.size();
	int A_l = A[0].size();
	int B_h = B.size();
	int B_l = B[0].size();

	if (A_l != B_h)
	{
		cout << "两矩阵维数无法相乘" << endl;
		exit(0);
	}

	vector<vector<double>> C= Create_Matrix(A_h, B_l);//创建矩阵C
	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < B_l; j++)
		{
			C[i][j] = 0;
			for (int k = 0; k < A_l; k++)
			{
				C[i][j] += A[i][k] * B[k][j];
			}
			if (abs(C[i][j]) < epsilon)
			{
				C[i][j] = 0;
			}
		}
	}
	return C;
}

数字乘矩阵

矩阵里的每一个数乘数字就好了

vector<vector<double>> Matrix_x_num(const vector<vector<double>>& A, double num)//数字乘矩阵
{
	int A_h = A.size();
	int A_l = A[0].size();

	vector<vector<double>> B = Create_Matrix(A_h, A_l);//创建矩阵B

	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < A_l; j++)
		{
			B[i][j] = num * A[i][j];
		}
	}
	return B;
}

矩阵上叠加、矩阵左右叠加

先判断两个矩阵的列相等，然后分别导入两个矩阵进行叠加

矩阵上叠加

vector<vector<double>> Superposition_on_matrix(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵A与矩阵B上下叠加获得新的矩阵C,并返回
{
	//判断矩阵的列是否相等
	int A_h = A.size();
	int A_l = A[0].size();
	int B_h = B.size();
	int B_l = B[0].size();

	if (A_l != B_l)
	{
		cout << "叠加的矩阵列数不相等" << endl;
		exit(0);
	}

	//创建
	vector<vector<double>> C = Create_Matrix(A_h + B_h, A_l);

	//将A传入
	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < A_l; j++)
		{
			C[i][j] = A[i][j];
		}
	}

	//将B传入
	for (int i = 0; i < B_h; i++)
	{
		for (int j = 0; j < B_l; j++)
		{
			C[i + A_h][j] = B[i][j];
		}
	}
	return C;
}

矩阵左右叠加

vector<vector<double>> L_R_Superposition_matrix(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵A与矩阵B左右叠加，获得新的矩阵C
{
	//判断矩阵的列是否相等
	int A_h = A.size();
	int A_l = A[0].size();
	int B_h = B.size();
	int B_l = B[0].size();

	if (A_h != B_h)
	{
		cout << "叠加的矩阵行数不相等" << endl;
		exit(0);
	}

	//创建叠加后的矩阵C
	vector<vector<double>> C = Create_Matrix(A_h, A_l + B_l);

	//将A传入
	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < A_l; j++)
		{
			C[i][j] = A[i][j];
		}
	}

	//将B传入
	for (int i = 0; i < B_h; i++)
	{
		for (int j = 0; j < B_l; j++)
		{
			C[i][j + A_l] = B[i][j];
		}
	}
	return C;
}

矩阵转置、矩阵旋转

这部分可以看我做过的题目

https://www.cnblogs.com/hcrzhi/p/17538056.html

矩阵转置

vector<vector<double>> Matrix_device(const vector<vector<double>>& A)//矩阵转置
{
	vector<vector<double>> AT = Create_Matrix(A[0].size(), A.size());//AT为转置矩阵
	int l = AT.size();
	int c = AT[0].size();
	for (int i = 0; i < l; i++)
	{
		for (int j = 0; j < c; j++)
		{
			AT[i][j] = A[j][i];
		}
	}
	return AT;
}

矩阵旋转

vector<vector<double>> Matrix_rotation(const vector<vector<double>>& A)//矩阵旋转
{
	int l = A.size();
	int c = A[0].size();
	//水平翻转
	for (int i = 0; i < l / 2; i++)
	{
		for (int j = 0; j < l; j++)
		{
			swap(A[i][j], A[l - i - 1][j]);//交换函数
		}
	}

	//对角线翻转
	for (int i = 0; i < l; i++)
	{
		for (int j = 0; j < i; j++)
		{
			swap(A[i][j], A[j][i]);
		}
	}
	return A;
}

矩阵求逆

这里用到了一个LU分解，大家可以去看一下：https://blog.csdn.net/qq_28972011/article/details/123935820

还有一个函数abs：https://blog.csdn.net/weixin_44006573/article/details/124596118?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522169081501216800180660357%2522%252C%2522scm%2522%253A%252220140713.130102334..%2522%257D&request_id=169081501216800180660357&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2~all~sobaiduend~default-1-124596118-null-null.142^v91^control_2,239^v12^control2&utm_term=abs%28&spm=1018.2226.3001.4187

vector<vector<double>> Matrix_inversion(const vector<vector<double>>& A)//矩阵求逆
{
	if (A.size() != A[0].size())
	{
		cout << "该矩阵无法求逆矩阵" << endl;
		exit(0);
	}
	int n = A.size();
	vector<vector<double>> inv_A = Create_Matrix(n, n);//逆矩阵
	vector<vector<double>> L = Create_Matrix(n, n);
	vector<vector<double>> U = Create_Matrix(n, n);
	vector<vector<double>> inv_L = Create_Matrix(n, n);
	vector<vector<double>> inv_U = Create_Matrix(n, n);//LU分解

	//L矩阵对角元素为1
	for (int i = 0; i < n; i++)
	{
		L[i][i] = 1;
	}

	//U矩阵第一行
	for (int i = 0; i < n; i++)
	{
		U[0][i] = A[0][i];
	}

	//L矩阵第一列
	for (int i = 0; i < n; i++)
	{
		U[0][i] = A[0][i];
	}

	//计算LU上下三角
	for (int i = 1; i < n; i++)
	{
		//计算U（i行j列）
		for (int j = i; j < n; j++)
		{
			double temp = 0;
			for (int k = 0; k < i; k++)
			{
				temp += L[i][k] * U[k][j];
			}
			U[i][j] = A[i][j] - temp;
			if (abs(U[i][j]) < epsilon)
			{
				U[i][j] = 0;
			}
		}
		//计算L（j行i列）
		for (int j = i; j < n; j++)
		{
			double temp = 0;
			for (int k = 0; k < i; k++)
			{
				temp += L[j][k] * U[k][i];
			}
			L[j][i] = 1.0 * (A[j][i] - temp) / U[i][i];
			if (abs(L[i][j]) < epsilon)
			{
				L[i][j] = 0;
			}
		}

	}

	//L U剩余位置设为0
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (i > j)
			{
				U[i][j] = 0;
			}
			else if (i < j)
			{
				L[i][j] = 0;
			}
		}
	}

	//LU求逆
	//求矩阵U的逆 
	for (int i = 0; i < n; i++)
	{
		inv_U[i][i] = 1 / U[i][i];// U对角元素的值，直接取倒数
		for (int k = i - 1; k >= 0; k--)
		{
			double s = 0;
			for (int j = k + 1; j <= i; j++)
			{
				s = s + U[k][j] * inv_U[j][i];
			}
			inv_U[k][i] = -s / U[k][k];//迭代计算，按列倒序依次得到每一个值，
			if (abs(inv_U[k][i]) < epsilon)
			{
				inv_U[k][i] = 0;
			}
		}
	}

	//求矩阵L的逆
	for (int i = 0; i < n; i++)
	{
		inv_L[i][i] = 1; //L对角元素的值，直接取倒数，这里为1
		for (int k = i + 1; k < n; k++)
		{
			for (int j = i; j <= k - 1; j++)
			{
				inv_L[k][i] = inv_L[k][i] - L[k][j] * inv_L[j][i];
				if (abs(inv_L[k][i]) < epsilon)
				{
					inv_L[k][i] = 0;
				}
			}
		}
	}

	inv_A = Matrix_mul(inv_U, inv_L);
	return inv_A;
}

矩阵输出

vector<vector<double>> Output_Matrix(const vector<vector<double>>& A)//矩阵输出
{
    int l = A.size();
	int c = A[0].size();
	cout << endl;
	for (int i = 0; i < l; i++)//输出矩阵
	{
		for (int j = 0; j < c; j++)
		{
			cout << A[i][j];
		}
		cout << endl;
	}
	cout << endl;
}

整合后（有bug）

#include<iostream>
#include<conio.h>
#include<vector>
using namespace std;

char option;//选项
const double epsilon = 1e-12;//小于该数判断为0

vector<vector<double>> v;//矩阵

vector<vector<double>> Create_Matrix(int h, int l)//创建 h行l列的矩阵，并将初始各值设定为0
{
	for (int i = 0; i < h; i++)
	{
		vector<double>v1(l, 0);
		v.push_back(v1);
	}
	return v;
}

vector<vector<double>> Matrix_add(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵相加
{
	int h = A.size();
	int l = A[0].size();

	vector<vector<double>> C;
	C = Create_Matrix(h, l);//创建矩阵C

	for (int i = 0; i < h; i++)
	{
		for (int j = 0; j < l; j++)
		{
			C[i][j] = A[i][j] + B[i][j];
			if (abs(C[i][j]) < epsilon)
			{
				C[i][j] = 0;
			}
		}
	}
	return C;
}

vector<vector<double>> Matrix_sub(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵相减
{
	int h = A.size();
	int l = A[0].size();

	vector<vector<double>> C;
	C = Create_Matrix(h, l);//创建矩阵C

	for (int i = 0; i < h; i++)
	{
		for (int j = 0; j < l; j++)
		{
			C[i][j] = A[i][j] - B[i][j];
			if (abs(C[i][j]) < epsilon)
			{
				C[i][j] = 0;
			}
		}
	}
	return C;
}

vector<vector<double>> Matrix_mul(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵相乘
{
	int A_h = A.size();
	int A_l = A[0].size();
	int B_h = B.size();
	int B_l = B[0].size();

	if (A_l != B_h)
	{
		cout << "两矩阵维数无法相乘" << endl;
		exit(0);
	}

	vector<vector<double>> C= Create_Matrix(A_h, B_l);//创建矩阵C
	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < B_l; j++)
		{
			C[i][j] = 0;
			for (int k = 0; k < A_l; k++)
			{
				C[i][j] += A[i][k] * B[k][j];
			}
			if (abs(C[i][j]) < epsilon)
			{
				C[i][j] = 0;
			}
		}
	}
	return C;
}

vector<vector<double>> Matrix_x_num(const vector<vector<double>>& A, double num)//数字乘矩阵
{
	int A_h = A.size();
	int A_l = A[0].size();

	vector<vector<double>> B = Create_Matrix(A_h, A_l);//创建矩阵B

	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < A_l; j++)
		{
			B[i][j] = num * A[i][j];
		}
	}
	return B;
}

vector<vector<double>> Superposition_on_matrix(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵A与矩阵B上下叠加获得新的矩阵C,并返回
{
	//判断矩阵的列是否相等
	int A_h = A.size();
	int A_l = A[0].size();
	int B_h = B.size();
	int B_l = B[0].size();

	if (A_l != B_l)
	{
		cout << "叠加的矩阵列数不相等" << endl;
		exit(0);
	}

	//创建
	vector<vector<double>> C = Create_Matrix(A_h + B_h, A_l);

	//将A传入
	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < A_l; j++)
		{
			C[i][j] = A[i][j];
		}
	}

	//将B传入
	for (int i = 0; i < B_h; i++)
	{
		for (int j = 0; j < B_l; j++)
		{
			C[i + A_h][j] = B[i][j];
		}
	}
	return C;
}

vector<vector<double>> L_R_Superposition_matrix(const vector<vector<double>>& A, const vector<vector<double>>& B)//矩阵A与矩阵B左右叠加，获得新的矩阵C
{
	//判断矩阵的列是否相等
	int A_h = A.size();
	int A_l = A[0].size();
	int B_h = B.size();
	int B_l = B[0].size();

	if (A_h != B_h)
	{
		cout << "叠加的矩阵行数不相等" << endl;
		exit(0);
	}

	//创建叠加后的矩阵C
	vector<vector<double>> C = Create_Matrix(A_h, A_l + B_l);

	//将A传入
	for (int i = 0; i < A_h; i++)
	{
		for (int j = 0; j < A_l; j++)
		{
			C[i][j] = A[i][j];
		}
	}

	//将B传入
	for (int i = 0; i < B_h; i++)
	{
		for (int j = 0; j < B_l; j++)
		{
			C[i][j + A_l] = B[i][j];
		}
	}
	return C;
}

vector<vector<double>> Matrix_device(const vector<vector<double>>& A)//矩阵转置
{
	vector<vector<double>> AT = Create_Matrix(A[0].size(), A.size());//AT为转置矩阵
	int l = AT.size();
	int c = AT[0].size();
	for (int i = 0; i < l; i++)
	{
		for (int j = 0; j < c; j++)
		{
			AT[i][j] = A[j][i];
		}
	}
	return AT;
}

vector<vector<double>> Matrix_rotation(const vector<vector<double>>& A)//矩阵旋转
{
	int l = A.size();
	int c = A[0].size();
	//水平翻转
	for (int i = 0; i < l / 2; i++)
	{
		for (int j = 0; j < l; j++)
		{
			swap(A[i][j], A[l - i - 1][j]);//交换函数
		}
	}

	//对角线翻转
	for (int i = 0; i < l; i++)
	{
		for (int j = 0; j < i; j++)
		{
			swap(A[i][j], A[j][i]);
		}
	}
	return A;
}

vector<vector<double>> Matrix_inversion(const vector<vector<double>>& A)//矩阵求逆
{
	if (A.size() != A[0].size())
	{
		cout << "该矩阵无法求逆矩阵" << endl;
		exit(0);
	}
	int n = A.size();
	vector<vector<double>> inv_A = Create_Matrix(n, n);//逆矩阵
	vector<vector<double>> L = Create_Matrix(n, n);
	vector<vector<double>> U = Create_Matrix(n, n);
	vector<vector<double>> inv_L = Create_Matrix(n, n);
	vector<vector<double>> inv_U = Create_Matrix(n, n);//LU分解

	//L矩阵对角元素为1
	for (int i = 0; i < n; i++)
	{
		L[i][i] = 1;
	}

	//U矩阵第一行
	for (int i = 0; i < n; i++)
	{
		U[0][i] = A[0][i];
	}

	//L矩阵第一列
	for (int i = 0; i < n; i++)
	{
		U[0][i] = A[0][i];
	}

	//计算LU上下三角
	for (int i = 1; i < n; i++)
	{
		//计算U（i行j列）
		for (int j = i; j < n; j++)
		{
			double temp = 0;
			for (int k = 0; k < i; k++)
			{
				temp += L[i][k] * U[k][j];
			}
			U[i][j] = A[i][j] - temp;
			if (abs(U[i][j]) < epsilon)
			{
				U[i][j] = 0;
			}
		}
		//计算L（j行i列）
		for (int j = i; j < n; j++)
		{
			double temp = 0;
			for (int k = 0; k < i; k++)
			{
				temp += L[j][k] * U[k][i];
			}
			L[j][i] = 1.0 * (A[j][i] - temp) / U[i][i];
			if (abs(L[i][j]) < epsilon)
			{
				L[i][j] = 0;
			}
		}

	}

	//L U剩余位置设为0
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (i > j)
			{
				U[i][j] = 0;
			}
			else if (i < j)
			{
				L[i][j] = 0;
			}
		}
	}

	//LU求逆
	//求矩阵U的逆 
	for (int i = 0; i < n; i++)
	{
		inv_U[i][i] = 1 / U[i][i];// U对角元素的值，直接取倒数
		for (int k = i - 1; k >= 0; k--)
		{
			double s = 0;
			for (int j = k + 1; j <= i; j++)
			{
				s = s + U[k][j] * inv_U[j][i];
			}
			inv_U[k][i] = -s / U[k][k];//迭代计算，按列倒序依次得到每一个值，
			if (abs(inv_U[k][i]) < epsilon)
			{
				inv_U[k][i] = 0;
			}
		}
	}

	//求矩阵L的逆
	for (int i = 0; i < n; i++)
	{
		inv_L[i][i] = 1; //L对角元素的值，直接取倒数，这里为1
		for (int k = i + 1; k < n; k++)
		{
			for (int j = i; j <= k - 1; j++)
			{
				inv_L[k][i] = inv_L[k][i] - L[k][j] * inv_L[j][i];
				if (abs(inv_L[k][i]) < epsilon)
				{
					inv_L[k][i] = 0;
				}
			}
		}
	}

	inv_A = Matrix_mul(inv_U, inv_L);
	return inv_A;
}

vector<vector<double>> Output_Matrix(const vector<vector<double>>& A)//矩阵输出
{
	cout << "\t是否输出矩阵" << endl;
	cout << "1.是" << endl;
	cout << "2.否" << endl;
	option = _getch();
	switch (option)
	{
	case '1':
	{
		int l = A.size();
		int c = A[0].size();
		cout << endl;
		for (int i = 0; i < l; i++)//输出矩阵
		{
			for (int j = 0; j < c; j++)
			{
				cout << A[i][j];
			}
			cout << endl;
		}
		cout << endl;
		break;
	}
	case '2':
		break;
	default:
		cout << "该选项不存在，请重新输入" << endl;
		break;
	}
}

int Matrix_Operations()//矩阵操作菜单
{
	cout << "\t请输入你的选项" << endl;
	cout << "1.矩阵相加" << endl;
	cout << "2.矩阵相减" << endl;
	cout << "3.矩阵相乘" << endl;
	cout << "4.数字乘矩阵" << endl;
	cout << "5.矩阵上叠加" << endl;
	cout << "6.矩阵左右叠加" << endl;
	cout << "7.矩阵转置" << endl;
	cout << "8.矩阵旋转" << endl;
	cout << "9.矩阵求逆" << endl;
	cout << "A.矩阵输出" << endl;
	cout << "B.创建矩阵\n" << endl;
	cout << "0.退出" << endl;
	option = _getch();
	switch (option)
	{
	case '1':
		Matrix_add(v, v);//矩阵相加
		break;
	case '2':
		Matrix_sub(v, v);//矩阵相减
		break;
	case '3':
		Matrix_mul(v, v);//矩阵相乘
	case '4':
		cout << "请输入数字" << endl;
		float num;
		cin >> num;
		Matrix_x_num(v, num);//数字乘矩阵
		break;
	case '5':
		Superposition_on_matrix(v, v);//矩阵上叠加
		break;
	case '6':
		L_R_Superposition_matrix(v, v);//矩阵左右叠加
		break;
	case '7':
		Matrix_device(v);//矩阵转置
		break;
	case '8':
		Matrix_rotation(v);//矩阵旋转
		break;
	case '9':
		Matrix_inversion(v);//矩阵求逆
		break;
	case 'A':
		Output_Matrix(v);//矩阵输出
		break;
	case 'B':
		cout << "请输入矩阵的行和列" << endl;
		int h, l;
		cin >> h >> l;
		Create_Matrix(h, l);//创建矩阵
		break;
	case '0':
		exit(0);
		break;
	default:
		cout << "该选项不存在，请重新输入" << endl;
		break;
	}
}

int main()
{
	while (1)
	{
		Matrix_Operations();//菜单界面
	}
	return 0;
}

　　

　

标签：Matrix,int,矩阵,C++,++,vector,例题,20230727,size
From： https://www.cnblogs.com/hcrzhi/p/17595215.html

相关文章

  • C++中不支持strdup(),使用_strdup()
    1.问题C4996 'strdup':ThePOSIXnameforthisitemisdeprecated.Instead,usetheISOCandC++conformantname:_strdup.Seeonlinehelpfordetails. Project1 G:\VS\Project1\Project1\MyString.cpp 802.原因错误C4996是一个编译器警告，它指示在代码中使用了一个......
  • c++11 call_once & once_flag
    #include<iostream>#include<mutex>usingnamespacestd;voidonce_fun(std::once_flag&&flag){std::call_once(flag,[](){cout<<"messagefromoncefun"<<endl;});}voidonce_fun1(s......
  • c++并发
    《c++并发编程第二版》，一方面翻译的挺拗口，一方面内容有点难度，看的很吃力，尤其到后面内存模型，无锁编程部分，看不下去了，仅记录部分内容。并发：采用任务切换的方式，一个任务运行一会在切换回另一个任务，在一个时间段内好像两个任务在同时进行并行：在多个CPU或多核CPU中同一时间多个任务同......
  • 递归例题
    //求最大公约数//#include<iostream>//usingnamespacestd;//intbig(inta,intb){// if(a%b==0){// returnb;// }else{// returnbig(b,a%b);// }//}//小青蛙跳台阶问题//intmain(intargc,char**argv){// inta,b;// cin>>a>>b;// cout<&l......
  • C++ | const的使用
    const基础用法用于声明一个不可再被修改的变量：constintnum=17;num=33; //Invalid,会报错用于指针首先，如果将一个指针用const声明为常量指针，那么这个指针的指向将不能被改变。但可以通过这个指针来修改被指向的对象：intdemo=17;int*constp=&demo; //......
  • C++ | 引用
    引用&使用引用intb=33,&a=b;b=17;cout<<a;//打印a为17使用引用声明变量类似于指针，声明的变量会同被引用的变量绑定，修改任何一方的数据都会使另一方的数据也改变。使用上的区别在于使用引用不需要*，引用可以理解为一个别名。在函数中使用引用传递参数将参......
  • C++ | 运算符重载
    运算符重载在类中的函数进行重载（成员函数）运算符重载用于重新定义运算符的作用，使用函数名称operatorOP作为函数名，其中OP为具体的运算符（如operator+）classTime{Timeoperator+(constTime&t);};Timea,b;Timec=a+b;在成员函数中重载的运算符，如+-等，默认左边......
  • C++初始化列表时，形参和实参名可以一样，编译器可以识别
    在这里初始化列表直接用age(age)即可,用this->age(age)反而会出错，C++不允许在成员初始化列表中使用this关键字来初始化类成员classPerson{public:Person(intage):age(age){}//Initializethe'age'memberdirectlyinthememberinitializerlistinta......
  • TreeCtrl 树的存储和还原 2， VC++
    TreeCrtl树的节点可以随意的添加，删除，移动。树保存到数据库中，并能从数据库读出还原。树节点的LPARAM存放ID，这个ID是数据表自增长主键。1.数据库中读出，按PID，SID顺序，这样读出的第一个为root2.读出的数据存入list3.创建root节点，把root放入map，从list中删除4.遍历list，pid，sid在map中找......
  • 从 C++到 Objective-C -－－－2
    从C++到Objective-C（7）：继承简单继承Objective-C也有继承的概念，但是不能多重继承。不过，它也有别的途径实现类似多重继承的机制，这个我们后面会讲到。C++Objective-CclassFoo:publicBar,{}@interfaceFoo:Bar//单继承//如果要同时“继承”Wiz，需要使用......

赞助商

阅读排行

网站主页关于我们联系我们网站地图

本网站内容转载自其他媒体，侵权联系[admin##ips99.com]。

Copyright © 2020-2023 IPS99 版权所有 IPS99