C++ 开发之实现操作符重载的实例
发布时间:2020-12-30 21:36:30 所属栏目:创业 来源:网络整理
导读:C++操纵符重载 实现结果图: 实例代码: Matrix.h #pragma once #include "vector" #include "iostream" #define rep(i,n) for(int i=1;i=n;i++) //宏界说for轮回,精简代码 using namespace std; class Matrix { public: //根基结构函数 Matrix(int Row=0,i
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main.c
#include "iostream"
#include "Matrix.h"
using namespace std;
int main(){
int row1,row2,col1,col2;
cout << "请输入第一个矩阵的行和列:n";
cin >> row1 >> col1;
Matrix m1(row1,col1);
cout << "请输入" << row1 << '*' << col1 << "个数:n";
cin >> m1;
cout << "输出矩阵的值:n";
cout << m1;
cout << "请输入第二个矩阵的行和列:n";
cin >> row2 >> col2;
Matrix m2(row2,col2);
cout << "请输入" << row2 << '*' << col2 << "个数:n ";
cin >> m2;
cout << "输出矩阵的值:n";
cout << m2;
if (col1 != row2)
cout << "这两个矩阵无法相乘n";
else
{
cout << "判定矩阵m1与m2是否相称:n";
if (m1==m2)
{
cout << "m1和m2相称:n";
}
else
{
cout << "m1和m2不相称:n";
}
cout << "m1拷贝结构m3矩阵功效输出:n";
Matrix m3(m1);
cout << m3;
cout << "m1赋值重载m4矩阵功效输出:n";
Matrix m4(m1.Row,m1.Column);
m4 = m1;
cout << m4;
cout << "m1*m2矩阵相乘输出m5:n";
Matrix m5(m1.Row,m2.Column);
m5 = m1*m2;
cout << m5;
cout << "矩阵m1*2输出m6:n";
Matrix m6(m1.Row,m1.Column);
m6 = m1*2;
cout << m6;
cout << "矩阵m1*0.5输出m7:n";
Matrix m7(m1.Row,m1.Column);
m7 = m1 * 0.5;
cout << m7;
cout << "m1*m2矩阵相乘输出m1:n";
m1 *= m2;
cout << m1;
cout << "m1矩阵前自增输出n";
cout << ++m1;
cout << "m1矩阵后自增输出n";
cout << m1++;
cout << "m1矩阵输出n";
cout << m1;
cout << "m1矩阵前自减输出n";
cout << --m1;
cout << "m1矩阵后自减输出n";
cout << m1--;
cout << "m1矩阵输出n";
cout << m1;
}
return 0;
}
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