实验三 矩阵类的设计与实现

实验三矩阵类的设计与实现一、实验目的通过上机实践，巩固课堂老师所讲关于矩阵设计的知识，通过反复的调试，找错，设计出矩阵。

二、实验过程#include<iostream.h>#include<stdlib.h>#include<math.h>#include<stdio.h>class XXXMatrix{private:int rows,columns;double **element;public:XXXMatrix(int rows=4,int columns=4);XXXMatrix(int *mat,int rows,int columns);XXXMatrix(const XXXMatrix& m);~XXXMatrix();int rintv(int n,double a[][4]);XXXMatrix& inverse(XXXMatrix& mat);XXXMatrix& transpose(double a[][4]);void init(int rows,int columns);double get(int i,int j);bool set(int i,int j,int value);friend ostream& operator <<(ostream& out,XXXMatrix &mat);XXXMatrix& operator +=(XXXMatrix& mat);XXXMatrix& operator=(const XXXMatrix& mat);bool operator==(const XXXMatrix& mat);bool operator!=(const XXXMatrix& mat);XXXMatrix& operator-(XXXMatrix& mat);XXXMatrix& operator-();XXXMatrix& operator*(XXXMatrix& mat);};void XXXMatrix::init(int rows,int columns) {if(rows<=0||columns<=0)throw"矩阵行或列数非正数异常";this->rows=rows;this->columns=columns;element=new double *[rows];for(int i=0;i<rows;i++){element[i]=new double[columns];for(int j=0;j<columns;j++)element[i][j]=0;}}XXXMatrix::XXXMatrix(int rows,int columns) {init(rows,columns);}XXXMatrix::XXXMatrix(int *mat,int rows,int columns){init(rows,columns);for(int i=0;i<rows;i++)for(int j=0;j<columns;j++)element[i][j]=*(mat+i*columns+j);}XXXMatrix::XXXMatrix(const XXXMatrix& m) {init(m.rows,m.columns);for(int i=0;i<m.rows;i++)for(int j;j<m.columns;j++)element[i][j]=m.element[i][j]; }XXXMatrix::~XXXMatrix(){for(int i=0;i<rows;i++)delete(element[i]);delete(element);}double XXXMatrix::get(int i,int j){if(i>=0&&i<rows&&j>=0&&j<columns)return element[i][j];}bool XXXMatrix::set(int i,int j,int value){if(i>=0&&i<rows&&j>=0&&j<columns){element[i][j]=value;return true;}return false;}ostream& operator<<(ostream& out,XXXMatrix& mat){out<<"矩阵XXXMatrix("<<mat.rows<<"x"<<mat.columns< <")"<<endl;for(int i=0;i<mat.rows;i++){for(int j=0;j<mat.columns;j++)out<<mat.element[i][j]<<" ";out<<"\n";}return out;}XXXMatrix& XXXMatrix:: operator=(const XXXMatrix& mat){if(this->rows==mat.rows&&this->column s==mat.columns){for(int i=0;i<mat.rows;i++)for(int j=0;j<mat.columns;j++)element[i][j]=mat.element[i][j];return *this;}throw"两个矩阵阶数不同，不能相加"; }bool XXXMatrix:: operator==(const XXXMatrix& mat){if(this->rows==mat.rows&&this->column s==mat.columns){for(int i=0;i<mat.rows;i++){for(int j=0;j<mat.columns;j++)if(this->element[i][j]!=mat.element[i][j]){break;return false;}elsereturn true;}}}bool XXXMatrix:: operator!=(const XXXMatrix& mat){if(this->rows==mat.rows&&this->column s==mat.columns){for(int i=0;i<mat.rows;i++){for(int j=0;j<mat.columns;j++)if(this->element[i][j]==mat.element[i][j]){break;return false;}elsereturn true;}}}XXXMatrix& XXXMatrix:: operator-( XXXMatrix& mat){if(this->rows==mat.rows&&this->column s==mat.columns){for(int i=0;i<mat.rows;i++)for(int j=0;j<mat.columns;j++)element[i][j]-=mat.element[i][j];return *this;}throw"两个矩阵阶数不同，不能相加"; }XXXMatrix& XXXMatrix::operator-(){for(int i=0;i<this->rows;i++)for(int j=0;j<this->columns;j++)this->element[i][j]=-this->element[i][j];return *this;throw"两个矩阵阶数不同，不能相加"; }XXXMatrix&XXXMatrix::operator*(XXXMatrix& mat){int i,j,k;double s;for(i=0;i<mat.rows;i++)for(j=0;j<mat.columns;j++){s=0.0;for(k=0;k<mat.columns;k++)s=s+mat.element[i][k]*this->element[k][j ];this->element[i][j]=s;}return *this;}XXXMatrix& XXXMatrix::inverse(XXXMatrix& mat){if(this->rows==mat.columns&&this->colu mns==mat.rows)for(int i=0;i<mat.columns;i++)for(int j=0;j<mat.rows;j++)this->element[i][j]=mat.element[j][i];return *this;}int XXXMatrix::rintv(int n,double a[][4]) {int *is,*js,i,j,k,u,v;double p,d;is=(int *) malloc(n*sizeof(int));js=(int *) malloc(n*sizeof(int));for(k=0;k<=n-1;k++){d=0.0;for(i=k;i<=n-1;i++)for(j=k;j<=n-1;j++){p=fabs(a[i][j]);if(p>d){d=p;is[k]=i;js[k]=j;}}if(d+1.0==1.0){free(is);free(js);printf("err**not inv\n");return(0);}if(is[k]!=k)for(j=0;j<=n-1;j++){p=a[k][j];a[k][j]=a[is[k]][j];a[is[k]][j]=p;}if(js[k]!=k)for(i=0;i<=n-1;i++){u=i*n+k;v=i*n+js[k];p=a[i][k];a[i][k]=a[i][js[k]];a[i][js[k]]=p;}a[k][k]=1.0/a[k][k];for(j=0;j<=n-1;j++)if(j!=k){u=k*n+j;a[k][j]=a[k][j]*a[k][k];}for(i=0;i<=n-1;i++)if(i!=k) for(j=0;j<=n;j++)if(j!=k){a[i][j]=a[i][j]-a[i][k]*a[k][j];} for(i=0;i<=n-1;i++)if(i!=k){a[i][k]=-a[i][k]*a[k][k];}}for(k=n-1;k>=0;k--){if(js[k]!=k)for(j=0;j<=n-1;j++){p=a[k][j];a[k][j]=a[js[k]][j];a[js[k]][j]=p;}if(is[k]!=k)for(i=0;i<n-1;i++){p=a[i][k];a[i][k]=a[i][is[k]];a[i][is[k]]=p;}}free(is);free(js);return (1);}XXXMatrix& XXXMatrix::transpose(double a[][4]){int q;q=rintv(4,a);for(int i=0;i<this->rows;i++)for(int j=0;j<this->columns;j++)this->element[i][j]=a[i][j];return *this;}XXXMatrix&XXXMatrix::operator+=(XXXMatrix& mat) {if(this->rows==mat.rows&&this->column s==mat.columns){for(int i=0;i<mat.rows;i++)for(int j=0;j<mat.columns;j++)element[i][j]+=mat.element[i][j];return *this;}throw"两个矩阵阶数不同，不能相加"; }int main(){const int M=3,N=4;static double a[4][4]={{0.5,0.0,0.0,0.0},{0.0,0.5,0.0,0.0},{0.0, 0.0,0.5,0.0},{0.0,0.0,0.0,0.5}};cout<<"待求逆矩阵为:"<<endl;for(int i=0;i<=3;i++){for(int j=0;j<=3;j++)cout<<a[i][j]<<" ";cout<<endl;}XXXMatrix B(3,4),C(4,3),D(4,4);D.transpose(a); //求逆cout<<"求逆后的矩阵为"<<D;int table1[M*N]={1,0,0,0,0,1};XXXMatrix mat1(table1,M,N); //矩阵1mat1.set(2,2,1);B=mat1;int table2[M][N]={1,2,3,4,5,6,7,8,9};XXXMatrix mat2(*table2,M,N); //矩阵2cout<<"矩阵P为"<<endl;cout<<mat1;cout<<"和P相等的矩阵Q为"<<endl; //赋值cout<<B;if(mat1==B)cout<<"两矩阵相等!!!"<<endl; //相等-B;cout<<"负的矩阵-Q"<<endl; //求负值cout<<B;cout<<"矩阵R为"<<endl;cout<<mat2;mat1+=mat2; //+=mat2-mat1;mat2-mat1;//相减cout<<"矩阵相加S=P+R\n"<<mat1;cout<<"矩阵相减R-S-S\n"<<mat2;if(mat1!=B)cout<<"两矩阵不再相等!!!"<<endl; //!=C.inverse(B);cout<<"转置"<<C; //转置return 0;}三、实验成果四、实验体会由于矩阵设计的成员函数较多，操作起来不是很容易，在上机过程中遇到很多困难，但是通过老师指导，经过多次调试，终于成功了。