用C++实现的一个简单的矩阵类（转载）
代码是转载的，粗略看了下，原始代码有内存泄露问题，于是修改了一下destructor，修改后的程序，通过了g++ -4.2 编译，应该没问题了。
有兴趣的同学可以下载。 :-)
/* code started. */
int Pow(int x, int y){
int ans, b, i;
if(y == 0)return 1;
if(y < 0) return -1;
for(i = y, b = 0; i > 0; i >>= 1) b++;
for(ans = 1, i = b - 1; i >= 0; i--){
if(y & (1 << i)){
ans = ans * ans * x;
}else{
ans = ans * ans;
}
}
return ans;
}
#include
#include
using namespace std;
template
class Matrix{
//class Matrix by felix021
//usage: Matrix
//如果typename不是short/int/long/long long
//那么取模运算需要包含math.h头文件(使用fmod函数)
//取模运算的代码仅在 + - * 三个函数内.
public:
T **data;
Matrix *temp;
int row, col, modnum;
//无参构造函数，没有给定矩阵的大小时，输出错误提示
Matrix(){
printf("Size parameter invalid!/n");
}
//构造函数， r是行数，c是列数，若给定m>0, 则在求和/差/积时对m取模
Matrix(int r, int c, int m = 0){
int i;
temp = NULL;
row = r, col = c;
data = new T *[r]; //分配一个 T *[r]类型的数组给data
for(i = 0; i < r; i++){
data[i] = new T [c]; //给data下的每个指针分配内存, 类型为 T [c]
}
if(m > 0) modnum = m;
else modnum = 0;
}
//析构函数，用于释放分配的内存
~Matrix(){
int i;
if(temp != NULL){delete temp; }
// modified by zkdkeen. guarantee no memory leakage.
for(i = 0; i < row; i++) delete [] data[i];
delete [] data;
}
//赋值，给第i行第j列的元素赋值为 a
void assign(int i, int j, const T a){
if(i < 0 || i >= row || j < 0 || j >= col) {
printf("Invalid row/column number./n");
return;
}else{
data[i][j] = a;
}
}
//取出元素，给第i行第j列的元素的值
T at(int i, int j){
if(i < 0 || i >= row || j < 0 || j >= col) {
printf("Invalid row/column number./n");
return;
}else{
return data[i][j];
}
}
//重载<<运算符，可用cout输出
friend inline ostream & operator<<(ostream &os, const Matrix &a){
int i, j;
for(i = 0; i < a.row; i++){
for(j = 0; j < a.col - 1; j++){
os << a.data[i][j] << " ";
}
os << a.data[i][j];
os << endl;
}
return os;
}
//重载>>运算符，可用cin输入
friend inline istream & operator>>(istream &is, const Matrix &a){
int i, j;
for(i = 0; i < a.row; i++){
for(j = 0; j < a.col; j++){
is >> a.data[i][j];
}
}
return is;
}
//重载=运算符，要求两个矩阵的大小相同
Matrix & operator = (const Matrix &a){
int i, j;
if(row != a.row || col != a.col){
printf("Unmatch Matrix!/n");
return *this;
}
for(i = 0; i < row; i++){
for(j = 0; j < col; j++){
data[i][j] = a.data[i][j];
if(modnum > 0){
data[i][j] = mod(data[i][j]);
}
}
}
return *this;
}
//重载+运算符，要求两个矩阵的大小相同
Matrix & operator + (const Matrix &a){
int i, j;
if(row != a.row || col != a.col){
printf("Unmatch Matrix!/n");
return *this;
}
if(temp != NULL) delete temp;
temp = new Matrix(row, col, modnum);
for(i = 0; i < row; i++){
for(j = 0; j < col; j++){
temp->data[i][j] = data[i][j] + a.data[i][j];
if(modnum > 0){
temp->data[i][j] = mod(temp->data[i][j]);
}
}
}
return *temp;
}
//重载-运算符，要求两个矩阵的大小相同
Matrix & operator - (const Matrix &a){
int i, j;
if(row != a.row || col != a.col){
printf("Unmatch Matrix!/n");
return *this;
}
if(temp != NULL) delete temp;
temp = new Matrix(row, col, modnum);
for(i = 0; i < row; i++){
for(j = 0; j < col; j++){
temp->data[i][j] = data[i][j] - a.data[i][j];
if(modnum > 0){
temp->data[i][j] = mod(temp->data[i][j] + modnum);
}
}
}
return *temp;
}
//重载*运算符，要求矩阵a的列数等于b的行数
Matrix & operator * (const Matrix &a){
int i, j, k;
T tmp;
if(col != a.row){
printf("Unmatch Matrix!/n");
return *this;
}
if(temp != NULL) delete temp;
temp = new Matrix(row, a.col, modnum);
for(i = 0; i < row; i++){
for(j = 0; j < a.col; j++){
tmp = 0;
for(k = 0; k < a.row; k++){
tmp += data[i][k] * a.data[k][j];
if(modnum > 0){
tmp = mod(tmp);
}
}
temp->data[i][j] = tmp;
}
}
return *temp;
}
//重载^运算符，要求矩阵的列数等于行数
Matrix & operator ^ (const int a){
int i, j;
if(row != col){
printf("No n*n matrix!/n");
return *this;
}
if(a < 0){
printf("Invalid a(%d)!/n", a);
return *this;
}
if(temp != NULL) delete temp;
temp = new Matrix(row, col, modnum);
for(i = 0; i < row; i++){ //单位方阵
for(j = 0; j < col; j++){
if(i == j )temp->data[i][j] = 1;
else temp->data[i][j] = 0;
}
}
for(i = a, j = 0; i > 0; i >>= 1) j++;
for(i = j - 1; i >= 0; i--){
if(a & (1 << i)){
*temp = *temp * *temp * (*this);
}else{
*temp = *temp * *temp;
}
}
return *temp;
}
//重载求模函数，对int, long, long long, float, double都有效
int mod(int i){
return i % modnum;
}
long mod(long i){
return i % modnum;
}
long long mod(long long i){
return i % modnum;
}
float mod(float i){
return fmod(i, modnum);
}
double mod(double i){
return fmod(i, modnum);
}
};
int main(){
int i, j;
Matrix
cout << "(cin/cout test)/nPlease input 2 (2*2) Matrix: /n";
cin >> t;
cout << t << endl;
t1 = t * t * t * t * t;
t2 = t ^ 5;
cout << "t^5 = /n" << t1 << endl << t2 << endl;
getchar();
for(i = 0; i < 5; i++){
for(j = 0; j < 10; j++){
a.assign(i, j, 1);
b.assign(j, i, 2);
}
}
c = a * b;
cout << "Matrix a:" << endl << a << endl;
cout << "Matrix b:" << endl << b << endl;
cout << "Matrix a * b:" << endl << c << endl;
// getchar();
return 0;
}
/* code ended. */