C++ 友元函数问题求这道题目的代码题目5（30分）：定义类X,Y,Z，使之满足以下几个条件：类X有么一个私有成员i，类

#include
class POINT
{
public;
Point(double a,double b){ x=a;y=b};
fiend double square(Point &a,Point &b);
private:
double x,y;
};
double square(Point &a,Point &b) { return sqr((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))};
void main()
{
Point p1(x1,y1),p2(x2,y2);
double l;
l=square(p1,p2);
}

1，编写一个程序，要求输入一个整数，并求出它的所有因子。
int * fnGetDivisor(int source)
{
int result[source];
memset(result,0,source);
for(int i =1,j=0;i

不用友元函数的话,可以使类B成为类A的友元类.然后类B中重载+操作符成员函数.
class A
{
...
public:
friend class B;
}
class B
{
...
public:
operator + (const A &a);
...
}

帮你实现，1,2,3 可否？ 后面的 4.3就简单了

哎，，，叫你求助我又不求助，无语其实我早写好了我的重载为-，如果你改+，就把之前的重载删除，就可以了

#include
#define MAX 10
using namespace std;
class matrix;
matrixoperator-(matrixb,const matrixc);
class matrix
{
private:
int row,col,(*a)[MAX];
public:
matrix(){}
matrix(int r,int c,int (*b)[MAX]):row(r),col(c),a(b){}
matrixoperator+(const matrixb)
{
for(int i=0;ifor(int j=0;ja[i][j]=a[i][j]+b.a[i][j];
return *this;
}
friend matrixoperator-(matrixb,const matrixc)
{
for(int i=0;ifor(int j=0;j b.a[i][j]=b.a[i][j]-c.a[i][j];
return b;
}
void show()
{
for(int i=0;i{
for(int j=0;jcout<cout<}
}
};

void main()
{
int row =2 ,col =3,a[2][MAX]={{1,2,3},{4,5,6}};
matrix mat1(row,col,a);
int b[2][MAX]={{4,3,2},{1,2,3}};
matrix mat2(row,col,b);
matrix mat3;
mat3 = mat1 + mat2;
cout<<"mat3"<mat3.show();
cout<<"mat1"<mat1.show();
cout<<"mat4"<matrix mat4;
mat4 = mat1 - mat2;
mat4.show();
}

Java的行不,C++的不记得了.
public class Circle
{
private static double PI=3.14
private double radius;
public Circle(double radius)
{
this.radius=radius;
}
public double Area()
{
return PI*radius*radius;
}
public double Perimeter()
{
return 2*PI*radius;
}
}

A错,友元函数不是成员函数,但要在类中声明

#include
class Circle
{
public:
Circle():radius(0){}
Circle(double r):radius(r){}
const double area() const
{
return M_PI * radius * radius;
}
const double perimeter() const
{
return 2 * M_PI * radius;
}
friend double area(Circle);
friend double perimeter(Circle);
private:
double radius;
};
double area(Circle c)
{
return M_PI * c.radius * c.radius;
}
double perimeter(Circle c)
{
return 2 * M_PI * c.radius;
}