下面程序的输出结果是【】。include include class point{double x; double y;下面程序的输出结果是【】。include ＜iostream.h＞include ＜math.h＞class point{double x;double y;public:point(double a, double b){x=a;y=b;}friend double distance(point a, point b) ;};double distance(point a, poin

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下面程序的输出结果是【】。include include class point{double x; double y;

下面程序的输出结果是【】。

include ＜iostream.h＞

include ＜math.h＞

class point

{

double x;

double y;

public:

point(double a, double b)

{

x=a;

y=b;

}

friend double distance(point a, point b) ;

};

double distance(point a, point b)

{

return sqrt ((a. x-b.x) * (a. x-b.x)+ (a. y-b. y) * (a. y-b. y) );

}

void main()

{

point p1(1,2);

point p2(5,2);

cout＜＜distance(p1,p2)＜＜end1;

}

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2.下面程序的输出结果是 ______。includevoidmain(){intx=20;int&y=x;cout下面程序的输出结果是 ______。 #include＜iostream.h＞ void main(){ int x=20； int &y=x； cout＜＜(x==y)＜＜“”； cout＜＜(&x==&y)＜＜end1： }A．20 20 20 20B．20 20C．1 1D．出错

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4.下面程序的输出结果是()。includeincludeusing namespace std;class point{p下面程序的输出结果是( )。 #include＜iostream＞ #include＜math.h＞ using namespace std; class point { private: double x; double y; public: point(double a,double b) { x=a; y=b; } friend double distances(point a,point b); }; double distances(point a,point b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } int main() { point p1(1,2); point p2(5,2); cout＜＜distances(p1,p2)＜＜end1; return 0; }A．2B．4C．8D．16

参考答案和解析

正确答案：4
4 解析：本题考核友元函数的应用。分析程序：类point中定义了两个私有成员x和y，以及一个友元函数distance。从而，函数 distance可以访问类point中的任何成员。在函数 distance中，返回值为sqrt((a. x-b. x)*(a. x-b. x)+ (a. y-b. y)*(a. y-b. y))。由此可知，函数distance的功能是计算a、b两点之间的距离。在土函数main中，先定义两点：p1(1，2)和p2(5，1)。然后调用函数distance计算两点之间的距离为4，所以程序最后输出为4。

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