变量x的值为5，x的地址为3000，若欲使point为指向x的指针变量，则下列赋值正确的是（ ）。 A: &x=5 B: *point=5 C: *point=3000 D: point=&x

8.5 阅读程序,写出运行结果 class Point{ int x=1; static int y=2; public void show(){System.out.println(x+","+y); }} public class tempDemo{ public static void main(String []args){ Point p1=new Point();p1.show(); p1.x=2;p1.y=3; Point p2=new Point();p2.show(); p1.show(); p1.y=5; p2.show(); }}

A point estimator will be unbiased if the A: expected value of the point estimator equals the value of the population parameter B: sample size is greater than 30 or np ³ 5 and n(1-p) ³ 5 C: sampling distribution is normally distributed D: value of the population parameter is known

盘点: break point|game point|match point|set point

【单选题】若有语句 int *point,a=4; 和 point=&a; 下面均代表地址的一组选项是 A. a,point,*&a B. &*a,&a,*point C. *&point,*point,&a D. &a,&*point ,point

【单选题】设Point为已定义的类,下面声明Point对象a语句正确的是() A. Point a=Point(); B. public Point a; C. Point a=new Point(); D. public Point a=new Point();

POP广告是（ ）。 A: point of product B: point of promotion C: point of purchasing D: point of point

下面为一个点类，重载了负号运算符，将原来点改为其关于原点的对称点。 请填空补充完整程序。 #include &#91;iostream.h&#93;[br][/br] class Point{ private: float x; float y; public: Point(){ x = 0; y = 0; } Point(float a, float b){ x = a; y = b; } // 显示坐标 void display() { cout << "X: " << x << " Y:" << y <<endl; } // 重载负运算符 -，取对称点 friend Point operator-(Point a) { Point r; （1） ； （2） ； return r; } }; void main() { Point p1(5, 6); ； // 取对称点 p1.display(); //p1中的x和y值分别为-5和-6 }

题 (1.0分) 题号:705 难度:中 第1章[br][/br]若有语句int *point,a=4;和 point=&a;下面均代表地址的一组选项是()。 A: a,point,*&a B: &*a,&a,*point C: *&point,*point,&a D: &a,&*point,point

As we were ( ) giving up hope, a letter arrived. A: in the point of B: to the point of C: on the point of D: for the point of