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All of the following may hinder a person's ability to transmit/receive a message except:
A Unfamiliarity with topic
B Withholding information
C Assuming
D Noise
E Preoccupation
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使用VC6打开考生文件夹下的工程test28_3。此工程包含一个test28_3.cpp,其中定义了类Person,但该类的定义并不完整。请按要求完成下列操作,将程序补充完整。(1)定义类Person的私有数据成员forename、surname 和 money,forename 和 surname都是char型的指针数据,money是double型的数据。请在注释“//**1**”之后添加适当的语句。(2)完成类Person的带三个参数的构造函数Person(char *f, char *s,double m),分别为forename和surname申请新的空间来存储参数f和s指针指向的内容,注意空间的大小,最后把参数m的值赋给money,请在注释“//**2**”之后添加适当的语句。(3)完成类Person的析构函数的定义,把forename和surname指向的空间释放,请在注释“//**3**”之后添加适当的语句。(4)完成类Person的成员函数display的定义,使其以格式“forname surname has money”的形式输出内容,请在注释“//**4*。”之后添加适当的语句。注意:除在指定位置添加语句之外,请不要改动程序中的其他内容。输出结果如下:Richard Berk has 1000.56源程序文件test28_3.cpp清单如下:include <iostream.h>include <string.h>class Person{private://** 1 **double money;public:Person(char *f, char *s, double m);~Person();void display();};Person::Person(char *f, char *s, double m){//** 2 **strcpy(forename, f);surname = new char[strlen(s)+1];strcpy(surname, s);money=m;}Person: :-Person (){//** 3 **}void Person:: display ( ){//** 4 **}void main ( ){Person p("Richard", "Berk", 1000.56);p.display ( );}
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A.receive;accept;choice
B.receive;choice;accept
C.accept:receive;choice
D.accept;choice;receive
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