进行简单功能性查询的简单代码

程序主要是一个查询功能了解cin cin.get cin.getlin cin.get(name,size)cin 进行读取一段字符通过接受回车结束cin.get 能接受包括回车的字符cin.getline 函数的作用是从输入流中读取一行字符,其用法与带3个参数的get函数类似。即
    cin.getline(字符数组(或字符指针), 字符个数n, 终止标志字符)//可以没有终止标志字符程序并没有对这个函数进行测试;cin.get(name,size）作用是在输入流中读取size的字数进行赋值给name测试不适合本程序因为程序的f接受的只是一个字符大材小用；

#include "stdafx.h"

#include<iostream>
#define size 20
using namespace std;
struct bop {
	char name[size];
	char title[size];
	char popname[size];
	int preference;
}pop[4] = { { "wimp","haha","fix",0 },{ "raki","youke","red",1 },{ "celia","lunduan","teacher",2 } ,{ "hopp","yahu","tianwen",3 } };;
int fun(char);
int main()
{
	cout << "输入a显示所有姓名\n输入b显示所有title\n输入c显示别称\n输入d目前不理解\n输入q退出" ;

	char f;

	cin>>f;
	while(f != ‘a‘ && f != ‘b‘ && f != ‘c‘ && f != ‘d ‘ && f != ‘q‘)

	{

		f = cin.get();
		cout << "choice again\n" <<  f << endl;
	}

	while (fun(f))
	{
		cout << "next choice" << endl;
		cin >> f;
	}
    return 0;
}
int fun(char f)
{
	int i;
	switch (f)
	{
	case ‘a‘:cout << "show name" << endl;
		for (i = 0; i<4; i++)
			cout << pop[i].name << endl; break;
	case ‘b‘:cout << "show title" << endl;
		for (i = 0; i<4; i++)
			cout << pop[i].title << endl; break;
	case ‘c‘:cout << "show bopname" << endl;
		for (i = 0; i<4; i++)
			cout << pop[i].popname << endl;
	case ‘d‘:cout << "show prefence" << endl;
		break;
	case ‘q‘:cout << "bye" << endl;
		return 0; break;
	}
	return 1;
}进行多次更改，刚开始无法解决输出流重复输出cout的文件经过分析发现cin.get不仅接受了字符还接受了回车才导致进行了两次循环