c++primerplus(第六版)编程题——第4章（复合类型）

声明：作者为了调试方便，每一章的程序写在一个工程文件中，每一道编程练习题新建一个独立文件，在主函数中调用,我建议同我一样的初学者可以采用这种方式，调试起来会比较方便。

(具体方式参见第3章模板)

1.编写一个程序，如下述输出示例所示的那样请求显示信息：（注意：该程序应该接受的名字包含多个单词，另外，程序将向下调整成绩，即向上调一个字母。假设用户请求A、B 或C，所以不用担心D和F之间的空档。）

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

void cprimerplus_exercise4_1()
{
string firstname, lastname;
char lettergrade;
int age;
cout << "What is you first name?";
getline( cin, firstname);

cout << "What is your last name?";
getline(cin ,lastname);

cout <<  "What letter grade do you deserve?";
cin >> lettergrade;

cout << "What is your age?";
cin >> age;

cout << "Name:" << lastname << "," << firstname << endl;
cout << "Grade:"<< char (lettergrade + 1) << endl;
cout << "Age:" << age << endl;
}

2.修改程序清单4.4，使用C++ string类而不是char数组。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
void cprimerplus_exercise4_2()
{
string name;
string dessert;
cout << "Enter your name:" << endl;
getline( cin, name);

cout << "Enter your favourite dessert:" << endl;
getline(cin ,dessert);

cout << "I have some delicious " << dessert << " for you," << name << endl;

}

3.编写一个程序，它要求用户首先输入其名，然后输入其姓；然后程序使用一个逗号和空格将姓和名组合起来，并储存和显示结果。请使用char数组和头文件cstring中的函数。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
void cprimerplus_exercise4_3()
{
char firstname[20];
char lastname[20];
cout << "Enter your first name:";
cin.getline( firstname, 20);

cout << "Enter your last name:";
cin.getline(lastname, 20);

cout << "Here is the information in a single string:" << strcat(strcat( firstname,","), lastname ) << endl;
}

4.编写一个程序，它要求用户首先输入其名，然后输入其姓；然后程序使用一个逗号和空格将姓和名组合起来，并储存和显示结果。请使用string对象和头文件string中的函数。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
void cprimerplus_exercise4_4()
{
string firstname;
string lastname;

cout << "Enter your first name:";
getline(cin, firstname);

cout << "Enter your last name:";
getline(cin, lastname);

cout << "Here is the information in a single string:"\
<<lastname + "," + firstname << endl;
}

5.结构CandyBar包含3个成员。第一个成员储存了糖块的品牌；第二个成员储存糖块的重量（可以有小数）；第三个成员储存了糖块的卡路里含量（整数）。请编写一个程序，声明这个结构，创建一个名为snack的CandyBar变量。并将其成员分别初始化为”Mocha Munch”、2.3和350。初始化在声明snack是进行。最后，程序显示变量snack变量的内容。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
void cprimerplus_exercise4_5()
{
struct CandyBar
{
char name[20];
double weight;
int calary;
};

CandyBar snack = {
"Mocha Munch",
2.3,
350
};
cout << "Name:" << snack.name << endl;
cout << "Weight:" << snack.weight << endl;
cout << "calary:" << snack.calary << endl;
}

6.结构CandyBar包含3个成员，如编程5所示，请编写一个程序，创建一个包含3个元素的CandyBar数组，并将他们初始化为所选择的值，然后显示每个结构的内容。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

void cprimerplus_exercise4_6()
{
struct CandyBar
{
char name[20];
double weight;
int calary;
};

CandyBar snack[3] = {
{"Mocha Munch", 2.3, 350},\
{"Mocha Munch", 2.3, 350},\
{"Mocha Munch", 2.3, 350},\
};
for (int i = 0; i < 3; i++)
{
cout << "Name:" << snack[i].name << endl;
cout << "Weight:" << snack[i].weight << endl;
cout << "calary:" << snack[i].calary << endl;
cout << endl;
}

}

7. William Wingate从事披萨饼分析服务。对于每个披萨饼，他都需要记录下列信息：
（1）披萨饼公司的名称，可以有多个单词组成。
（2）披萨饼的直径。
（3）披萨饼的重量。
请设计一个能够储存这些信息的结构，并编写一个使用着这种结构变量的程序。程序将请求用户输入上述信息，然后显示这些信息。请使用cin（或其他方法）和cout。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

void cprimerplus_exercise4_7()
{
struct pizzamessage{
string name;
double diameter;
double weight;
};

pizzamessage message;
cout << "Please input the message of pizza!"<<endl;
cout << "Name:";
getline( cin, message.name);

cout <<"Diameter:";
cin >> message.diameter;

cout << "Weight:";
cin >> message.weight;

cout << "the message of the record is:"<< endl;
cout << "Name:" << message.name;
cout << "Diameter:" << message.diameter;
cout << "Weight:" << message.weight;
}

8.完成编程7，但使用new来为结构分配内存，而不是声明一个结构变量。另外，让程序在请求输入披萨饼公司名称之前输入披萨饼的直径。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

void cprimerplus_exercise4_8()
{
struct pizzamessage{
string name;
double diameter;
double weight;
};

int num;
cout << "How many pieces of pizzamessage you want to be record?";
cin >> num;
cin.get();

pizzamessage *message = new pizzamessage[num];

for (int i = 0; i < num; i++)
{
cout << "Please input the pizza message of the " << i + 1 << "th record:"<< endl;
cout << "Name:";
getline(cin,message[i].name);

cout <<"Diameter:";
cin >> message[i].diameter;

cout << "Weight:";
cin >>message[i].weight;
cout << endl;

cin.get();

}

for (int i = 0; i < num; i++)
{
cout << "the message of the "<< i+1 << "record is:"<< endl;
cout << "Name:" << message[i].name << endl;
cout << "Diameter:" << message[i].diameter << endl;
cout << "Weight:" << message[i].weight << endl;
cout << endl;
}

}

9.完成编程6，但使用new来动态分配数组，而不是声明一个包含3个元素的CandyBar数组。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

void cprimerplus_exercise4_9()
{
struct CandyBar
{
char name[20];
double weight;
int calary;
};

cout << "How many do you want to create?";
int num;
cin >> num;
cin.get();

CandyBar *candies = new CandyBar[num];

for (int i = 0; i < num; i++ )
{
cout << "please input the message:" << endl;
cout << "name:";
cin >> candies[i].name;

cout << "weight:";
cin >> candies[i].weight;

cout << "calary:";
cin >> candies[i].calary;
}
cout << endl;

for (int i = 0; i < num; i++)
{
cout << "the message:" << endl;
cout << "name:" << candies[i].name << endl;
cout << "weight:" << candies[i].weight << endl;
cout << "calary:" << candies[i].calary << endl;

cout << endl;
}
}

10. 编写一个程序，让用户输入三次40码跑的成绩（也可以让用户输入），并显示次数和平均成绩。请使用一个array对象来储存数据（也可以使用数组）。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

void cprimerplus_exercise4_10()
{
cout << "How many times do you want to record?"<< endl;
int num;
cin >> num;

double * times = new double[num];

for (int i = 0; i < num; i++)
{
cout << i+1 << " th input:" << endl;
cin >> times[i];
}

double sum = 0;
for (int i = 0; i < num; i++)
{
sum +=times[i];
}

cout << "you have " << num << "records" << endl;
cout << " your average score is " << sum / num << endl;
}