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.编写一个小程序，读取键盘输入，直到遇到@符号为止，并回显输入（数字除外），同时将大写字符转换为小写，将小写字符转换为大写（别忘了cctype函数系列）。
#include <iostream>
#include <cctype>
using namespace std;
void cprimerplus_exercise_6_1()
{
char letter;
cout << "Please input letters:";
cin >> letter;
cin.get(); while ( letter != '@')
{
if (isdigit(letter))
{
cin.get(letter);
}
else
{
if (islower(letter))
{
letter = toupper(letter);
}else if (isupper(letter))
{
letter = tolower(letter);
} cout << letter;
cin >> letter;
cin.get();
} } }
2.编写一个程序，最多将10个donation值读入到一个double数组中（如果你愿意，也可以使用模板类array）。程序遇到非数字输入时将结束输入，并报告这些数字的平均值以及数组中有多少个数字大于平均值。
#include <iostream>
#include <cctype>
using namespace std;
void cprimerplus_exercise_6_2()
{
double donation[10];
double sum = 0.0;
double average = 0.0;
double tmp;
int i = 0;
int cnt = 0; while ( cin >> tmp && i < 10)
{
donation[i] = tmp;
sum +=donation[i];
++i;
} if ( i != 0)
{
average = sum / i;
} for (int j = 0; j < i; j++)
{
if (donation[j] > average)
{
++cnt;
}
} cout << "The average is:" << average << endl;
cout << "There are " << cnt << " numbers are above the average!"<< endl; }
3.编写一个菜单驱动程序的雏形。该程序显示一个提供4个选项的菜单——每个选项用一个字母标记。如果用户使用有效选项之外的字母进行响应，程序将提示用户输入一个有效的字母，直到用户这样做为止。然后，该程序使用一条switch语句，根据用户的选择执行一个简单的操作。
#include <iostream> using namespace std;
void cprimerplus_exercise_6_3()
{
cout << "Please enter one of the following choices:" << endl\
<<"c) carnivore \tp) pianist \nt) tree \tg) game" << endl; cout << "Please enter a c, p, t, or g:"; char letter;
cin >> letter; while ( letter != 'c' && letter != 'p' && letter != 't' && letter != 'g')
{
cout << "Please enter a c, p, t, or g:";
cin >> letter;
}
switch (letter)
{
case 'c':
cout << "A maple is a carnivore";
break;
case 'p':
cout << "A maple is a pianist";
case 't':
cout << "A maple is a tree";
break;
case 'g':
cout << "A maple is a game";
break;
default:
break;
}
}
4. 加入Benevolent Order of Programmer后，在BOP大会上，人们便可以通过加入者的真实姓名、头衔或秘密BOP姓名来了解他，请编写一个程序，可以使用真实姓名、头衔、秘密姓名或成员偏好来列出成员。编写该程序时，请使用下面结构
//Benevolent Order of Programmers name structure
Struct bop
{
char fullname[strsize];
char title[strsize];
char bopname[strsize];
int preference;
};
该程序创建一个由上述结构组成的小型数组，并将其初始化为适当的值，另外，该程序使用一个循环让用户在下面的选项中进行选择：
A. Display by name B. Display by title C. Display by bopname
D. Display by preference Q. quit
#include <iostream> using namespace std;
void cprimerplus_exercise_6_4()
{
const int strsize = 20;
const int mennum = 3;
struct bop
{
char fullname[strsize]; //real name
char title[strsize]; //job title
char bopname[strsize]; //secret BOP name
int preference; // 0 = fullname, 1 = title, 2 = bopname
}; bop member[mennum] = {
{ "thm", "leader", "sb", 0},
{ "cgf", "sb", "sb", 1},
{ "th", "dsb", "ssb", 2} };
cout << "Enter your choice!";
char ch;
cin.get(ch);
while (cin >> ch && ch != 'q')
{
switch (ch)
{
case 'a':
for (int i = 0; i < mennum; i++)
cout << member[i].fullname << endl;
break;
case 'b':
for (int i = 0; i < mennum; i++)
cout << member[i].title << endl;
break;
case 'c':
for (int i = 0; i < mennum; i++)
cout << member[i].bopname << endl;
break;
case 'd':
for (int i = 0; i < mennum; i++)
{
if (member[i].preference == 0)
{
cout << member[i].fullname << endl;
}else if (member[i].preference = 1)
{
cout << member[i].title << endl;
}else if (member[i].preference = 2)
{
cout << member[i].bopname << endl;
}
}
break;
default:
break;
}
cout << "Next choice:";
}
cout << "Bye!\n"; }
5. 在neutronia王国，货币单位是tvarp，收入所得税的计算方式如下：
5000 tvarps：不收税
5001~15000 tvarps：10%
15001~35000 tvarps：15%
35000 tvarps以上：10%
如：收入38000 tvarps，所得税：5000*0.0+10000*0.1+20000*0.15+3000*0.2；
#include <iostream> using namespace std;
void cprimerplus_exercise_6_5()
{
cout << "please input your income:";
double income, revenue;
while (cin >> income && income >= 0)
{
if (income <= 5000)
{
revenue = 0.0;
}else if ( income > 5000 && income <= 15000)
{
revenue = (income - 5000) * 0.1;
}else if( income > 15000 && income < 35000)
{
revenue = 5000 * 0.00 + 10000 * 0.10 + (income -20000) * 0.15;
}else if( income > 35000)
{
revenue = 5000 * 0.00 + 10000 * 0.10 + + 20000 * 0.15 + (income -35000) * 0.15;
} cout << "your revenue is:" << revenue << endl;
cout << "please enter your income:"; }
}
6.编写一个程序，记录捐助给“维护合法权利团体”的资金，该程序要求用户输入捐献者数目，然后要求用户输入每一个捐献者的姓名和款项，这些信息都被存储在一个动态分配的结构数组中，每个结构有两个成员；用来储存姓名的字符数组（或string对象）和用来存储款项的double成员。读取所有的数据后，程序将显示所有捐款超过10000的捐款者的姓名以及捐款的数额。该列表前应包含一个标题，指出下面的捐款者是重要捐款人（Grand Patrons）。然后，程序将列出其他的捐款者，该列表要以Patrons开头。如果某类别没有捐款人，则程序将打印单词“none”。该程序只显示这两种类别，而不进行排序。
#include <iostream>
#include <string> using namespace std;
void cprimerplus_exercise_6_6()
{
int num;
cout << "please input the donate num:";
cin >> num;
cin.get(); struct patron
{
string name;
double money;
}; patron *ps = new patron[num]; for (int i = 0; i < num; ++i)
{
cout <<"please input the "<< i+1 <<"th patron name:";
getline(std::cin, ps[i].name);
cout << "please input the " << i+1 << "th patron money:";
cin >> ps[i].money;
cin.get();
}
int cnt = 0, snt = 0;
cout << "Grand Patrons:" << endl ;
for (int i = 0; i < num; i++)
{
if (ps[i].money >= 10000)
{
cout << ps[i].name << '\t' <<ps[i].money << endl;
++cnt;
} }
if ( cnt == 0)
{
cout << "none";
} cout << "\nPatrons:" << endl;
for (int i = 0; i < num; i++)
{
if (ps[i].money < 10000)
{
cout << ps[i].name << '\t' << ps[i].money << endl;
++snt;
} }
if ( snt == 0)
{
cout << "none";
} delete []ps; }
7.编写一个程序，他每次读取一个单词，直到用户只输入q。然后，该程序指出有多少单词以元音打头，有多少个单词以辅音打头，还有多少单词不属于这两类。为此，方法之一是使用isalpha()来区分以字母和其他字符打头的单词，然后对于通过salpha()测试的单词，使用if或switch语句来确定哪些是以元音打头。
#include <iostream>
#include <string>
#include <ctype.h> using namespace std;
void cprimerplus_exercise_6_7()
{
string word;
cout << "Enter words (q to quit):";
int cnt = 0;
int vowel = 0, constant = 0, other = 0;
while (cin >> word && !word.empty())
{
if(isalpha(word[0]))
{
if (word[0] == 'q' && word.length() == 1)
break;
else if( word[0] == 'a' || word[0] == 'e' || word[0] == 'i' || word[0] == 'o' || word[0] == 'u')
{
++vowel;
}else
{
++constant;
}
}else
{
++other;
}
} cout << vowel << " words beginning with vowels "<< endl;
cout << constant <<" words beginning with constants " << endl;
cout << other <<" others" << endl;
}
8.编写一个程序，它打开一个文件夹，逐个字符的读取该文件，直到到达文件末尾，然后指出该文件中包含多少个字符。
#include <iostream>
#include <fstream>
#include <cstdlib> using namespace std;
void cprimerplus_exercise_6_8()
{
char ch;
int sum = 0;
ifstream inFile;
inFile.open("1.txt");
if( !inFile.is_open())
{
cout << "Could not open the file!\n";
cout << "Program terminating.\n";
exit(EXIT_FAILURE);
} inFile >> ch;
while (inFile.good())
{
++sum;
inFile >> ch;
}
if (inFile.eof())
{
cout << "End of file reached.\n";
}else if(inFile.fail())
{
cout << "Input terminated by data mismatch.\n";
}else
{
cout << "Input terminated for unknown reason.\n";
}
cout << "there are " << sum << " chars in this file!\n";
}
9.完成编程练习6，但从文件中读取所需的信息。该文件的第一项应为捐款人数，余下的内容应为成对的行。在每一对中，第一行为捐款人姓名，第二行为捐款数额。及该文件类似于下面：
4
Sam Stone
2000
Freida Flass
10050
Tammy Tubbs
5000
Rich Raptor
5500
#include <iostream>
#include <fstream>
#include <string>
#include <cstdlib>
using namespace std;
const int SIZE = 60; void cprimerplus_exercise_6_9()
{
char filename[SIZE];
ifstream inFile;
cout << "Enter the name of the file: ";
cin.getline(filename,SIZE);
inFile.open(filename);
if (!inFile.is_open())
{
cout << "Could not open the file" << filename << endl;
cout << "Program terminating.\n";
exit(EXIT_FAILURE);
} struct patron
{
string name;
double money;
}; int num, cnt = 0, snt = 0;
inFile >> num;
inFile.get(); patron *ps = new patron[num];
for (int i = 0; i < num; i++)
{
getline(inFile, ps[i].name);
inFile >> ps[i].money;
inFile.get();
}
cout << "Grand patrons:\n";
for (int i = 0; i < num; i++)
{
if (ps[i].money >= 10000)
{
cout << ps[i].name << '\n' << ps[i].money << endl;
++cnt;
}
} if ( cnt == 0)
{
cout << "none";
} cout << "\nPatrons:" << endl;
for (int i = 0; i < num; i++)
{
if (ps[i].money < 10000)
{
cout << ps[i].name << '\t' << ps[i].money << endl;
++snt;
} }
if ( snt == 0)
{
cout << "none";
} delete []ps; inFile.close();
}