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; } }
//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"; }
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(); }
c++primerplus(第六版)编程题——第6章（分支语句和逻辑运算符）,布布扣,bubuko.com
c++primerplus(第六版)编程题——第6章（分支语句和逻辑运算符）
原文：http://www.cnblogs.com/bramely/p/3853092.html