以下程序由 Teddy van Jerry （我自己）编写并运行，基本保证正确性。（有时可能会为优化程序超前使用某些内容）
Contents
Before we comb through the codes
这一章内容仍然非常好理解，我学习花了一天半（也不是一直在学C++）。主要原因是内容在之前大多已经出现过，因而再次介绍时就尤其轻松了。
Review:
【C++ Primer(5th Edition) Exercise】练习程序 - Chapter1（第一章）
【C++ Primer(5th Edition) Exercise】练习程序 - Chapter2（第二章）
【C++ Primer(5th Edition) Exercise】练习程序 - Chapter3（第三章）
【C++ Primer(5th Edition) Exercise】练习程序 - Chapter4（第四章）
TIP:标有(C++/11)者为C++/11标准下可以使用的题，若为老版本应加以修改。
Exercise 5.3
#include <iostream>
using namespace std;
int main()
{
int sum = 0, val = 1;
while (val <= 10)
sum += val, ++val; // equivalent to sum += val, val++;
cout << sum << endl;
return 0;
}
其实，也可以简写成这样：
#include <iostream>
using namespace std;
int main()
{
int sum = 0, val = 1;
while (val <= 10)
sum += (val++);
cout << sum << endl;
return 0;
}
Exercise 5.5
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
unsigned short score;
while (cin >> score && score <= 100)
{
string grade;
vector<string> vec{
"F","D","C","B","A","A++" };
if (score < 60) grade = vec[0];
else
{
grade = vec[(score - 50) / 10];
if (score % 10 < 3 && score != 100)
grade += "-";
else
{
if (score % 10 > 7)
grade += "+";
}
}
cout << "The grade is " << grade << endl;
}
return 0;
}
注意：第17行score != 100很重要，要不然输入100会输出A++-的。
Exercise 5.6
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
unsigned short score;
while (cin >> score && score <= 100)
{
string grade;
vector<string> vec{
"F","D","C","B","A","A++" };
grade = (score < 60) ? vec[0] : vec[(score - 50) / 10];
grade += (score % 10 > 7) ? "+" : ((score % 10 < 3 && score != 100) ? "-" : "");
cout << "The grade is " << grade << endl;
}
return 0;
}
虽然比5.5简洁，但可读性大大下降。
Exercise 5.9
#include <iostream>
using namespace std;
int main()
{
char ch;
unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0;
while (cin >> ch)
{
if (ch == 'a') aCnt++;
if (ch == 'e') eCnt++;
if (ch == 'i') iCnt++;
if (ch == 'o') oCnt++;
if (ch == 'u') uCnt++;
}
cout << "The number of 'a' is " << aCnt << ".\n";
cout << "The number of 'e' is " << eCnt << ".\n";
cout << "The number of 'i' is " << iCnt << ".\n";
cout << "The number of 'o' is " << oCnt << ".\n";
cout << "The number of 'u' is " << uCnt << ".\n";
return 0;
}
Exercise 5.10
Way I:
#include <iostream>
using namespace std;
int main()
{
char ch;
unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0;
while (cin >> ch)
{
switch (ch)
{
case 'a': case 'A':
++aCnt;
break;
case 'e': case 'E':
++eCnt;
break;
case 'i': case 'I':
++iCnt;
break;
case 'o': case 'O':
++oCnt;
break;
case 'u': case 'U':
++uCnt;
break;
}
}
cout << "The number of 'a' is " << aCnt << ".\n";
cout << "The number of 'e' is " << eCnt << ".\n";
cout << "The number of 'i' is " << iCnt << ".\n";
cout << "The number of 'o' is " << oCnt << ".\n";
cout << "The number of 'u' is " << uCnt << ".\n";
return 0;
}
Way II:
#include <iostream>
using namespace std;
int main()
{
char ch;
unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0;
while (cin >> ch)
{
ch = tolower(ch);
switch (ch)
{
case 'a':
++aCnt;
break;
case 'e':
++eCnt;
break;
case 'i':
++iCnt;
break;
case 'o':
++oCnt;
break;
case 'u':
++uCnt;
break;
}
}
cout << "The number of 'a' is " << aCnt << ".\n";
cout << "The number of 'e' is " << eCnt << ".\n";
cout << "The number of 'i' is " << iCnt << ".\n";
cout << "The number of 'o' is " << oCnt << ".\n";
cout << "The number of 'u' is " << uCnt << ".\n";
return 0;
}
Exercise 5.11
注：此处若不使用 Line 8 的 noskipws 将无法达到目的。
#include <iostream>
using namespace std;
int main()
{
char ch;
unsigned aCnt = 0, eCnt = 0, iCnt = 0, oCnt = 0, uCnt = 0, spaceCnt = 0, tabCnt = 0, newlineCnt = 0;
while (cin >> noskipws >> ch) // 'noskipws' means 'cin' doesn't omit space.
{
ch = tolower(ch);
switch (ch)
{
case 'a':
++aCnt;
break;
case 'e':
++eCnt;
break;
case 'i':
++iCnt;
break;
case 'o':
++oCnt;
break;
case 'u':
++uCnt;
break;
case ' ':
++spaceCnt;
break;
case '\t':
++tabCnt;
break;
case '\n':
++newlineCnt;
break;
}
}
cout << "The number of 'a' is " << aCnt << ".\n";
cout << "The number of 'e' is " << eCnt << ".\n";
cout << "The number of 'i' is " << iCnt << ".\n";
cout << "The number of 'o' is " << oCnt << ".\n";
cout << "The number of 'u' is " << uCnt << ".\n";
cout << "The number of Space is " << spaceCnt << ".\n";
cout << "The number of Tab is " << tabCnt << ".\n";
cout << "The number of Newline is " << newlineCnt << ".\n";
return 0;
}
noskipws 详见 C++ Primer 17.5.1 (English Version Page 760)
Exercise 5.12
同样要用 noskipws。
#include <iostream>
using namespace std;
int main()
{
char ch;
char last_char = '0';
unsigned ffCnt = 0, flCnt = 0, fiCnt = 0;
while (cin >> noskipws >> ch) // 'noskipws' means 'cin' doesn't omit space.
{
if (last_char == 'f')
{
switch (ch)
{
case 'f':
++ffCnt;
break;
case 'l':
++flCnt;
break;
case 'i':
++fiCnt;
break;
}
}
last_char = ch;
}
cout << "The number of 'ff' is " << ffCnt << ".\n";
cout << "The number of 'fl' is " << flCnt << ".\n";
cout << "The number of 'fi' is " << fiCnt << ".\n";
return 0;
}
Exercise 5.14
程序及启示见我的博客 调试 C++ Primer Exercise 5.14 的心得体会。
Exercise 5.17
开头部分解释详见我的博客 关于 C++中 输入多行不定数量数字 的思考。
#include <iostream>
#include <vector>
#include <string>
#include <sstream> // std::istringstream
using namespace std;
int main()
{
vector<int> vint1, vint2;
string num1, num2;
cout << " First vector:";
getline(cin, num1);
istringstream is1(num1);
int i;
while (is1 >> i)
vint1.push_back(i);
cout << "Second vector:";
getline(cin, num2);
istringstream is2(num2);
int j;
while (is2 >> j)
vint2.push_back(j);
bool result = 0;
if (vint1 == vint2) result = 1;
else
{
if (vint1.size() == vint2.size()) result = 0;
else
{
if (vint1.size() > vint2.size()) exchange(vint1, vint2); // now vint1 is shorter than vint2
unsigned s = 0;
for (auto k = 0; k != vint1.size(); k++)
{
if (vint1[k] == vint2[k]) ++s;
}
if (s == vint1.size()) result = 1;
}
}
if (!result)
cout << "false" << endl;
else cout << "true" << endl;
return 0;
}
或者使用头文件 cinvec.h，见 关于 C++中 输入多行不定数量数字 的思考。
关于编写头文件函数的一些心得见 调试 cinvec.h （C++头文件定义函数）的心得体会。
#include <iostream>
#include <vector>
#include <string>
#include "cinvec.h"
using namespace std;
int main()
{
string num1, num2;
cout << " First vector:";
vector<int>vint1 = cinvec_int(num1);
cout << "Second vector:";
vector<int>vint2 = cinvec_int(num2);
bool result = 0;
if (vint1 == vint2) result = 1;
else
{
if (vint1.size() == vint2.size()) result = 0;
else
{
if (vint1.size() > vint2.size()) exchange(vint1, vint2); // now vint1 is shorter than vint2
unsigned s = 0;
for (auto k = 0; k != vint1.size(); k++)
{
if (vint1[k] == vint2[k]) ++s;
}
if (s == vint1.size()) result = 1;
}
}
if (!result)
cout << "false" << endl;
else cout << "true" << endl;
return 0;
}
Exercise 5.19
#include <iostream>
#include <string>
using namespace std;
int main()
{
string rsp;
do
{
string str1, str2;
cout << "Please enter two strings: ";
cin >> str1 >> str2;
string result = (str1 == str2) ? ("equal to ") : ((str1 > str2) ? "greater than " : "smaller than ");
cout << str1 << " is " << result << str2 << "\nMore? Enter yes(Y) or no(N): ";
cin >> rsp;
} while (!rsp.empty() && rsp[0] != 'n' && rsp[0] != 'N');
return 0;
}
Exercise 5.20
#include <iostream>
#include <string>
using namespace std;
int main()
{
string word_now;
string word_last = "";
string word_final = "";
while (cin >> word_now)
{
if (word_now == word_last)
{
word_final = word_now;
break;
}
else word_last = word_now;
}
if (word_final.empty()) cout << "No word was repeated." << endl;
else cout << word_final << " repeats twice." << endl;
return 0;
}
Exercise 5.21
#include <iostream>
#include <string>
using namespace std;
int main()
{
string word_now;
string word_last = "";
string word_final = "";
while (cin >> word_now)
{
if (islower(word_now[0]) || !isalpha(word_now[0]))
{
word_last = word_now;
continue;
}
if (word_now == word_last)
{
word_final = word_now;
break;
}
else word_last = word_now;
}
if (word_final.empty()) cout << "No word was repeated." << endl;
else cout << word_final << " repeats twice." << endl;
return 0;
}
Exercise 5.22
do
{
int sz = get_size();
}
while (sz <= 0);
Exercise 5.23
#include <iostream>
using namespace std;
int main()
{
int a1, a2;
cin >> a1 >> a2;
double quotient = static_cast<double>(a1) / a2;
cout << quotient << endl;
return 0;
}
Example 1:
6 2
Output:
3
Example 2:
5 3
Output:
1.66667
Example 3:
2 0
Output:
inf
Example 4:
0 0
Output:
-nan(ind)
Example 5:
-3 0
Output:
-inf
Exercise 5.24
Exercise 5.25
#include <iostream>
#include <stdexcept>
#include <string>
using namespace std;
int main()
{
int a1, a2;
while (cin >> a1 >> a2)
{
try
{
if (a2 == 0) throw runtime_error("Divisor cannot be zero!");
double quotient = static_cast<double>(a1) / a2;
cout << quotient << endl;
}
catch (runtime_error err)
{
cout << err.what() << "\nTry Again? Enter yes(Y) or no(N)." << endl;
string c;
cin >> c;
if (!cin || c[0] == 'n' || c[0] == 'N')
break;
}
}
return 0;
}
Input & Output:
