C++ primer plus(第六版)编程练习答案第7章C++的编程模块

Dear 丶 2024-03-24 08:34 7阅读 0赞

## 一、程序清单 ##

## calling.cpp ##

// calling.cpp -- defining, prototyping, and calling a function
    #include <iostream>
    
    void simple();    // function prototype
    
    int main()
    {
    	using namespace std;
    	cout << "main() will call the simple() function:\n";
    	simple();     // function call
    	cout << "main() is finished with the simple() function.\n";
    	// cin.get();
    	return 0;
    }
    
    // function definition
    void simple()
    {
    	using namespace std;
    	cout << "I'm but a simple function.\n";
    }

输出结果：

main() will call the simple() function:
    I'm but a simple function.
    main() is finished with the simple() function.

## protos.cpp ##

// protos.cpp -- using prototypes and function calls
    #include <iostream>
    void cheers(int);       // prototype: no return value
    double cube(double x);  // prototype: returns a double
    int main()
    {
    	using namespace std;
    	cheers(5);          // function call
    	cout << "Give me a number: ";//输入一个数字
    	double side;
    	cin >> side;
    	double volume = cube(side);    // function call V=side*side*side
    	cout << "A " << side << "-foot cube has a volume of ";
    	cout << volume << " cubic feet.\n";
    	cheers(cube(2));    // prototype protection at work   2*2*2
    	// cin.get();
    	// cin.get();
    	return 0;
    }
    
    void cheers(int n)//输出cheers的个数
    {
    	using namespace std;
    	for (int i = 0; i < n; i++)//输出8次
    		cout << "Cheers! ";
    	cout << endl;
    }
    
    double cube(double x)
    {
    	return x * x * x;//返回立方值
    }

输出结果：

Cheers! Cheers! Cheers! Cheers! Cheers!
    Give me a number: 5
    A 5-foot cube has a volume of 125 cubic feet.
    Cheers! Cheers! Cheers! Cheers! Cheers! Cheers! Cheers! Cheers!

## twoarg.cpp ##

// twoarg.cpp -- a function with 2 arguments
    #include <iostream>
    using namespace std;
    void n_chars(char c, int n);
    int main()
    {
    	int times;
    	char ch;
    
    	cout << "Enter a character: ";
    	cin >> ch;
    	while (ch != 'q')        // q to quit
    	{
    		cout << "Enter an integer: ";
    		cin >> times;
    		n_chars(ch, times); // function with two arguments
    		cout << "\nEnter another character or press the"
    			" q-key to quit: ";
    		cin >> ch;
    	}
    	cout << "The value of times is " << times << ".\n";
    	cout << "Bye\n";
    	// cin.get();
    	// cin.get();
    	return 0;
    }
    
    void n_chars(char c, int n) // displays c n times
    {
    	while (n-- > 0)         // continue until n reaches 0
    		cout << c;
    }

输出结果：

Enter a character: w
    Enter an integer: 50
    wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww
    Enter another character or press the q-key to quit: a
    Enter an integer: 20
    aaaaaaaaaaaaaaaaaaaa
    Enter another character or press the q-key to quit: q
    The value of times is 20.
    Bye

## lotto.cpp ##

假设必须从51个数中选取6个，而获奖的概率为1/R，则R的计算公式为：

![watermark_type_d3F5LXplbmhlaQ_shadow_50_text_Q1NETiBA6ZW_5rKZ5pyJ6IKl6bG8_size_20_color_FFFFFF_t_70_g_se_x_16][]

// lotto.cpp -- probability of winning
    #include <iostream>
    // Note: some implementations require double instead of long double
    long double probability(unsigned numbers, unsigned picks);
    int main()
    {
    	using namespace std;
    	double total, choices;//total表示
    	cout << "Enter the total number of choices on the game card and\n"
    		"the number of picks allowed:\n";
    	while ((cin >> total >> choices) && choices <= total)
    	{
    		cout << "You have one chance in ";
    		//total相当于:51*50*49*48*47*46 choice相当于6*5*4*3*2*1
    		cout << probability(total, choices);      // compute the odds 
    		cout << " of winning.\n";
    		cout << "Next two numbers (q to quit): ";
    	}
    	cout << "bye\n";
    	// cin.get();
    	// cin.get();
    	return 0;
    }
    
    // the following function calculates the probability of picking picks
    // numbers correctly from numbers choices
    long double probability(unsigned numbers, unsigned picks)
    {
    	long double result = 1.0;  // here come some local variables
    	long double n;
    	unsigned p;
    
    	for (n = numbers, p = picks; p > 0; n--, p--)
    		result = result * n / p;//n相当于:51*50*49*48*47*46 p相当于6*5*4*3*2*1
    	return result;
    }

执行结果：

Enter the total number of choices on the game card and
    the number of picks allowed:
    49 6
    You have one chance in 1.39838e+07 of winning.
    Next two numbers (q to quit): 51 6
    You have one chance in 1.80095e+07 of winning.
    Next two numbers (q to quit): 38 6
    You have one chance in 2.76068e+06 of winning.
    Next two numbers (q to quit): q
    bye

## arrfun1.cpp ##

// arrfun1.cpp -- functions with an array argument
    #include <iostream>
    const int ArSize = 8;
    int sum_arr(int arr[], int n);        // prototype
    int main()
    {
    	using namespace std;
    	int cookies[ArSize] = { 1,2,4,8,16,32,64,128 };
    	// some systems require preceding int with static to
    	// enable array initialization
    
    	int sum = sum_arr(cookies, ArSize);
    	cout << "Total cookies eaten: " << sum << "\n";
    	// cin.get();
    	return 0;
    }
    
    // return the sum of an integer array
    int sum_arr(int arr[], int n)
    {
    	int total = 0;
    
    	for (int i = 0; i < n; i++)
    		total = total + arr[i];//计算数组中元素的总和
    	return total;
    }

执行结果：

Total cookies eaten: 255

##  **arrfun2.cpp** ##

// arrfun2.cpp -- functions with an array argument
    #include <iostream>
    const int ArSize = 8;
    int sum_arr(int arr[], int n);
    // use std:: instead of using directive
    int main()
    {
    	int cookies[ArSize] = { 1,2,4,8,16,32,64,128 };
    	//  some systems require preceding int with static to
    	//  enable array initialization
    
    	std::cout << cookies << " = array address, ";
    	//  some systems require a type cast: unsigned (cookies)
    
    	std::cout << sizeof cookies << " = sizeof cookies\n";//sizeof arr=4 sizeof cookies=4*8=32
    	int sum = sum_arr(cookies, ArSize);
    	std::cout << "Total cookies eaten: " << sum << std::endl;
    	sum = sum_arr(cookies, 3);        // a lie 通过告诉该函数有3个元素，可以让它计算前三个元素的总和
    	std::cout << "First three eaters ate " << sum << " cookies.\n";
    	sum = sum_arr(cookies + 4, 4);    // another lie 
    	//由于cookies是第一个元素的地址，因此cookies+4计算的是第5个元素的地址
    	//所以sum将计算5 6 7 8个元素的总和
    	std::cout << "Last four eaters ate " << sum << " cookies.\n";
    	// std::cin.get();
    	return 0;
    }
    
    // return the sum of an integer array
    int sum_arr(int arr[], int n)
    {
    	int total = 0;
    	std::cout << arr << " = arr, ";
    	// some systems require a type cast: unsigned (arr)
    
    	std::cout << sizeof arr << " = sizeof arr\n";
    	for (int i = 0; i < n; i++)
    		total = total + arr[i];
    	return total;
    }

执行结果：

009FFAC0 = array address, 32 = sizeof cookies
    009FFAC0 = arr, 4 = sizeof arr
    Total cookies eaten: 255
    009FFAC0 = arr, 4 = sizeof arr
    First three eaters ate 7 cookies.
    009FFAD0 = arr, 4 = sizeof arr
    Last four eaters ate 240 cookies.

##  **arrfun3.cpp** ##

执行结果：

// arrfun3.cpp -- array functions and const
    #include <iostream>
    const int Max = 5;
    
    // function prototypes
    int fill_array(double ar[], int limit);
    void show_array(const double ar[], int n);  // don't change data
    void revalue(double r, double ar[], int n);
    
    int main()
    {
    	using namespace std;
    	double properties[Max];
    
    	int size = fill_array(properties, Max);
    	show_array(properties, size);
    	if (size > 0)
    	{
    		cout << "Enter revaluation factor: ";
    		double factor;
    		while (!(cin >> factor))    // bad input
    		{
    			cin.clear();
    			while (cin.get() != '\n')
    				continue;
    			cout << "Bad input; Please enter a number: ";
    		}
    		revalue(factor, properties, size);//factor表示revalue中的r,size表示组数
    		show_array(properties, size);//输出Property #1: $..... ,size表示组数
    	}
    	cout << "Done.\n";
    	// cin.get();
    	// cin.get();
    	return 0;
    }
    
    int fill_array(double ar[], int limit)
    {
    	using namespace std;
    	double temp;
    	int i;
    	for (i = 0; i < limit; i++)
    	{
    		cout << "Enter value #" << (i + 1) << ": ";
    		cin >> temp;//输入数据
    		if (!cin)    // bad input
    		{
    			cin.clear();
    			while (cin.get() != '\n')
    				continue;
    			cout << "Bad input; input process terminated.\n";
    			break;
    		}
    		else if (temp < 0)     // signal to terminate 当小于0时,终止
    			break;
    		ar[i] = temp;
    	}
    	return i;
    }
    
    // the following function can use, but not alter,
    // the array whose address is ar
    void show_array(const double ar[], int n)
    {
    	using namespace std;
    	for (int i = 0; i < n; i++)
    	{
    		cout << "Property #" << (i + 1) << ": $";//Property #1: $....Property #2: $.... Property #3: $....
    		cout << ar[i] << endl;//输出fill_array中的数据
    	}
    }
    
    // multiplies each element of ar[] by r
    void revalue(double r, double ar[], int n)
    {
    	for (int i = 0; i < n; i++)
    		ar[i] *= r;//相当于Property #1: $200000 然后200000*1.2=240000
    }

Enter value #1: 100000
    Enter value #2: 80000
    Enter value #3: 222000
    Enter value #4: 240000
    Enter value #5: 118000
    Property #1: $100000
    Property #2: $80000
    Property #3: $222000
    Property #4: $240000
    Property #5: $118000
    Enter revaluation factor: 0.8
    Property #1: $80000
    Property #2: $64000
    Property #3: $177600
    Property #4: $192000
    Property #5: $94400
    Done.
    
    Enter value #1: 200000
    Enter value #2: 84000
    Enter value #3: 160000
    Enter value #4: -2  
    //当值<0时候终止
    Property #1: $200000
    Property #2: $84000
    Property #3: $160000
    Enter revaluation factor: 1.2
    Property #1: $240000
    Property #2: $100800
    Property #3: $192000
    Done.

## arrfun4.cpp ##

// arrfun4.cpp -- functions with an array range
    #include <iostream>
    const int ArSize = 8;
    int sum_arr(const int * begin, const int * end);
    int main()
    {
    	using namespace std;
    	int cookies[ArSize] = { 1,2,4,8,16,32,64,128 };
    	//  some systems require preceding int with static to
    	//  enable array initialization
    
    	int sum = sum_arr(cookies, cookies + ArSize);
    	cout << "Total cookies eaten: " << sum << endl;//1+2+4+8+16+32+64+128=255
    	sum = sum_arr(cookies, cookies + 3);        // first 3 elements 1+2+4=7
    	cout << "First three eaters ate " << sum << " cookies.\n";
    	sum = sum_arr(cookies + 4, cookies + 8);    // last 4 elements 16+32+64+128=240
    	cout << "Last four eaters ate " << sum << " cookies.\n";
    	// cin.get();
    	return 0;
    }
    
    // return the sum of an integer array
    int sum_arr(const int * begin, const int * end)
    {
    	const int * pt;
    	int total = 0;
    
    	for (pt = begin; pt != end; pt++)
    		total = total + *pt;//将pt设置为指向最后一个元素(begin指向的位置)的指针
    	//并将*pt(元素的值)加入到total中
    	return total;
    }

执行结果：

Total cookies eaten: 255
    First three eaters ate 7 cookies.
    Last four eaters ate 240 cookies.

## strgfun.cpp ##

// strgfun.cpp -- functions with a string argument
    #include <iostream>
    unsigned int c_in_str(const char * str, char ch);
    int main()
    {
        using namespace std;
        char mmm[15] = "minimum";    // string in an array
    // some systems require preceding char with static to
    // enable array initialization
    
        char *wail = "ululate";    // wail points to string
    
        unsigned int ms = c_in_str(mmm, 'm');
        unsigned int us = c_in_str(wail, 'u');
        cout << ms << " m characters in " << mmm << endl;
        cout << us << " u characters in " << wail << endl;
        // cin.get();
        return 0;
    }
    
    // this function counts the number of ch characters
    // in the string str
    unsigned int c_in_str(const char * str, char ch)
    {
        unsigned int count = 0;
    
        while (*str)        // quit when *str is '\0'
        {
            if (*str == ch)
                count++;
            str++;        // move pointer to next char
        }
        return count; 
    }

执行结果：

Total cookies eaten: 255
    First three eaters ate 7 cookies.
    Last four eaters ate 240 cookies.

## strgback.cpp ##

// strgback.cpp -- a function that returns a pointer to char
    #include <iostream>
    char * buildstr(char c, int n);     // prototype
    int main()
    {
    	using namespace std;
    	int times;
    	char ch;
    
    	cout << "Enter a character: ";
    	cin >> ch;//输入字符串
    	cout << "Enter an integer: ";
    	cin >> times;//输入次数
    	char *ps = buildstr(ch, times);
    	cout << ps << endl;
    	delete[] ps;                   // free memory 清空内存
    	ps = buildstr('+', 20);         // reuse pointer
    	cout << ps << "-DONE-" << ps << endl;//++++++++++++++++++++-DONE-++++++++++++++++++++
    	delete[] ps;                   // free memory
    	// cin.get();
    	// cin.get();
    	return 0;
    }
    
    // builds string made of n c characters
    char * buildstr(char c, int n)
    {
    	char * pstr = new char[n + 1];
    	pstr[n] = '\0';         // terminate string
    	while (n-- > 0)
    		pstr[n] = c;        // fill rest of string 
    	//请求分配n+1个字节的内存来存储该字符串
    	//并将最后一个字节设置为空值字符,然后从后向前对数组进行填充
    	return pstr;
    }

执行结果：

Enter a character: V
    Enter an integer: 46
    VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV
    ++++++++++++++++++++-DONE-++++++++++++++++++++

## travel.cpp ##

//  -- using structures with functions
    #include <iostream>
    struct travel_time
    {
    	int hours;
    	int mins;
    };
    const int Mins_per_hr = 60;
    
    travel_time sum(travel_time t1, travel_time t2);
    void show_time(travel_time t);
    
    int main()
    {
    	using namespace std;
    	travel_time day1 = { 5, 45 };    // 5 hrs, 45 min
    	travel_time day2 = { 4, 55 };    // 4 hrs, 55 min
    
    	travel_time trip = sum(day1, day2);
    	cout << "Two-day total: ";
    	show_time(trip);
    
    	travel_time day3 = { 4, 32 };
    	cout << "Three-day total: ";
    	show_time(sum(trip, day3));
    	// cin.get();
    
    	return 0;
    }
    
    travel_time sum(travel_time t1, travel_time t2)
    {
    	travel_time total;
    
    	total.mins = (t1.mins + t2.mins) % Mins_per_hr;//相当于(45+55)%60=1
    	total.hours = t1.hours + t2.hours +
    		(t1.mins + t2.mins) / Mins_per_hr;//5+4+1=10h
    	return total;
    }
    
    void show_time(travel_time t)
    {
    	using namespace std;
    	cout << t.hours << " hours, "
    		<< t.mins << " minutes\n";
    }

执行结果：

Two-day total: 10 hours, 40 minutes
    Three-day total: 15 hours, 12 minutes

## strctfun.cpp ##

// strctfun.cpp -- functions with a structure argument
    #include <iostream>
    #include <cmath>
    
    // structure declarations
    struct polar
    {
    	double distance;      // distance from origin
    	double angle;         // direction from origin
    };
    struct rect
    {
    	double x;             // horizontal distance from origin
    	double y;             // vertical distance from origin
    };
    
    // prototypes
    polar rect_to_polar(rect xypos);
    void show_polar(polar dapos);
    
    int main()
    {
    	using namespace std;
    	rect rplace;
    	polar pplace;
    
    	cout << "Enter the x and y values: ";
    	while (cin >> rplace.x >> rplace.y)  // slick use of cin
    	{
    		pplace = rect_to_polar(rplace);//
    		show_polar(pplace);
    		cout << "Next two numbers (q to quit): ";
    	}
    	cout << "Done.\n";
    	return 0;
    }
    
    // convert rectangular to polar coordinates
    polar rect_to_polar(rect xypos)
    {
    	using namespace std;
    	polar answer;
    
    	answer.distance =
    		sqrt(xypos.x * xypos.x + xypos.y * xypos.y);//30*30+40*40开根
    	answer.angle = atan2(xypos.y, xypos.x);//arctan(3/4)
    	return answer;      // returns a polar structure
    }
    
    // show polar coordinates, converting angle to degrees
    void show_polar(polar dapos)
    {
    	using namespace std;
    	const double Rad_to_deg = 57.29577951;
    
    	cout << "distance = " << dapos.distance;
    	cout << ", angle = " << dapos.angle * Rad_to_deg;
    	cout << " degrees\n";
    }

执行结果：

Enter the x and y values: 30 40
    distance = 50, angle = 53.1301 degrees
    Next two numbers (q to quit): 100 -100
    distance = 141.421, angle = -45 degrees
    Next two numbers (q to quit): -100 100
    distance = 141.421, angle = 135 degrees
    Next two numbers (q to quit): q
    Done.

## strctptr.cpp ##

// strctptr.cpp -- functions with pointer to structure arguments
    #include <iostream>
    #include <cmath>
    
    // structure templates
    struct polar
    {
        double distance;      // distance from origin
        double angle;         // direction from origin
    };
    struct rect
    {
        double x;             // horizontal distance from origin
        double y;             // vertical distance from origin
    };
    
    // prototypes
    void rect_to_polar(const rect * pxy, polar * pda);
    void show_polar (const polar * pda);
    
    int main()
    {
        using namespace std;
        rect rplace;
        polar pplace;
    
        cout << "Enter the x and y values: ";
        while (cin >> rplace.x >> rplace.y)
        {
            rect_to_polar(&rplace, &pplace);    // pass addresses
            show_polar(&pplace);        // pass address
            cout << "Next two numbers (q to quit): ";
        }
        cout << "Done.\n";
        return 0;
    }
    
    // show polar coordinates, converting angle to degrees
    void show_polar (const polar * pda)
    {
        using namespace std;
        const double Rad_to_deg = 57.29577951;
    
        cout << "distance = " << pda->distance;
        cout << ", angle = " << pda->angle * Rad_to_deg;
        cout << " degrees\n";
    }
    
    // convert rectangular to polar coordinates
    void rect_to_polar(const rect * pxy, polar * pda)
    {
        using namespace std;
        pda->distance =
            sqrt(pxy->x * pxy->x + pxy->y * pxy->y);
        pda->angle = atan2(pxy->y, pxy->x); 
    }

执行结果：

Enter the x and y values: 30 40
    distance = 50, angle = 53.1301 degrees
    Next two numbers (q to quit): -100 100
    distance = 141.421, angle = 135 degrees
    Next two numbers (q to quit): q
    Done.

## topfive.cpp ##

// topfive.cpp -- handling an array of string objects
    #include <iostream>
    #include <string>
    using namespace std;
    const int SIZE = 5;
    void display(const string sa[], int n);
    int main()
    {
    	string list[SIZE];     // an array holding 5 string object
    	cout << "Enter your " << SIZE << " favorite astronomical sights:\n";
    	for (int i = 0; i < SIZE; i++)
    	{
    		cout << i + 1 << ": ";
    		getline(cin, list[i]);
    	}
    
    	cout << "Your list:\n";
    	display(list, SIZE);
    	// cin.get();
    
    	return 0;
    }
    
    void display(const string sa[], int n)
    {
    	for (int i = 0; i < n; i++)
    		cout << i + 1 << ": " << sa[i] << endl;
    }

执行结果：

Enter your 5 favorite astronomical sights:
    1: Orion Nebula
    2: M13
    3: Saturn
    4: Jupiter
    5: Moon
    Your list:
    1: Orion Nebula
    2: M13
    3: Saturn
    4: Jupiter
    5: Moon

## arrobj.cpp ##

//arrobj.cpp -- functions with array objects
    #include <iostream>
    #include <array>
    #include <string>
    const int Seasons = 4;
    const std::array<std::string, Seasons> Snames =
    { "Spring", "Summer", "Fall", "Winter" };
    
    void fill(std::array<double, Seasons> * pa);
    void show(std::array<double, Seasons> da);
    int main()
    {
    	std::array<double, 4> expenses;
    	fill(&expenses);
    	show(expenses);
    	// std::cin.get();
    	// std::cin.get();
    	return 0;
    }
    
    void fill(std::array<double, Seasons> * pa)
    {
    	for (int i = 0; i < Seasons; i++)
    	{
    		std::cout << "Enter " << Snames[i] << " expenses: ";
    		std::cin >> (*pa)[i];
    	}
    }
    
    void show(std::array<double, Seasons> da)
    {
    	double total = 0.0;
    	std::cout << "\nEXPENSES\n";
    	for (int i = 0; i < Seasons; i++)
    	{
    		std::cout << Snames[i] << ": $" << da[i] << '\n';
    		total += da[i];
    	}
    	std::cout << "Total: $" << total << '\n';
    }

执行结果：

Enter Spring expenses: 212
    Enter Summer expenses: 256
    Enter Fall expenses: 208
    Enter Winter expenses: 244
    
    EXPENSES
    Spring: $212
    Summer: $256
    Fall: $208
    Winter: $244
    Total: $920

执行结果：

Counting down ... 4
    Counting down ... 3
    Counting down ... 2
    Counting down ... 1
    Counting down ... 0
    0: Kaboom!
    1: Kaboom!
    2: Kaboom!
    3: Kaboom!
    4: Kaboom!

## ruler.cpp ##

// ruler.cpp -- using recursion to subdivide a ruler
    #include <iostream>
    const int Len = 66;
    const int Divs = 6;
    void subdivide(char ar[], int low, int high, int level);
    int main()
    {
    	char ruler[Len];
    	int i;
    	for (i = 1; i < Len - 2; i++)
    		ruler[i] = ' ';
    	ruler[Len - 1] = '\0';
    	int max = Len - 2;
    	int min = 0;
    	ruler[min] = ruler[max] = '|';
    	std::cout << ruler << std::endl;
    	for (i = 1; i <= Divs; i++)
    	{
    		subdivide(ruler, min, max, i);
    		std::cout << ruler << std::endl;
    		for (int j = 1; j < Len - 2; j++)
    			ruler[j] = ' ';  // reset to blank ruler
    	}
    	// std::cin.get();
    
    	return 0;
    }
    
    void subdivide(char ar[], int low, int high, int level)
    {
    	if (level == 0)
    		return;
    	int mid = (high + low) / 2;
    	ar[mid] = '|';
    	subdivide(ar, low, mid, level - 1);
    	subdivide(ar, mid, high, level - 1);
    }

执行结果：

|                                                               |
    |                               |                               |
    |               |               |               |               |
    |       |       |       |       |       |       |       |       |
    |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |
    | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
    |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

## recur.cpp ##

// recur.cpp -- using recursion
    #include <iostream>
    void countdown(int n);
    
    int main()
    {
    	countdown(4);           // call the recursive function
    	// std::cin.get();
    	return 0;
    }
    
    void countdown(int n)
    {
    	using namespace std;
    	cout << "Counting down ... " << n << endl;
    	if (n > 0)
    		countdown(n - 1);     // function calls itself
    	cout << n << ": Kaboom!\n";
    }

执行结果：

Counting down ... 4
    Counting down ... 3
    Counting down ... 2
    Counting down ... 1
    Counting down ... 0
    0: Kaboom!
    1: Kaboom!
    2: Kaboom!
    3: Kaboom!
    4: Kaboom!

## arfupt.cpp ##

// arfupt.cpp -- an array of function pointers
    #include <iostream>
    // various notations, same signatures
    const double * f1(const double ar[], int n);
    const double * f2(const double[], int);
    const double * f3(const double *, int);
    
    int main()
    {
    	using namespace std;
    	double av[3] = { 1112.3, 1542.6, 2227.9 };
    
    	// pointer to a function
    	const double *(*p1)(const double *, int) = f1;
    	auto p2 = f2;  // C++0x automatic type deduction
    	// pre-C++0x can use the following code instead
    	// const double *(*p2)(const double *, int) = f2;
    	cout << "Using pointers to functions:\n";
    	cout << " Address  Value\n";
    	cout << (*p1)(av, 3) << ": " << *(*p1)(av, 3) << endl;
    	cout << p2(av, 3) << ": " << *p2(av, 3) << endl;
    
    	// pa an array of pointers
    	// auto doesn't work with list initialization
    	const double *(*pa[3])(const double *, int) = { f1,f2,f3 };
    	// but it does work for initializing to a single value
    	// pb a pointer to first element of pa
    	auto pb = pa;
    	// pre-C++0x can use the following code instead
    	// const double *(**pb)(const double *, int) = pa;
    	cout << "\nUsing an array of pointers to functions:\n";
    	cout << " Address  Value\n";
    	for (int i = 0; i < 3; i++)
    		cout << pa[i](av, 3) << ": " << *pa[i](av, 3) << endl;
    	cout << "\nUsing a pointer to a pointer to a function:\n";
    	cout << " Address  Value\n";
    	for (int i = 0; i < 3; i++)
    		cout << pb[i](av, 3) << ": " << *pb[i](av, 3) << endl;
    
    	// what about a pointer to an array of function pointers
    	cout << "\nUsing pointers to an array of pointers:\n";
    	cout << " Address  Value\n";
    	// easy way to declare pc 
    	auto pc = &pa;
    	// pre-C++0x can use the following code instead
       // const double *(*(*pc)[3])(const double *, int) = &pa;
    	cout << (*pc)[0](av, 3) << ": " << *(*pc)[0](av, 3) << endl;
    	// hard way to declare pd
    	const double *(*(*pd)[3])(const double *, int) = &pa;
    	// store return value in pdb
    	const double * pdb = (*pd)[1](av, 3);
    	cout << pdb << ": " << *pdb << endl;
    	// alternative notation
    	cout << (*(*pd)[2])(av, 3) << ": " << *(*(*pd)[2])(av, 3) << endl;
    	// cin.get();
    	return 0;
    }
    
    // some rather dull functions
    
    const double * f1(const double * ar, int n)
    {
    	return ar;
    }
    const double * f2(const double ar[], int n)
    {
    	return ar + 1;
    }
    const double * f3(const double ar[], int n)
    {
    	return ar + 2;
    }

执行结果：

Using pointers to functions:
     Address  Value
    006FF9B4: 1112.3
    006FF9BC: 1542.6
    
    Using an array of pointers to functions:
     Address  Value
    006FF9B4: 1112.3
    006FF9BC: 1542.6
    006FF9C4: 2227.9
    
    Using a pointer to a pointer to a function:
     Address  Value
    006FF9B4: 1112.3
    006FF9BC: 1542.6
    006FF9C4: 2227.9
    
    Using pointers to an array of pointers:
     Address  Value
    006FF9B4: 1112.3
    006FF9BC: 1542.6
    006FF9C4: 2227.9

## 二、复习题 ##

1.请看下面两个计算空格和换行符数目的代码片段：

//version 1
    while(cin.get(ch))   //quit on eof
    {  
        if(ch==' ')
             spaces++;
        if(ch=='\n')
             newlines++;
    }
    
    //version 2
    while(cin.get(ch))   //quit eof
    {
        if(ch==' ')
             spacees++;
        else if(ch=='\n')    
             newlines++;
    }

第二个格式比第一个格式好在哪里？

第二个版本比第一个版本效率更高，因为在第一个中对于每个字符都需要判断两次，而在第二个版本中，如果字符为空格，在经过if判断确定为空格后，该字符肯定不是换行符，第二个else if的判断直接跳过，节省判断时间。

2.在程序清单6.2中，用ch+1替换++ch将发生什么情况？

> 程序清单6.2 ifelse.cpp
> 
>     // ifelse.cpp -- using the if else statement
>     #include <iostream>
>     int main()
>     {
>        char ch;
>        std::cout << "Type, and I shall repeat.\n";
>        std::cin.get(ch);
>        while (ch != '.')
>        {
>            if (ch == '\n')
>                 std::cout << ch; // done if newline
>            else
>                 std::cout << ++ch; // done otherwise
>            std::cin.get(ch);
>     }
>        // try ch + 1 instead of ++ch for interesting effect
>        std::cout << "\nPlease excuse the slight confusion.\n";
>        // std::cin.get();
>        // std::cin.get();
>        return 0;
>     }
> 
> \++ch的数据类型依旧是char型，但对于char+1最终类型为int型，因此当++ch换成ch+1后，输出的是数字。

3.请认真考虑下面的程序：

#include<iostream>
    using namespace std;
    int main()
    {
        char ch;
        int ct1,ct2;
        ct1=ct2=0;
        while((ch=cin.get())!='$')
        {
            cout<<ch;
            ct1++;
            if(ch='$')
                  ct2++;
             cout<<ch;
        }
        cout<<"ct1="<<ct1<<",ct2="<<ct2<<"\n";
        return 0;
    }

假设输入如下（请在每行末尾按回车键）：

`Hi!`  
`Send $10 or $20 now!`

则输出将是什么（还记得吗，输入被缓冲）？

输入输出结果为

Hi!
    H$i$!$
    $Send $10 or $20 now!
    S$e$n$d$ $ct1=9,ct2=9

由于程序中使用的是`ch='$'`,所以每次循环该if条件内代码都执行一次，因此ct1与ct2相等。同时在输入`Hi!`之后键入回车，因此在读取回车之后，打印出来，光标换行，紧接着打印$。

4.创建表示下述条件的逻辑表达式：

**a.weight大于或等于115，但小于125。**  
**b.ch为q或Q。**  
**c.x为偶数，但不是26.**  
**d.x为偶数，但不是26。**  
**e.donation为1000-2000或guest为1。**  
**f.ch是小写字母或大写字母（假设小写字母是依次编码的，大写字母也是依次编码的，但在大小写字母间编码是不连续的）。**

//a
    weight>=115 && weight<125   
    
    //b
    ch=='q' || ch=='Q'
    
    //c
    x%2==0 && x!=26
    
    //d
    x%2==0 && x%26!=0
    
    //e
    donation>=1000 && donation<=20000 || guest==1
    
    //f
    (ch>='a' && ch<='z') || (ch<='Z' && ch>='A')

5.在英语中，"I will not not speak（我不会不说） "的意思与"I will speak（我要说）"相同。在c++中，!!x是否与x相同呢？

对于bool变量而言，!!x与x是相同的，但对于其他类型变量不一定相同，例如!!5=1，!!5≠5。

6.创建一个条件表达式，其值为变量的绝对值。也就是说，如果变量x为正，则表达式的值为x；但如果x为负，则表达式的值为-x--这是一个正值。

x>=0 ? x : -x;

7.用switch改写下面的代码片段：

if(ch=='A')
        a_grade++;
    else if(ch=='B')
        b_grade++;
    else if(ch=='C')
        c_grade++;
    else if(ch=='D')
        d_grade++;
    else
        f_grade++;

改写后代码：

switch(ch)
    {
        case 'A' : a_grade++;
                        break;
        case 'B' : b_grade++;
                        break;
        case 'C' : c_grade++;
                        break;
        case 'D' : d_grade++;
                        break;
        default :   f_grade++;
                        break;
    }

8.对于程序清单6.10，与使用字符（如a和c）表示菜单选项和case标签有何优点呢？（提示：想一想用户输入q和输入5的情况。）

> 程序清单6.10 switch.cpp
> 
>     // switch.cpp -- using the switch statement
>     #include <iostream>
>     using namespace std;
>     void showmenu(); // function prototypes
>     void report();
>     void comfort();
>     int main()
>     {
>        showmenu();
>        int choice;
>        cin >> choice;
>        while (choice != 5)
>        {
>            switch (choice)
>            {
>            case 1: cout << "\a\n";
>                break;
>            case 2: report();
>                break;
>            case 3: cout << "The boss was in all day.\n";
>                break;
>            case 4: comfort();
>                break;
>            default: cout << "That's not a choice.\n";
>            }
>            showmenu();
>            cin >> choice;
>        }
>        cout << "Bye!\n";
>        return 0;
>     }
>     void showmenu()
>     {
>        cout << "Please enter 1, 2, 3, 4, or 5:\n"
>            "1) alarm 2) report\n"
>            "3) alibi 4) comfort\n"
>            "5) quit\n";
>     }
>     void report()
>     {
>        cout << "It's been an excellent week for business.\n"
>            "Sales are up 120%. Expenses are down 35%.\n";
>     }
>     void comfort()
>     {
>        cout << "Your employees think you are the finest CEO\n"
>            "in the industry. The board of directors think\n"
>            "you are the finest CEO in the industry.\n";
>     }
> 
> 使用数字作为菜单选项和case标签，限定了用户只有输入数字的时候才能有效，若用户错误的输入非整数类型，导致程序被挂起。而使用字符作为菜单选项和case标签，当用户输入错误类型，程序能正确通过default部分提示用户输入错误，用户体验更加友好，提高了程序的容错性和健壮性。

9.请看下面的代码片段

int line = 0;
    char ch;
    while(cin.get(ch))
    {
        if(ch=='Q')
            break;
        if(ch!='\n')
            continue;
        line++;
    }

重写后代码：

int line = 0;
    char ch;
    while(cin.get(ch) && ch!='Q')
    {
        if(ch=='\n')
            line++;
    }

## 三、编程练习 ##

1.编写一个程序，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字符，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;
    
    int main()
    {
        cout << "Please enter characters (type @ to stop):";
        char ch;
        cin.get(ch);
        while (ch != '@')
        {
            if (islower(ch))
                ch = toupper(ch);
            else if (isupper(ch))
                ch = tolower(ch);
            if (!isdigit(ch))
                cout << ch;
            cin.get(ch);
        }
        system("pause");
        return 0;
    }

2.编写一个程序，最多将10个donation值读入一个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<array>
    using namespace std;
    int main()
    {
        array<double,10>donations;
        double sum=0;
        cout<<"Please enter donation(non-number to stop):";
        int i;
        double donation;
        for(i=0;(i<10)&&(cin>>donation);++i)
        {
            donations[i]=donation;
            sum=+donations[i];
        }    
    
        double avg=sum/(i+1);
        cout<<"Average:"<<avg<<endl;
        cout<<"Number larger than average:";
        for(int j=0;j<i;j++)
         {
            if(donations[j]>avg)
                cout<<donations[j]<<" ";
         }   
        cout<<endl;
        return 0;
    }

3.编写一个菜单驱动程序的雏形。该程序显示一个提供四个选项的菜单--每个选项用一个字母表标记。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有效的字母，直到用户这样选择为止。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操作。该程序的运行情况如下：

Please enter one of the following choices:
    c) carnivore         p) pianist
    t) tree                  g)game
    f
    Please enter a c,p,t or g:q
    Please enter a c,p,t or g:t
    A maple is a tree.

代码实现：

#include <iostream>
    using namespace std;
    int main()
    {
        cout<<"Please enter one of the following choices:"<<endl;
        cout<<"c) carnivore         p) pianist"<<endl;
        cout<<"t) tree              g) game"<<endl;
        char ch;
        cin>>ch;
        while(ch!='c' && ch!='p' && ch!='t' && ch!='g')
        { 
            cout<<"Please enter a c,p,t or g:";
            cin>>ch;
        }   
        switch (ch)
        {
            case 'c':
                cout<<"A maple is a carnivore."<<endl;
                break;
            case 'p':
                cout<<"A maple is a pianist."<<endl;
                break;
            case 't':
                cout<<"A maple is a tree."<<endl;
                break;
            case 'g':
                cout<<"A maple is a game."<<endl;
                break;
            default:
                break;
        }
        return 0;
    }

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（她）。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，请使用下面的结构：

//Benevolent Order of Programmers name structure
    struct bop{
        char fullname[strsize];     //real name
        char title[strzie];              //job title
        char bopname[strsize];   //secret BOP name
        int preference;                //0=fullname,1=title,2=bopname
    };

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.diaplay by title
    c.display by bopname     d.display by preference
    q.quit

注意，“display by preference”并不意味显示成员的偏好，而是意味着根据成员的偏好来列出成员。例如，如果偏好号为1，则选择d将显示程序员的头衔。该程序的运行情况如下：

Benevolent Order of Programmers Report
    a. display by name         b. display by title
    c. display by bopname      d. display by preference
    q. quit
    Enter your choice: a
    Wimp Macho
    Raki Rhodes
    Celia Laiter
    Hoppy Hipman
    Pat Hand
    Next choice: d
    Wimp Macho
    Junior Programmer
    MIPS
    Analyst Trainee
    LOOPY
    Next choice: q
    Bye!

代码实现 ：

#include <iostream>
    
    using namespace std;
    
    const int strsize = 20;
    struct bop
    {
        char fullname[strsize];
        char title[strsize];
        char bopname[strsize];
        int preference;         //0 = fullname, 1 = title, 2 = bopname
    };
    
    int main()
    {
        int programmerNum = 5;
        bop programmers[programmerNum]={
       {"Wimp Macho", "Junior Programmer", "WM",0},
        {"Raki Rhodes", "Junior Programmer", "RR", 1},
        {"Celia Laiter", "Junior Programmer", "MIPS", 2},
        {"Hoppy Hipman", "Analyst Trainee", "HH", 1},
        {"Pat Hand", "Junior Programmer", "LOOPY", 2}};
        cout << "Benevolent Order of Programmers Report" << endl;
        cout << "a. display by name          b. display by title" << endl;
        cout << "c. display by bopname       d. display by preference" << endl;
        cout << "q. quit" << endl;
        cout << "Enter your choice: ";
        char choice;
        while(cin.get(choice))
        {
            if(choice == 'a')
            {
                for(int i = 0; i < programmerNum; i++)
                {
                    cout << programmers[i].fullname << endl;
                }
            }
            else if(choice == 'b')
            {
                for(int i = 0; i < programmerNum; i++)
                {
                    cout << programmers[i].title << endl;
                }
            }
            else if(choice == 'c')
            {
                for(int i = 0; i < programmerNum; i++)
                {
                    cout << programmers[i].bopname << endl;
                }
            }
            else if(choice == 'd')
            {
                for(int i = 0; i < programmerNum; i++)
                {
                    if(programmers[i].preference == 0)
                    {
                        cout << programmers[i].fullname << endl;
                    }
                    else if(programmers[i].preference == 1)
                    {
                        cout << programmers[i].title << endl;
                    }
                    else
                    {
                        cout << programmers[i].bopname << endl;
                    }
                }
            }
            else if(choice == 'q')
            {
                cout << "Bye!" << endl;
                break;
            }
            else
            {
                if(choice == '\n')
                {
                    continue;
                }
                cout << "input error, please re-enter!" << endl;
            }
            cout << "Next choice: ";
        }
    
        return 0;
    }

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: 20%`

例如,收入为38000 tvarps时,所得税为5000×0.00+10000×0.10+2000×0.15+3000×0.20,即 4600tvarps。请编写一个程序，使用循环来要求用户输入收入，并报告所得税，当用户输入负数或非数字时，循环将结束。

代码实现：

#include<iostream>
    
    using namespace std;
    
    int main()
    {
        cout << "Please enter your income(if you enter negative number or non-numberic, it will exit.) : ";
        double income, tax;
        while(cin >> income)
        {
            cout<< income <<endl;
            if(income > 35000)
            {
                tax = 10000 * 0.10 + 20000 * 0.15 + (income - 35000) * 0.20;
            }
            else if(income >= 15001)
            {
                tax = (income - 3500) * 0.15 +10000 * 0.10;
            }
            else if(income >= 5001)
            {
                tax = (income - 5000) * 0.10;
            }
            else if(income >= 0)
            {
                tax = 0;
            }
            else
            {
                cout << "Enter negative number, exiting soon..." << endl;
                break;
            }
            cout << "your should pay tax : " << tax << " tvarps." << endl;
            cout << "Please enter another income(if you enter negative number or non-numberic, it will exit.) : ";
        }
        cout << "Exited, Bye !" << endl;
        return 0;
    }

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赠者数目，然后要求用户输入每一个捐献者的姓名和款项。这些信息被存储在一个动态分配的结构数组中.每个数据结构有两个成员：用来存储姓名的字符串数组(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的捐款着姓名及其捐款数额。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Partons开头。如果某种类别没有捐款者，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。

代码实现：

#include<iostream>
    #include<string>
    
    using namespace std;
    
    struct DonationInfo{
        string name;
        double amount;
    };
    
    void donationInfoDisplay(DonationInfo donationInfo[], int num, bool isGrandPatrons)
    {
        int countNum = 0;
        for(int i = 0 ; i < num ; i++)
        {
            if(isGrandPatrons)
            {
                if(donationInfo[i].amount > 10000)
                {
                    countNum++;
                    cout << donationInfo[i].name  << endl;
                }
            }
            else
            {
                if(donationInfo[i].amount <= 10000)
                {
                    countNum++;
                    cout << donationInfo[i].name  << endl;
                }
            }
        }
        if(countNum == 0)
            cout << "none" << endl;
    }
    
    int main()
    {
        cout << "Please enter the number of donors :";
        int num;
        while(!(cin >> num))
        {
            cin.clear();
            cin.sync();
            cin.ignore();
            cout << "***[WARNING]:Please enter positive integer.***" << endl;
            cout << "Please enter the number of donors :";
        }
        DonationInfo* donationInfo = new DonationInfo[num];
        for(int i = 0; i < num; i++)
        {
            cout << "enter the donor name :";
            cin >> donationInfo[i].name;
            cout << "enter the donation amount :";
            cin >> donationInfo[i].amount;
        }
        cout << "******Grand Partons******" << endl;
        donationInfoDisplay(donationInfo, num, true);
        cout << "******Partons******" << endl;
        donationInfoDisplay(donationInfo, num, false);
        return 0;
    }

7.编写一个程序,它每次读取一个单词，直到用户只输入q。然后,该程序指出有多少个单词以元音打头，有多少个单词以辅音打头，还有多少个单词不属于这两类。为此，方法之一是，使用isalpha()来区分以字母和其他字母打头的单词，然后对于通过isalpha()测试的单词，使用if或switch语句来确定哪些以元音打头。该程序的运行情况如下：

`Enter words (q to quit):`  
`The 12 awesome oxem ambled`  
`quiety across 15 meters of lawn. q`  
`5 words beginning with vowels`  
`4 words beginning with consonants`  
`2 others`

代码实现：

#include<iostream>
    #include<string>
    using namespace std;
    
    int main()
    {
        cout << "Enter words (q to quit) :" << endl;
        string word;
        int vowelNum = 0;
        int consonantNum = 0;
        int othersNum = 0;
        while(cin >> word)
        {
            if(word == "q")
            {
                break;
            }
            else
            {
                if(isalpha(word[0]))
                {
                    switch(word[0])
                    {
                        case 'a':
                        case 'e':
                        case 'i':
                        case 'o':
                        case 'u':
                        case 'A':
                        case 'E':
                        case 'I':
                        case 'O':
                        case 'U':
                            vowelNum++;
                            break;
                        default:
                            consonantNum++;
                    }
                }
                else
                {
                    othersNum++;
                }
            }
        }
        cout << vowelNum << " words beginning with vowels" << endl;
        cout << consonantNum << " words beginning with consonants" << endl;
        cout << othersNum << " others" <<endl;
        return 0;
    }

8.编写一个程序，他打开一个文件，逐个字地读取该文件，直到到达文件末尾，然后指出该文件中包含多少个字符。

代码实现：

#include<iostream>
    #include<fstream>
    
    using namespace std;
    
    int main()
    {
        ifstream txtSource;
        txtSource.open("test.txt", ios::in);
        if(!txtSource.is_open())
        {
            cout << "File failed to open !" << endl;
            exit(EXIT_FAILURE);
        }
        char ch;
        int characterNum = 0;
        while(txtSource >> ch)
        {
            characterNum++;
        }
        cout << "The file contains " << characterNum << " characters" << endl;
        return 0;
    }

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。即该文件类似于下面：

代码实现：

#include<iostream>
    #include<string>
    #include<fstream>
    
    using namespace std;
    
    struct DonationInfo{
        string name;
        double amount;
    };
    
    void donationInfoDisplay(DonationInfo donationInfo[], int num, bool isGrandPatrons)
    {
        int countNum = 0;
        for(int i = 0 ; i < num ; i++)
        {
            if(isGrandPatrons)
            {
                if(donationInfo[i].amount > 10000)
                {
                    countNum++;
                    cout << donationInfo[i].name  << endl;
                }
            }
            else
            {
                if(donationInfo[i].amount <= 10000)
                {
                    countNum++;
                    cout << donationInfo[i].name  << endl;
                }
            }
        }
        if(countNum == 0)
            cout << "none" << endl;
    }
    
    int main()
    {
        cout << "Read file content..." << endl;
        ifstream txtDonationInfo;
        txtDonationInfo.open("test.txt", ios::in);
        if(!txtDonationInfo.is_open())
        {
            cout << "File failed to open !" << endl;
            exit(EXIT_FAILURE);
        }
        int num = 0;
        (txtDonationInfo >> num).get();
        DonationInfo* donationInfo = new DonationInfo[num];
        for(int i = 0; i < num; i++)
        {
            txtDonationInfo >> donationInfo[i].name;
            txtDonationInfo >> donationInfo[i].amount;
        }
        cout << "******Grand Partons******" << endl;
        donationInfoDisplay(donationInfo, num, true);
        cout << "******Partons******" << endl;
        donationInfoDisplay(donationInfo, num, false);
        return 0;
    }

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