C++ primer Plus（第六版）_课后编程练习_第八章

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本文分享了C++ Primer Plus第六版第八章的编程练习，包括题8.1至题8.7。作者强调题8.4和8.5的重要性，特别是题8.5中因中文括号引起的bug，提醒读者在编程时要特别留意此类细节。

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题8.1：

#include <iostream>
#include <ctime>
using namespace std;

void ShowStr(const char * str, int print_time = 1);

int main()
{
	char str[100];
	int time;
	cout << "Enter the string: ";
	cin.getline(str, 100);
	cout << "Enter the time of show: ";
	cin >> time;

	ShowStr(str, time);
	cout << endl;

	ShowStr(str);

	//以下是时延和回车读取，防止看不到结果
	clock_t delay = 50 * CLOCKS_PER_SEC;   
	clock_t start = clock();
	while (clock() - start <delay)
		;
	cin.get();  
	cin.get();  
	return 0;
}

void ShowStr(const char * str, int print_time)    //记住定义函数的时候不写默认参数，而是在原型上默认
{
	if(print_time > 0)   //输入0次的时候则不显示
	{
		cout << str << " / ";
		print_time--;
		ShowStr(str, print_time);
	}
}

题8.2：

#include <iostream>
#include <ctime>
using namespace std;

struct CandyBar
{
	char name[128];
	double weight;
	int calorie;
};

void fill(CandyBar & candy_n, char * name_n, double weight_n, int calorie_n);
void show(const CandyBar & candy_n);

int main()
{
	CandyBar candy1;
	fill(candy1, "Millennium Munch", 2.85, 350);
	show(candy1);

	//以下是时延和回车读取，防止看不到结果
	clock_t delay = 50 * CLOCKS_PER_SEC;   
	clock_t start = clock();
	while (clock() - start <delay)
		;
	cin.get();  
	cin.get();  
	return 0;
}

void fill(CandyBar & candy_n, char * name_n, double weight_n, int calorie_n)
{
	strcpy(candy_n.name, name_n);    //实验证明C++使用strcpy函数是不用包含头文件<cstring>的
	candy_n.weight = weight_n;
	candy_n.calorie = calorie_n;
}

void show(const CandyBar & candy_n)
{
	cout << "Name: " << candy_n.name << endl;
	cout << "Weight: " << candy_n.weight << endl;
	cout << "Calorie: " << candy_n.calorie << endl;
}

题8.3：

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

void to_upper(string & srt);

int main()
{
	string your_string;
	cout << "Enter a string (q to quit): ";
	getline(cin, your_string);   //记住getline会读掉回车，但是在读数字之后用一定要先把回车读掉

	while (your_string != "q")
	{
		to_upper(your_string);
		cout << your_string << endl;

		cout << "Next string (q to quit): ";
		getline(cin, your_string);
	}


	//以下是时延和回车读取，防止看不到结果
	clock_t delay = 50 * CLOCKS_PER_SEC;   
	clock_t start = clock();
	while (clock() - start <delay)
		;
	cin.get();  
	cin.get();  
	return 0;
}

void to_upper(string & srt)
{
	for (int i = 0; i < srt.length(); i++)
	{
		srt[i] = toupper(srt[i]);
	}
}

题8.4：（这里要注意）

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

struct stringy
{
	char * str;
	int ct;
};

void set(stringy & your_stringy, char * your_string);
void show(const stringy & your_stringy, int time = 1);
void show(const char * ar, int time = 1);

int main()
{
	stringy beany;
	char testing[] = "Reality isn't what it used to be.";
	set(beany, testing);

	show(beany);
	show(beany, 2);

	testing[0] = 'D';
	testing[1] = 'u';

	show(testing);
	show(testing, 2);
	show("Done!");


	//以下是时延和回车读取，防止看不到结果
	clock_t delay = 50 * CLOCKS_PER_SEC;   
	clock_t start = clock();
	while (clock() - start <delay)
		;
	cin.get();  
	cin.get();  
	return 0;
}

void set(stringy & your_stringy, char * your_string)
{
    int string_length = strlen(your_string);
    your_stringy.str = new char[string_length + 1];  //可以理解为给your_stringy.str这个指针定义是什么样的指针
                                                     //一定要用[]而不是()不然会有问题，我看别人写的这里就有bug
                                                     //这里我认为是函数局部作用new，所以不用delete
    strcpy(your_stringy.str, your_string);
    your_stringy.ct = string_length;
}

void show(const stringy & your_stringy, int time)
{
	for (int i = 0; i < time; i++)
        cout << your_stringy.str << endl;
}

void show(const char * ar, int time)
{
	for (int i = 0; i < time; i++)
        cout << ar << endl;
}

题8.5：（这道题我因为一个中文下输入的括号导致bug，我找了半天才找到啊，所以提醒在座各位一定要注意哈哈哈！！！）

#include <iostream>
#include <ctime>
using namespace std;

template <typename AnyType>
AnyType maxn(AnyType *ar);

int main()
{
	int int_array[5] = {1, 2, 5, 4, 3};
	double double_array[5] = {2.4, 4.1, 3.1, 5.1, 5.4};

	cout << maxn(int_array) << endl;
	cout << maxn(double_array) << endl;

	//以下是时延和回车读取，防止看不到结果
	clock_t delay = 50 * CLOCKS_PER_SEC;   
	clock_t start = clock();
	while (clock() - start <delay)
		;
	cin.get();  
	cin.get();  
	return 0;
}

template <typename AnyType>
AnyType maxn(AnyType  *ar)
{
	AnyType max = ar[0];
	for (int i = 1; i < 5; i++)
	{
		if (ar[i] > max)
			max = ar[i];
	}
	return max;
}

题8.6：

#include <iostream>
#include <ctime>
using namespace std;

template <typename AnyType>
AnyType maxn(AnyType *ar, int size);

template <> const char * maxn(const char *str[], int n);

int main()
{
	int int_array[6] = { 43, 235, 54, 232, 123, 65 };
    double double_array[4] = { 32.1, 453.2, 53.3, 67.4 };
    const char * str_array[5] = { "A", "AB", "ABC", "ABCDE", "ABCD" }; //str_array[5]是五个字符串的指针数组
                                                                       //可以看做str_array[0]="A"
                                                                       //可以看做是二维数组

    int int_max = maxn(int_array, 6);
    double double_max = maxn(double_array, 4);
    const char * length_max_str = maxn(str_array, 5);

    cout << "max of int array: " << int_max << endl;
    cout << "max of double array: " << double_max << endl;
    cout << "max length string of string array: " << length_max_str << endl;

	//以下是时延和回车读取，防止看不到结果
	clock_t delay = 50 * CLOCKS_PER_SEC;   
	clock_t start = clock();
	while (clock() - start <delay)
		;
	cin.get();  
	cin.get();  
	return 0;
}

template <typename AnyType>
AnyType maxn(AnyType *ar, int size)
{
    AnyType max = ar[0];
    for (int i = 0; i < size; i++)
    {
        if (max < ar[i])
        {
            max = ar[i];
        }
    }
    return max;
}

template <> const char * maxn(const char *str[], int n)
{
    const char * ps = str[0];

    for (int i = 0; i < n; i++)
    {
        if (strlen(ps) < strlen(str[i]))    //注意这里比较的不是地址而是对应的字符串！！！！
        {
            ps = str[i];
        }
    }
    return ps;
}

题8.7：

#include <iostream>
#include <ctime>
using namespace std;

struct debts
{
	char name[50];
	double amount;
};

template <typename AnyType>
AnyType SumArray(AnyType ar[], int n);
template <typename AnyType>
AnyType SumArray(AnyType * ar[], int n);

int main()
{
	int thing[6] = {13, 31, 103, 301, 310, 130};
    debts mr_E[3] =
	{
		{"A", 2400.0},
		{"B", 1300.0},
		{"C", 1800.0}
	};
	double * pd[3];

	for (int i = 0; i <3; i++)
		pd[i] = &mr_E[i].amount;

	cout << "Sum of thing: " << SumArray(thing, 6) << endl;
	cout << "Sum of debts: " << SumArray(pd, 3) << endl;

	//以下是时延和回车读取，防止看不到结果
	clock_t delay = 50 * CLOCKS_PER_SEC;   
	clock_t start = clock();
	while (clock() - start <delay)
		;
	cin.get();  
	cin.get();  
	return 0;
}

template <typename AnyType>
AnyType SumArray(AnyType ar[], int n)
{
	AnyType sum =0;
	for (int i = 0; i < n; i++)
		sum += ar[i];
	return sum;
}

template <typename AnyType>
AnyType SumArray(AnyType * ar[], int n)
{
	AnyType sum =0;
	for (int i = 0; i < n; i++)
		sum += *ar[i];
	return sum;
}