C++PrimerPlus 第七章 函数-C++的编程模块（编程练习含答案）

1、编写一个程序，不断要求用户输入两个数，直到其中的一个为0。对于每两个数，程序将使用一个函数来计算它们调和平均数，并将结果返回给main()，而后者将报告结果。调和平均数指的是倒数平均值的倒数，计算公式如下：

        调和平均数 = 2.0 * x * y / (x + y)

答案：

#include<iostream>
using namespace std;
double average(double, double);
int main()
{
double x, y;
cout << "Enter two numbers: ";
while (cin >> x >> y)
{
if (x == 0 || y == 0)
break;
double ave = average(x, y);
cout << "Average: " << ave << endl;
cout << "Enter two numbers: ";
}
return 0;
}
double average(double x, double y)
{
double ave;
ave = 2.0 * x * y / (x + y);
return ave;
}

2、编写一个程序，要求用户输入最多10个高尔夫成绩，并将其存储在一个数组中。程序允许用户提早结束输入，并在一行上显示所有成绩，然后报告平均成绩。请使用3个数组处理函数来分别进行输入、显示和计算平均成绩。

答案：

#include <iostream>
using namespace std;
int input(double grade[], int limit);
void show(double grade[], int num);
double compute(double grade[], int num);
int main() {
const int limit = 10;
double socre[limit];
int num = input(socre, limit);
show(socre, num);
cout << "平均值为：" << compute(socre, num) << endl;
system("pause");
return 0;
}
int input(double grade[], int limit) {
int i;
double sc;
for (i = 0; i < limit; i++) {
cout << "enter the scores of golf(q to quit!):";
cin >> sc;
if (!cin) {
cin.clear();
while (cin.get() != 'n')
continue;
cout << "end entering!n";
break;
}
grade[i] = sc;
}
return i;
}
void show(double grade[], int num) {
for (int i = 0; i < num; i++)
cout << grade[i] << endl;
}
double compute(double grade[], int num) {
double sum = 0;
for (int i = 0; i < num; i++)
sum += grade[i];
return double(sum / num);
}

3、下面是一个结构声明：

        struct box

        {

                char maker[40];

                float height;

                float width;

                float length;

                float volume;

        };

a. 编写一个函数，按值传递box结构，并显示每个成员的值。

b. 编写一个函数，传递box结构的地址，并将volume成员设置为其他三维长度的乘积。

c. 编写一个使用这两个函数的简单程序。

答案：

#include <iostream>
using namespace std;
struct box
{
char maker[40];
float height;
float width;
float length;
float volume;
};
void ShowBox(box);
float VolumeBox(box*);
int main() {
cout << "Please Enter The Box Information:n";
box b;
cout << "Enter the maker: ";
cin >> b.maker;
cout << "Enter the height: ";
cin >> b.height;
cout << "Enter the width: ";
cin >> b.width;
cout << "Enter the length: ";
cin >> b.length;
b.volume = VolumeBox(&b);
ShowBox(b);
return 0;
}
void ShowBox(box b) {
cout << "maker: " << b.maker << endl;
cout << "height: " << b.height << endl;
cout << "width: " << b.width << endl;
cout << "length: " << b.length << endl;
cout << "volume: " << b.volume << endl;
}
float VolumeBox(box* b) {
b->volume = b->height * b->width * b->length;
return b->volume;
}

4、许多州的彩票发行机构都使用如程序清单7.4所示的简单彩票玩法的变体。在这些玩法中，玩家从一组被称为域号码（field number）的号码中选择几个。例如，可以从域号码1~47中选择5个号码；还可以从第二区间（如1~27）选择一个号码（称为特选号码）。要赢得头奖，必须正确猜中所有的号码。中头奖的几率是选中所有域号码的几率与选中特选号码几率的乘积。例如，在这个例子中，中头奖的几率是从47个号码中正确选取5个号码的几率与从27个号码中正确选择1个号码的几率的乘积。请修改程序清单7.4，以计算中得这种彩票头奖的几率。

答案：

#include<iostream>
using namespace std;
long double probability(unsigned number, unsigned numbers);
int main()
{
double total, choices, temp;
cout << "Enter the total number of choices on the game card andn";
cout << "the number of picks allowed and the second section(<= total number):" << endl;
while ((cin >> total >> choices >> temp) && choices <= total && temp <= total)
{
cout << "you have one chance in ";
cout << probability(total, choices) * probability(temp, 1);
cout << " of winning.n";
cout << "Next two numbers (q to quit): ";
}
cout << "byen";
return 0;
}
long double probability(unsigned number, unsigned numbers)
{
long double result = 1.0;
long double n;
unsigned p;
for (n = number, p = numbers; p > 0; n--, p--)
result = result * n / p;
return 1.0 / result;
}

5、定义一个递归函数，接受一个整数参数，并返回该参数的阶乘。前面讲过，3的阶乘写作3!，等于3*2!，依此类推；而0!被定义为1。通用的计算公式是，如果n大于零，则n! = n*(n-1)!。在程序中对该函数进行测试，程序使用循环让用户输入不同的值，程序将报告这些值的阶乘。

答案：

#include <iostream>
using namespace std;
int factorial(int x);
int main() {
int n;
while (true) {
cout << "Please Enter n: ";
cin >> n;
while (!cin) {
cin.clear();
while (cin.get() != 'n')
continue;
}
cout << "The factorial of " << n << " is: " << factorial(n) << endl;
}
return 0;
}
int factorial(int x) {
int n;
if (x == 0) {
n = 1;
}
else {
n = x * factorial(x - 1);
}
return n;
}

6、编写一个程序，它使用下列函数：

Fill_array()将一个double数组的名称和长度作为参数。它提示用户输入double值，并将这些值存储到数组中。当数组被填满或用户输入了非数字时，输入将停止，并返回实际输入了多少个数字。

Show_array()将一个double数组的名称和长度作为参数，并显示该数组的内容。

Reverse_array()将一个double数组的名称和长度作为参数，并将存储在数组中的值的顺序反转。

程序将使用这些函数来填充数组，然后显示数组；反转数组，然后显示数组；反转数组中除第一个和最后一个元素之外的所有元素，然后显示数组。

答案：

#include<iostream>
using namespace std;
const int SIZE = 10;
int Fill_array(double[], int);
void Show_array(double[], int);
void Reverse_array(double[], int);
int main()
{
double ar[SIZE];
int i = Fill_array(ar, SIZE);
cout << "Show Array:n";
Show_array(ar, i);
cout << "Reverse Array:n";
Reverse_array(ar, i);
Show_array(ar, i);
return 0;
}
int Fill_array(double ar[], int n) {
int i = 0;
double number;
cout << "Enter " << n << " numbers: " << endl;
while (i<n && cin >> number) {
ar[i++] = number;
}
return i;
}
void Show_array(double ar[], int n) {
for (int i = 0; i < n; i++) {
cout << ar[i] << " ";
}
cout << endl;
}
void Reverse_array(double ar[], int n) {
double temp;
for (int i = 0; i < n / 2; i++) {
temp = ar[i];
ar[i] = ar[n - 1 - i];
ar[n - 1 - i] = temp;
}
}

7、修改程序清单7.7中的3个数组处理函数，使之使用两个指针参数来表示区间。fill_array()函数不返回实际读取了多少个数字，而是返回一个指针，该指针指向最后被填充的位置；其他的函数可以将该指针作为第二个参数，以标识数据结尾。

答案：

#include<iostream>
using namespace std;
const int Max = 5;
double* fill_array(double* ar, double* limit);
void show_array(const double* ar, double* size);
void revalue(double r, double* ar, double* size);
int main()
{
double properties[Max];
double* size = fill_array(properties, properties + Max);
show_array(properties, size);
if (size > 0)
{
cout << "Enter revaluation factor: ";
double factor;
while (!(cin >> factor))
{
cin.clear();
while (cin.get() != 'n')
continue;
cout << "Bad input; Please enter a number: ";
}
revalue(factor, properties, size);
show_array(properties, size);
}
cout << "Done.n";
return 0;
}
double* fill_array(double* ar, double* limit)
{
double temp;
int i = 0;
double* p;
for (p = ar; p < limit; p++,i++)
{
cout << "Enter value #" << (i + 1) << ": ";
cin >> temp;
if (!cin)
{
cin.clear();
while (cin.get() != 'n')
continue;
cout << "Bad input; input process terminated.n";
break;
}
else if (temp < 0)
break;
*p = temp;
}
return p;
}
void show_array(const double* ar, double* size)
{
for (int i = 0; ar < size; ar++,i++)
{
cout << "Property #" << (i + 1) << ": $";
cout << *ar << endl;
}
}
void revalue(double r, double* ar, double* size)
{
for (int i = 0; ar < size; ar++,i++)
*ar = (*ar) * r;
}

8、在不使用array类的情况下完成程序清单7.15所做的工作。编写两个这样的版本：

a. 使用const char *数组存储表示季度名称的字符串，并使用double数组存储开支。

b. 使用const char *数组存储表示季度名称的字符串，并使用一个结构，该结构只有一个成员——一个用于存储开支的double数组。这种设计与使用array类的基本设计类似。

答案：

a.

#include<iostream>
using namespace std;
const int Seasons = 4;
const char* Snames[Seasons] = { "Spring", "Summer","Fall","Winter" };
void fill(double ar[], const int n);
void show(double ar[], const int n);
int main()
{
double arPay[Seasons];
fill(arPay,Seasons);
show(arPay,Seasons);
return 0;
}
void fill(double ar[], const int n)
{
for (int i = 0; i < n; i++)
{
cout << "Enter " << Snames[i] << " expenses";
cin >> ar[i];
}
}
void show(double ar[], const int n)
{
double total = 0.0;
cout << "nEXPENSESn";
for (int i = 0; i < n; i++)
{
cout << Snames[i] << ": $" << ar[i] << endl;
total += ar[i];
}
cout << "Total Expenses: $" << total << endl;
}

b.

#include<iostream>
using namespace std;
const int Seasons = 4;
const char* Snames[Seasons] = { "Spring", "Summer","Fall","Winter" };
struct pay
{
double arPay[4];
};
void fill(pay *p, const int n);
void show(pay *p, const int n);
int main()
{
pay* expenses = new pay;
fill(expenses,Seasons);
show(expenses,Seasons);
return 0;
}
void fill(pay *p, const int n)
{
for (int i = 0; i < n; i++)
{
cout << "Enter " << Snames[i] << " expenses";
cin >> p->arPay[i];
}
}
void show(pay *p, const int n)
{
double total = 0.0;
cout << "nEXPENSESn";
for (int i = 0; i < n; i++)
{
cout << Snames[i] << ": $" << p->arPay[i] << endl;
total += p->arPay[i];
}
cout << "Total Expenses: $" << total << endl;
}

9、这个练习让您编写处理数组和结构的函数。下面是程序的框架，请提供其中描述的函数，以完成该程序。

#include<iostream>
using namespace std;
const int SLEN = 30;
struct student
{
char fullName[SLEN];
char hobby[SLEN];
int ooplevel;
};
// getinfo() has two arguments: a pointer to the first element of
// an array of student structures and an int representing the
// number of elements of the array. The function solicits and
// stores data about students. It terminates input upon filling
// the array or upon encountering a blank line for the student
// name. The function returns the actual number of array elements
// filled.
int getinfo(student pa[], int n);
//display1() takes a student structures as an argument
//and displays its contents
void display1(student st);
//display2() takes the address of student structure as an
//argument and displays the structure's contents
void display2(const student* ps);
//display3() takes the address of the first element of an array
//of student structures and the number of array elements as
//arguments and displays the contents of the structures
void display3(const student pa[], int n);
int main()
{
cout << "Enter class size: ";
int class_size;
cin >> class_size;
while (cin.get() != 'n')
continue;
student* ptr_stu = new student[class_size];
int entered = getinfo(ptr_stu, class_size);
for (int i = 0; i < entered; i++)
{
display1(ptr_stu[i]);
display2(&ptr_stu[i]);
}
display3(ptr_stu, entered);
delete[] ptr_stu;
cout << "Donen";
return 0;
}

答案：

#include<iostream>
using namespace std;
const int SLEN = 30;
struct student
{
char fullName[SLEN];
char hobby[SLEN];
int ooplevel;
};
int getinfo(student pa[], int n);
void display1(student st);
void display2(const student* ps);
void display3(const student pa[], int n);
int main()
{
cout << "Enter class size: ";
int class_size;
cin >> class_size;
while (cin.get() != 'n')
continue;
student* ptr_stu = new student[class_size];
int entered = getinfo(ptr_stu, class_size);
for (int i = 0; i < entered; i++)
{
display1(ptr_stu[i]);
display2(&ptr_stu[i]);
}
display3(ptr_stu, entered);
delete[] ptr_stu;
cout << "Donen";
return 0;
}
int getinfo(student pa[], int n)
{
for (int i = 0; i < n; i++)
{
cout << "Enter the information of NO." << i + 1 << " student:n";
cout << "Fullnam: ";
cin.getline(pa[i].fullName, SLEN);
cout << "Hobby: ";
cin.getline(pa[i].hobby, SLEN);
cout << "OOPLevel: ";
cin >> pa[i].ooplevel;
cin.get();
}
return n;
}
void display1(student st)
{
cout << "Fullname: " << st.fullName << endl;
cout << "Hobby: " << st.hobby << endl;
cout << "OOPLevel: " << st.ooplevel << endl;
}
void display2(const student* ps)
{
cout << "FullName: " << ps->fullName << endl;
cout << "Hobby: " << ps->hobby << endl;
cout << "OOPLevel: " << ps->ooplevel << endl;
}
void display3(const student pa[], int n)
{
for (int i = 0; i < n; i++)
{
cout << "Fullname: " << pa[i].fullName << endl;
cout << "Hobby: " << pa[i].hobby << endl;
cout << "OOPLevel: " << pa[i].ooplevel << endl;
}
}

10、设计一个名为calculate()的函数，它接受两个double值和一个指向函数的指针，而被指向的函数接受两个double参数，并返回一个double值。calculate()函数的类型也是double，并返回被指向的函数使用calculate()的两个double参数计算得到的值。例如，假设add()函数的定义如下：

        double add(double x, double y)

        {

                return x + y;

        }

则下述代码中的函数调用将导致calculate()把2.5和10.4传递给add()函数，并返回add()的返回值（12.9）：

        double q = calculate(2.5, 10.4, add);

请编写一个程序，它调用上述两个函数和至少另一个与add()类似的函数。该程序使用循环来让用户成对地输入数字。对于每对数字，程序都使用calculate()来调用add()和至少一个其他的函数。如果读者爱冒险，可以尝试创建一个指针数组，其中的指针指向add()样式的函数，并编写一个循环，使用这些指针连续让calculate()调用这些函数。提示：下面是声明这种指针数组的方式，其中包含三个指针：

        double (*pf[3])(double, double);

可以采用数组初始化语法，并将函数名作为地址来初始化这样的数组。

答案：

#include<iostream>
using namespace std;
double add(double, double);
double subtract(double, double);
double multiply(double, double);
double(*p[3])(double, double) = { add,subtract,multiply };
double calculate(double, double, double(*pf)(double, double));
int main()
{
double x, y;
cout << "Enter two numbers: n";
while (cin >> x >> y) {
cout << x << " + " << y << " = " << calculate(x, y, p[0]) << endl;
cout << x << " - " << y << " = " << calculate(x, y, p[1]) << endl;
cout << x << " * " << y << " = " << calculate(x, y, p[2]) << endl;
cout << "Enter two numbers: n";
}
cout << "Done!n";
return 0;
}
double calculate(double n, double m, double(*pf)(double a, double b)) {
return(*pf)(n, m);
}
double add(double x, double y) {
return x + y;
}
double subtract(double x, double y) {
return x - y;
}
double multiply(double x, double y) {
return x * y;
}

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