C++ Primer Plus(第六版)第七章 函数--C++的编程模块复习题
7.13 编程练习
{
int x, y = 0;
cout << "请输入两个整数：" << endl;
while (cin>>x>>y)
{
if (x == 0 || y == 0)
{
cout << "输入不能为零 " << endl;
system("pause");
return 0;
}
//调用调和平均数函数
double avgnum = funcAvgnum(x,y);
cout << "avgnum = " << avgnum << endl;
}
system("pause");
return 0;
}
#include
double cal_average(double x1, double x2);
int main()
{
using namespace std;
double num1, num2, result;
cout << "Enter two number(enter 0 to quit): ";
cin >> num1 >> num2;
while(num1 * num2 != 0)
{
if(!cin)
{
cin.clear();
while (cin.get() != '\n')
continue;
cout << "Bad input, process will terminate.\n";
break;
}
result = cal_average(num1, num2);
cout << "The calculate reuslt is " << result << ".\n";
cout << "Enter next two number(enter 0 to quit): ";
cin >> num1 >> num2;
}
cout << "Bye\n";
return 0;
}
double cal_average(double x, double y)
{
double aver;
aver = 2.0 * x * y / (x + y);
return aver;
}
void showInfo(const double* arr, int size);
double mathInfo(const double* arr, int size);
#define MAXNUM 10
{
cout << "请输入最多10个高尔夫成绩，提前结束请输入-1" << endl;
//输入函数,地址形式
//确定数组
double golfnum[MAXNUM];
int countnum = funcInput(golfnum, MAXNUM);
//显示函数
//此处函数参数可以设置成const,不被修改
showInfo(golfnum, countnum);
//计算函数
double avg = mathInfo(golfnum, countnum);
cout << countnum << "个平均数为：" << avg << endl;
return 0;
}
int funcInput(double *arr,int size)
{
double num;
int count = 0;
for (int i = 0; i < MAXNUM; i++)
{
cout << "第" << i+1 << "个成绩：" << endl;
cin >> num;
if (num == -1)
{
return count;
}
arr[i] = num;
count++;
}
return count;
}
void showInfo(const double* arr, int size)
{
if (0 == size)
{
cout << "没有数据进行展示，请验证后再操作，感谢；" << endl;
return;
}
for (int i = 0; i < size; i++)
{
cout << "第" << i + 1 << "个成绩为：" << arr[i] << endl;
}
return;
}
double mathInfo(const double* arr, int size)
{
double totalnum = 0;
double avgnum = 0;
if (0 == size)
{
cout << "没有数据进行计算，请验证后再操作，感谢；" << endl;
return avgnum;
}
for (int i = 0; i < size; i++)
{
totalnum += arr[i];
}
avgnum = totalnum / size;
return avgnum;
}
转载答案:
// ex2.cpp -- calculate the average grade of golf
#include
const int Max = 10;
int input_ar(double *grades, int Max);
void show(double grade[], int n);
double cal_av(double grade[], int n);
int main()
{
using namespace std;
double grade[Max];
double average_g;
int len = input_ar(grade,Max);
show(grade,len);
average_g = cal_av(grade,len);
cout << "Average grade = " << average_g << endl;
return 0;
}
int input_ar(double *grades, int n)
{
using namespace std;
int count = 0;
cout << "Enter the grade of golf(enter q to quit): ";
while(count < n && cin >> grades[count])
{
count++;
if(count == 10)
break;
cout <<"Next grade(enter q to quit): ";
}
return count;
}
void show(double grade[], int n)
{
using namespace std;
cout << "All grades: ";
for(int i = 0; i < n; i++)
cout << grade[i] << " ";
cout << endl;
}
double cal_av(double grade[], int n)
{
double sum = 0;
for(int i = 0; i < n; i++)
sum += grade[i];
return sum / n;
}
float showAddrBoxInfo(box *add);
//a．编写一个函数，按值传递box结构，并显示每个成员的值。
showBoxInfo(mybox);
//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。
box yourbox = { "foot",20,20,30 };
float yourvolume = showAddrBoxInfo(&yourbox);
cout << "your volume = " << yourvolume << endl;
cout << "your volume = " << yourbox.volume <
void showBoxInfo(box mybox)
{
cout << "make:" << mybox.make << endl;
cout << "height:" << mybox.height << endl;
cout << "width:" << mybox.width << endl;
cout << "length:" << mybox.length << endl;
cout << "volume:" << mybox.volume << endl;
}
float showAddrBoxInfo(box* add)
{
cout << "your height:" << add->height << endl;
cout << "your length:" << add->length << endl;
cout << "your width:" << add->width << endl;
add->volume = add->height * add->width * add->length;
return add->volume;
}
转载答案：
// ex3.cpp -- using struct with function
#include
struct box
{
char maker[40];
float height;
float width;
float length;
float volume;
};
void show(box b);
void set_volume(box *);
int main()
{
using namespace std;
box b1;
cout << "Enter the maker: ";
cin.getline(b1.maker,40);
cout << "Enter the box of hight: ";
cin >> b1.height;
cout << "Enter the box of width: ";
cin >> b1.width;
cout << "Enter the box of length: ";
cin >> b1.length;
set_volume(&b1);
show(b1);
return 0;
}
void show(box b)
{
using namespace std;
cout << "Maker: " << b.maker << endl;
cout << "Hight: " << b.height << endl;
cout << "Width: " << b.width << endl;
cout << "length: " << b.length << endl;
cout << "Volume: " << b.volume << endl;
}
void set_volume(box *pb)
{
pb->volume = pb->height * pb->width * pb->length;
}
#include
long double cal_property(int r1, int n, int r2);
int main()
{
using namespace std;
int range1, range2, num;
cout << "Enter the number of field number: ";
cin >> range1;
cout << "Enter the number of choose field number: ";
cin >> num;
cout << "Enter the number of specific number: ";
cin >> range2;
long double property = cal_property(range1, num, range2);
cout << "The property of get money is one in " << property << endl;
return 0;
}
long double cal_property(int r1, int n, int r2)
{
long double result = 1.0;
for(int i = r1,j = n; j > 0; i--, j--)//conditon write wrongly
result = result * i / j;
result *= r2;
return result;
}
int total = 1;
int cinnum;
while (cin >> cinnum)
{
total = 1;
int rcuresult = recursiveFunc(cinnum);
cout << "递归结果为：" << rcuresult << endl;
if (cinnum == 0)
{
system("pause");
return 0;
}
}
{
if (num > 0)
{
total = num * recursiveFunc(num - 1);
}
return total;
}
#include
long long cal_factorial(int n);
int main()
{
using namespace std;
int num;
long long factorial;
cout << "Enter the number(enter q to quit): ";
while(cin >> num)
{
factorial = cal_factorial(num);
cout << num << "! = " << factorial << endl;
cout << "Enter next number(enter q to quit): ";
}
cout << "Done\n";
return 0;
}
long long cal_factorial(int n)
{
long long result = 1;
if(n > 1)
result = n * cal_factorial(n-1);
else if(n == 1 || n ==0)
result *=1;
return result;
}
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。当数组被填满或者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,并将存储在该数组中的值顺序反转。
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。
转载答案：
本题填充函数用while循环和if语句；显示函数用for循环即可；反转函数使用for循环，关键时for(int i = 0, j = len-1; i < j; i++,j--) // the last number of array is array[len-1]
// ex6.cpp -- using function to deal with array
#include
const int Size = 5;
int Fill_array(double [], int len);
void Show_array(double arr[], int len);
void Reverse_array(double *, int len);
int main()
{
using namespace std;
double grade[Size];
int len = Fill_array(grade, Size);
Show_array(grade, len);
Reverse_array(grade, len);
cout << "Reverse array:\n";
Show_array(grade, len);
Reverse_array(grade + 1, len -2);
cout << "Reverse part of array:\n";
Show_array(grade,len);
return 0;
}
int Fill_array(double arr[], int len)
{
using namespace std;
int count = 0;
cout << "Enter a number(enter q to quit): ";
while(cin >> arr[count])
{
count++; // the position of count++ is important
if(count >= len)
break;
cout << "Enter next number(enter q to quit): ";
}
return count;
}
void Show_array(double arr[], int len)
{
using namespace std;
for(int i = 0; i < len; i++)
cout << arr[i] << " ";
cout << endl;
}
void Reverse_array(double *par, int len)
{
double temp;
for(int i = 0, j = len-1; i < j; i++,j--)
{
temp = par[i];
par[i] = par[j];
par[j] = temp;
}
}
7. 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。
转载答案：
首先从函数声明开始改，改完开始改函数定义，改完函数定义改函数调用。
代码如下：
//ex7.cpp -- array functions and point
#include
const int Max = 5;
double * fill_arr(double *, double *);
void show_arr(double *, double *);
void revalue(double r, double *, double *);
int main()
{
using namespace std;
double property[Max];
double * ptend = fill_arr(property, property + Max);
show_arr(property, ptend);
if(ptend - property > 0)
{
cout << "Enter the revaluation facctor: ";
double factor;
while(!(cin >> factor))
{
cin.clear();
while(cin.get() != '\n')
continue;
cout << "Bad input; Please enter a number: ";
}
revalue(factor, property, ptend);
show_arr(property, ptend);
}
cout << "Done.\n";
cin.get();
cin.get();
return 0;
}
double * fill_arr(double * begin, double * end)
{
using namespace std;
double temp;
double * pt;
int count = 0;
for(pt = begin; pt != end; pt++) // for condition is pt != end, not pt == end
{
cout << "Enter value #" << (count + 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;
*pt = temp;
count++;
}
return pt;
}
void show_arr(double * begin, double * end)
{
double *pt;
int i = 0;
for(pt = begin; pt != end; pt++)
std::cout << "Property #" << (i + 1)
<< " : $" << *pt << std::endl;
i++;
}
void revalue(double r, double * begin, double * end)
{
double *pt;
for(pt = begin; pt != end; pt++)
*pt *= r;
}
8. 在不使用array类的情况下完成程序清单7.15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。
答案转载：
本题编写程序时，使用const char *数组存储表示季度的字符串，刚开始一直编写不出来，想复杂了，以为需要一个二维的字符数组才可以表示，实际上声明一个char *类型的数组，存储四个字符串；该代码:const char *Sname[4] = {"Spring", "Summer", "Winter"};
调用显示时有两种方法，分别是：cout << Sname[i];cout << *(Sname + i)
a的代码如下：
// ex8_a.cpp -- functions with double array(C++11)
#include
#include
using namespace std;
const int Seasons = 4;
const char* Sname[Seasons] =
{"Spring", "Summer", "Fall", "Winter"};
void fill(double *,int n);
void show(double [], int);
int main()
{
double expense[Seasons];
fill(expense, Seasons);
cout << "Your expense list:\n";
show(expense,Seasons);
return 0;
}
void fill(double *pta, int n)
{
double temp;
int count = 0;
cout << "Enter your expense of every season, enter q to quit.\n";
while(count < n)
{
cout << Sname[count] << ": " ;
cin >> temp;
if(!cin)
{
cin.clear();
while(cin.get() != '\n')
continue;
cout << "Bad input, process will terminated.";
break;
}
else if(temp < 0)
break;
*(pta + count) = temp;
count ++;
}
}
void show(double arr[], int n)
{
for (int i = 0; i < n; i++)
{
cout << Sname[i] << ": " << arr[i] << endl;
}
}
b.代码如下：
// ex8_b.cpp -- functions with double structure(C++11)
#include
#include
const int Seasons = 4;
const char* Sname[Seasons] =
{"Spring", "Summer", "Fall", "Winter"};
struct Expense
{
double expense[Seasons];
};
using namespace std;
void fill(double *,int n);
void show(double [], int);
int main()
{
Expense ex;
fill(ex.expense, Seasons);
cout << "Your expense list:\n";
show(ex.expense,Seasons);
return 0;
}
void fill(double *pta, int n)
{
double temp;
int count = 0;
cout << "Enter your expense of every season, enter q to quit.\n";
while(count < n)
{
cout << *(Sname + count) << ": " ;
cin >> temp;
if(!cin)
{
cin.clear();
while(cin.get() != '\n')
continue;
cout << "Bad input, process will terminated.";
break;
}
else if(temp < 0)
break;
*(pta + count) = temp;
count ++;
}
}
void show(double arr[], int n)
{
for (int i = 0; i < n; i++)
{
cout << Sname[i] << ": " << arr[i] << endl;
}
}
9. 这个练习让您编写处理数组和结构的函数。下面时程序框架，请提供其中描述的函数，完成该程序。
// ex9.cpp -- wirte the function
#include
using namespace std;
const int SLEN = 30;
struct student
{
char fullname[SLEN];
char hobby[SLEN];
int ooplevel;
};
// getinfo() has tow arguments: a pointer to the first element of
// an array of student structure and an int representing the
// number og element 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 structure as an argument
// and diplay its contents
void display1(student st);
// display2() takes the address of student structure as an
// argument and diplay the structure's contents
void display2(student * ps);
// display3() takes the address of the first element of an array
// of student structure as and the number of array elements as
// arguments and diplay the structure's contents
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 << "Done\n";
return 0;
}
转载答案：
本题实现几个函数的功能不难，在实现输入姓名为空格跳出程序时，出现了问题，原因是使用char字符串数组用==与空行字符串比较，因此不能实现，后想起应该使用strcmp()函数，于是实现该功能。
代码如下：
// ex9.cpp -- wirte the function
#include
#include
using namespace std;
const int SLEN = 30;
struct student
{
char fullname[SLEN];
char hobby[SLEN];
int ooplevel;
};
// getinfo() has tow arguments: a pointer to the first element of
// an array of student structure and an int representing the
// number og element 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 structure as an argument
// and diplay its contents
void display1(student st);
// display2() takes the address of student structure as an
// argument and diplay the structure's contents
void display2(student * ps);
// display3() takes the address of the first element of an array
// of student structure as and the number of array elements as
// arguments and diplay the structure's contents
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 << "Done\n";
return 0;
}
int getinfo(student pa[], int n)
{
int count = 0;
for(int i = 0; i < n; i ++)
{
cout << "Student #" << i+1 << ":\n";
cout << "Enter the name(enter blank line to quit): ";
cin.getline(pa[i].fullname, SLEN);
if(!strcmp(pa[i].fullname," ")) //
break;
count++;
cout << "Enter the hobby: ";
cin.getline(pa[i].hobby,SLEN);
cout << "Enter the opplevel: ";
cin >> pa[i].ooplevel;
cin.get();
}
return count;
}
void display1(student st)
{
cout << "Fullname: " << st.fullname << endl;
cout << "Hobby: " << st.hobby << endl;
cout << "Ooplevel: " << st.ooplevel << endl;
}
void display2(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 << "Student #" << i + 1 << ":\n";
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的输入数字。对于每对数字，程序都是用calculate()来调用add()和至少一个其他函数。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，使用这些指针函数连续让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);
转载答案：
// ex10.cpp -- using point to function
#include
double calculate(double , double , double (*p) (double , double ));
double add(double x, double y);
double sub(double x, double y);
double multi(double x, double y);
double div(double x, double y);
int main()
{
using namespace std;
double (*pf[4]) (double , double) = {add, sub, multi, div};
char op[4] = {'+','-','*','/'};
double num1, num2;
cout << "Enter two number(enter q to quit): ";
while(cin >> num1 >> num2)
{
for(int i = 0; i < 4; i++)
{
double result = calculate(num1,num2, pf[i]);
cout << num1 << " "<< op[i] << " " << num2 << " = " << result << endl;
}
cout << "Enter next two number(enter q to quit): ";
}
cout << "Done\n";
return 0;
}
double calculate(double x, double y, double (*p) (double , double ))
{
return (*p)(x,y); // return p(x,y) is right
}
double add(double x, double y)
{
return x + y;
}
double sub(double x, double y)
{
return x - y;
}
double multi(double x, double y)
{
return x * y;
}
double div(double x, double y)
{
return x / y;
}
写在最后：
为什么最后几道题只有转载答案了，因为实在写不动了，坚持一件事真的好难，一旦放松了警惕，就会变得懒散，不愿意思考，C++技术博大精深，现在才看到第七章，之前每次都没有坚持下来，但愿这次能走的再远一些。