函數與對象： 沒啥内容 直接上示例吧
#include <iostream>
#include <string>
using namespace std;
const int SIZE = 5;
void display(const string sa[], int n);
int main()
{
string list[SIZE];
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);
return 0;
}
void display(const string sa[], int n)
{
for (int i =0; i < n; i++)
cout << i + 1 << ": " << sa[i] << endl;
}
函數與array
#include <iostream>
#include <string>
#include <array>
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, Seasons> expenses;
fill(&expenses);
show(expenses);
return 0;
}
void fill(std::array<double, Seasons>* pa)
{
using namespace std;
for (int i = 0; i < Seasons; i++)
{
cout << "Enter " << Snames[i] << " expenses: ";
cin >> (*pa)[i];
}
}
void show(std::array<double, Seasons> da)
{
using namespace std;
double total = 0.0;
cout << "\nEXPENSES\n";
for (int i = 0; i < Seasons; i++)
{
cout << Snames[i] << ": $" << da[i] << endl;
total += da[i];
}
cout << "Total Expenses: $" << total << endl;
}
留意 傳入的是位址還是array的值
遞歸：
簡單遞歸調用：
#include <iostream>
void countdown(int n);
int main()
{
countdown(4);
return 0;
}
void countdown(int n)
{
using namespace std;
cout << "Counting down ... " << n << " (n at " << &n << ")" << endl;
if ( n > 0)
countdown(n-1);
cout << n << ": Kaboom!" << " ( n at " << &n << ")" << endl;
}
複雜一丢丢的遞歸調用：
二分法畫标尺
#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] = ' ';
}
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);
}
輸出結果
函數指針：
double pam (int)
函數名：位址
函數名()：函數的傳回值
double *pf(int) 表示的是一個傳回double類型指針的函數
double（*pf)(int) 表示 函數的指針
将函數指派給函數指針 傳回類型一定要與指針類型相同
pf = pam
通過函數指針來調用函數：
double y = (*pf)(5) 或者 pf(5)
示例：
#include <iostream>
double betsy(int);
double pam(int);
void estimate(int lines, double (*pf)(int));
int main()
{
using namespace std;
int code;
cout << "How many lines of code do you need? ";
cin >> code;
cout << "Here's Betsy's estimate:\n";
estimate(code , betsy);
cout << "Here's Pam's estimate:\n";
estimate(code, pam);
return 0;
}
double betsy(int lns)
{
return 0.05 * lns;
}
double pam(int lns)
{
return 0.03 * lns + 0.0004 * lns * lns;
}
void estimate(int lines, double (*pf)(int))
{
using namespace std;
cout << lines << " lines will take ";
cout << (*pf)(lines) << " hour(s)\n";
}
奇怪 為毛沒有顔色 算了不管他
嗯 深入函數指針（變态來一波）
const double* f1(const double ar[], int n)
const double* f2(const double [], int n)
const double* f3(const double*, int n)
這三個函數聲明是同一個意思
改成用函數指針表示（直接初始化）：
const double* (*p1)(const double*, int n) = f1;
可以使用 auto自動推斷類型
auto p2 = f2;
那麼 如下調用函數都是OK的 (*p1)(av, 2) , p2(av, 2) (av是一個double數組）
然後：
造門搞一個函數指針數組 指向這3個函數
const double* (*pa[3])(const double*, int) = {f1, f2, f3};
這裡表示3個指針元素的函數數組 指向3個函數并傳回const double指針
這裡不能使用auto 因為auto不能用于初始化清單 隻能用于單值初始化
但可以這樣 auto pb = pa;
是以 現在pb和pa 都是指向函數指針數組第一個元素的指針（f1）
那麼 ， 如何用呢？
比如使用數組第二個元素
const double* px = pa[1[(av, 3)(傳參)
或者 (*pb[1])(av, 3)(我感覺 (*(pb + 1))(av, 3)也行）
如果要擷取傳回資料的值
const double x = *pa[1[(av, 3)
const double x = *(*pb[1])(av, 3)
然後再來一個 建立一個指向該函數指針數組的指針
阿西吧~
自動的辦法： auto pc = &pa;
手動的辦法：
const double* (*(*pd)[3])(const double *, int) = &pa;
解釋下
（*pd)[3]表示的是一個數組指針， 指向3個函數指針的數組
*(*pd)[3]表示的是該數組的指針，
const double* 表示的是函數傳回的資料類型 因為函數指針必須要跟函數的傳回值類型一緻
那麼在解引用方面
*pd表示的是該函數指針數組，* (*pd)[i](av, 3) 表示的是第i元素所指向的函數（要傳參）的傳回值資料(const double *)
*(*(*pd)[i])(av, 3) 是得到傳回值資料的值(const double)
然後針對數組指針：
**&pa == *pa == pa[0]
因為&pa代表的是整個數組的位址
卧槽 終于理順了 有點想吐了
示例：
#include <iostream>
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};
const double* (*p1)(const double*, int) = f1;
auto p2 = 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;
const double* (*pa[3])(const double*, int) = {f1, f2, f3};
auto pb = pa;
cout << "\nUsing an array of pointers to functions:\n";
cout << " Address Vlue\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;
cout << "\nUsing pointers to an array of pointers:\n";
cout << " Address Value\n";
auto pc = &pa;
cout << (*pc)[0](av, 3) << ": " << *(*pc)[0](av, 3) << endl;
const double* (*(*pd)[3])(const double*, int) = &pa;
const double* pdb = (*pd)[1](av, 3);
cout << pdb << ": " << *pdb << endl;
cout << (*(*pd)[2])(av, 3) << ": " << *(*(*pd)[2])(av, 3) << endl;
return 0;
}
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;
}
除了使用auto來簡化定義變量外還可以使用typedef
typedef const double* (*p_fun)(const double *, int)
p_fun = f1
p_fun pa[3] = {f1, f2, f3}
p_fun (*pd)[3] = &pa
第七章完結。我腦袋可能要休息下
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