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  • 【C++】函数和指针

    最近在看C++ primer plus,感觉函数与指针这一章难点比较多,记写笔记,加强理解.

    From C++ Primer Plus: Chapter 7 Function:C++ Programming Modules 

    1. 如何声明函数指针?

    和函数原型类似: 需要声明指针指向函数的返回值和参数列表

    double pam(int); //参数为int 类型,返回值为double 类型的函数
    double (*pf);(int)  //指向参数为int类型,返回值为double 类型的指针
    pf = pam;   //函数名代表了函数的地址
    
    double x = pam(4); //函数名调用
    double x = (*pf)(4); //指针调用
    double x = pf(4); //C++也允许将指针名当作函数名使用

    2. C++ 11 自动类型推断

    const double * f1(const double *, int);
    
    const double * (*p1)(const double *, int); //p1 poitns to f1
    auto p2 = f1; //C++11 automatic type deduction,p2 points to f1 as well

    3. 将指针名当作函数名使用

    //前面函数为double *类型,cout第一部分返回double指针,第二部分返回double指针指向的值
    cout<<(*p1)(av,3)<<":"<<*(*p1)(av,3)<<endl;
    //和上面的cout一样只不过是使用函数指针名来调用函数
    cout<<p2(av,3)<<":"<<*p2(av,3)<<endl;

    4.  函数指针数组

    const double *(*pa[3]) (const double *,int) = {f1,f2,f3}; //创建函数指针数组
    //通过指针调用函数,得到返回的指针
    const double *px = pa[0](av,3); //call by pointer as if it were a function name
    const double *py = (*pa[0])(av,3); //正常调用
    
    //得到函数返回指针指向的值
    double x = *pa[0](av,3);
    double x = *(*pa[0])(av,3);

    5. 指向指针数组的指针

    指针数组和数组指针的区别

    *pd[3] //an array of 3 pointers
    (*pd)[3] //a pointer to an array of three elements

     指向数组的指针

    1 auto pc = &pa;   //&pa是整个数组的地址, pa是数组第一个元素首地址
    2  
    3 const double * (*(*pd)[3])(const double *,  int ) = &pa; //和第一个等价
    4 
    5 **&pa = *pa = pa[0]

     代码:

     1 //arfupt.cpp -- an array of function pointers
     2 #include<iostream>
     3 //various notations,same signatures
     4 const double *f1(const double ar[],int n);
     5 const double *f2(const double [],int);
     6 const double *f3(const double *,int);
     7 
     8 int main()
     9 {
    10     using namespace std;
    11     double av[3] = {1112.3,1542.6,2227.9};
    12 
    13     //pointer to a function
    14 
    15     const double *(*p1)(const double *,int) = f1;
    16     auto p2 = f2;//C++ 11 utomatic  type deduction
    17     //pre-C++11 can use the following code instead
    18     //const double *(*p2)(const double *,int) = f2;
    19     cout<<"Using pointers to functions:
    ";
    20     cout<<"Address Value
    ";
    21     cout<<(*p1)(av,3)<<":"<<*(*p1)(av,3)<<endl;
    22     cout<<p2(av,3)<<":"<<*p2(av,3)<<endl;
    23 
    24     //pa an array of pointers
    25     //auto doesn't work with list initialization
    26     const double *(*pa[3])(const double *,int) = {f1,f2,f3};
    27     //pb a pointer to first element of pa
    28     auto pb = pa;
    29     // pre-C++11 can use the following code instead
    30     // const double *(**pb)(const double *, int) = pa;
    31     cout<<"
    Using an array of pointers to functions:
    ";
    32     cout<<"Address Value
    ";
    33     for(int i = 0;i < 3; i++)
    34         cout<<pa[i](av,3)<<":"<<*pa[i](av,3)<<endl;
    35     cout<<"
    Using a pointer to a pointer to a function:
    ";
    36     cout<<"Address Value
    ";
    37     for(int i = 0;i < 3; i++)
    38         cout<<pb[i](av,3)<<":"<<*pb[i](av,3)<<endl;
    39 
    40     //what about a pointer to an array of function pointers
    41     cout<<"
    Using pointers to an array of pointers:
    ";
    42     cout<<"Address Value
    ";
    43     //easy way to declare pc
    44     auto pc = &pa;
    45     // pre-C++11 can use the following code instead
    46     // const double *(*(*pc)[3])(const double *, int) = &pa;
    47     cout<<(*pc)[0](av,3)<<":"<<*(*pc)[0](av,3)<<endl;
    48     //hard way to declare pd
    49     const double *(*(*pd)[3])(const double *,int) = &pa;
    50     //store return value in pdb
    51     const double *pdb = (*pd)[1](av,3);
    52     cout<<pdb<<":"<<*pdb<<endl;
    53     //alternative notation
    54     cout<<(*(pd)[2])(av,3)<<":"<<*(*(*pd)[2])(av,3)<<endl;
    55 }
    56 
    57 const double * f1(const double * ar, int n)
    58 {
    59 return ar;
    60 }
    61 const double * f2(const double ar[], int n)
    62 {
    63 return ar+1;
    64 }
    65 const double * f3(const double ar[], int n)
    66 {
    67 return ar+2;
    68 }
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  • 原文地址:https://www.cnblogs.com/vincentcheng/p/7252700.html
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