从指向函数指针数组的指针中存储返回值如何工作？

Posted 2023-02-21

【中文标题】从指向函数指针数组的指针中存储返回值如何工作？

【英文标题】：How does Storing a return value in a pointer from a pointer -to- array of function pointers work?

【发布时间】：2018-10-20 23:54:26

【问题描述】：

此示例 (7.19) 摘自 Stephen Prata 的 C++ Primer 6th ed。

#include <iostream>
// various notations, same signatures 
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 ;

// pointer to function 
const double *(*p1)(const double *, int) = f1; 
auto p2 = f2; // C++11 automatic type deduction 
// pre-C++11 can use the following code instead 
// const double *(*p2) (const double *, int ) = 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; 

// pa an array of pointers 
// auto doesn't work with list initialization 
const double *(*pa[3])(const double *, int) =  f1, f2, f3 ;
// but it does work for initializing to a single value 
// pb a pointer to first element of pa 
auto pb = pa;
// pre-C++11 can use the following code instead 
// const double *(**pb) (const double *, int) = pa; 
cout << "\nUsing an array of pointers to functions:\n"; 
cout << " Address  Value\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;

// what about a pointer to an array of function pointers 
cout << "\nUsing pointers to an array of pointers:\n";
cout << " Address  Value\n";
// easy way to declare pc 
auto pc = &pa; 
// pre-C++11 can use the following code instead 
// const double *(*(*pc)[3])(const double *, int) = &pa;
cout << (*pc)[0](av, 3) << ": " << *(*pc)[0](av, 3) << endl; 
// hard way to declare pd 
const double *(*(*pd)[3]) (const double *, int) = &pa;
// store return value in pb 
const double * pdb = (*pd[1]) (av, 3);
cout << pdb << ": " << *pdb << endl; 
// alternative notation 
cout << (*(*pd)[2])(av, 3) << ": " << *(*(*pd[2]))(av, 3) << endl; 
// cin.get();
return 0;


// some rather dull functions 

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;

指针*pdb是怎么回事？指针是占用整个数组还是仅占用一个元素？当程序运行时，在声明 *pdb 的地方抛出异常。谁能告诉我里面发生了什么？

【问题讨论】：

“Stephen Prata 的 C++ Primer 第 6 版。” - Prata没有写过“C++ Primer” - 他写过“C++ Primer Plus”，一个非常糟糕的学习资源。

代码看起来确实不太好。

有什么更好的学习建议吗？

真正的“C++ 入门”？由实际从事 C++ 实现工作的人员撰写。

【参考方案1】：

pdb 和其他指针一样只是一个指针。它由函数调用返回到f2。

因为它只是一个指针，它指向一个元素。取决于您调用的函数及其合约，是否知道后面是否还有元素。