如何将 return 值存储在从指针到函数指针数组的指针中？

Question

这个例子 (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只占用一个元素？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？

Answer 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，所以它指向一个元素。这取决于你，你调用的函数和它的合同，知道之后是否还有元素。

如何将 return 值存储在从指针到函数指针数组的指针中？

How does Storing a return value in a pointer from a pointer -to- array of function pointers work?

c++

pointers

function-pointers