深入探讨函数指针
2019-07-02 本文已影响0人
Luliang
最近复习了一下C++ 的函数指针部分,在此记录一下,例子来源与《C++ premier plus》,理解后还是不难的。
// arfupt.cpp -- an array of function pointers
#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 };
// 函数指针
const double *(*p1)(const double *, int) = f1;
auto p2 = f2; // 自动类型推导
// pre-C++0x 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是一个函数指针数组
const double *(*pa[3])(const double *, int) = { f1,f2,f3 };
// but it does work for initializing to a single value
// pb 是一个指向pa 第一个元素的指针
auto pb = pa;
// 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;
}
// 指向函数数组指针的指针
cout << "\nUsing pointers to an array of pointers:\n";
cout << " Address Value\n";
// easy way to declare pc
auto pc = &pa;
// pre-C++0x 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 pdb
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;
}