C++：函數指針數組

Question

假設我們有 2 個函數

foo() { cout << "Hello"; }
foo2() { cout << " wolrd!"; }

我如何創建一個指針數組（比如a ， b ）， a指向foo() ， b指向foo2() ？ 我的目標是將這些指針存儲在數組 A 中，然后遍歷 A 以執行這些函數。

Answer 1

您可以按如下方式使用類型化函數指針：

using FunPtrType = void(*)();

FunPtrType arr[]{&foo,  &foo2};
// or
std::array<FunPtrType, 2> arr2{&foo,  &foo2};

// ... do something with the array of free function pointers
// example
for(auto fun: arr2)
    fun();

Answer 2

有一個簡單的實現：

#include <iostream>
#include <vector>
using namespace std;
// Defining test functions
void a(){cout<<"Function A"<<endl;}
void b(){cout<<"Function B"<<endl;}

int main()
{
    /*Declaring a vector of functions 
      Which return void and takes no arguments.
    */
    vector<void(*)()> fonc;
    //Adding my functions in my vector
    fonc.push_back(a);
    fonc.push_back(b);
    //Calling with a loop.
    for(int i=0; i<2; i++){
        fonc[i]();
    }
    return 0;
}

Answer 3

這些天不需要 typedef，只需使用auto 。

#include <iostream>
void foo1() { std::cout << "Hello"; }
void foo2() { std::cout << " world!"; }
auto foos = { &foo1, &foo2 };
int main() { for (auto foo : foos) foo(); }

Answer 4

有兩種等效的方法可以執行您想要的操作：

方法一

#include <iostream>
void foo() 
{ 
    std::cout << "Hello";
}
void foo2() 
{ 
    std::cout << " wolrd!"; 
    
}


int main()
{
   
    void (*a)() = foo;// a is a pointer to a function that takes no parameter and also does not return anything
    
    void (*b)() = foo2;// b is a pointer to a function that takes no parameter and also does not return anything
    
    
    //create array(of size 2) that can hold pointers to functions that does not return anything and also does not take any parameter
    void (*arr[2])() = { a, b};
    
    arr[0](); // calls foo 
    
    arr[1](); //calls foo1
    
    return 0;
}

方法1可以在這里執行。

方法二

#include <iostream>
void foo() 
{ 
    std::cout << "Hello";
}
void foo2() 
{ 
    std::cout << " wolrd!"; 
    
}


int main()
{
   
    //create array(of size 2) that can hold pointers to functions that does not return anything
    void (*arr[2])() = { foo, foo2};
    
    arr[0](); // calls foo 
    
    arr[1](); //calls foo1
    
    return 0;
}

方法2可以在這里執行。