Calling a function through pointer to an array of function pointers

Question

I am trying to understand the syntax to call the function through a pointer to an array of function pointers. I have array of function pointers FPTR arr[2], and a pointer to this array FPTR (vptr)[2] . But it gives me an error when trying to call through the pointer to an array

typedef int (*FPTR)();
int func1(){
        cout<<"func1() being called\n";
}
int func2(){
        cout<<"fun2() being called\n";
}

    FPTR arr[2] = {&func1,&func2};

    FPTR (*vptr)[2];
    vptr=&arr;

    cout<<"\n"<<vptr[0]<<endl;
    cout<<"\n"<<vptr[0]()<<endl;  // ERROR  when trying to call the first function

vptr is pointer, so first you should dereference it before accesing elements of array pointed to by it: (*vptr)[0](). — rafix07
– rafix07, Commented Sep 9, 2019 at 5:11
I had tried that but i couldn't understand why (*vptr)[0] prints 1 always no matter for [0] or [1] or any other element of an array. Sameways, cout<<func1 or cout<<&func1 also prints 1 always — anurag86
– anurag86, Commented Sep 9, 2019 at 5:14
If you want to print pointer you need to cast to void* otherwise ostream::operator<<(bool) is called. cout<<"\n"<< (void*)(*vptr)[0]<<endl; — rafix07
– rafix07, Commented Sep 9, 2019 at 5:21

ikh · Accepted Answer · 2019-09-09 05:17:01Z

5

vptr is pointer to an array, so you must dereference it to use the array.

#include <iostream>
using std::cout;
using std::endl;

typedef int (*FPTR)();
int func1(){
        cout<<"func1() being called\n";
        return 0;
}
int func2(){
        cout<<"fun2() being called\n";
        return 2;
}

int main(){
    FPTR arr[2] = {&func1,&func2};

    FPTR (*vptr)[2];
    vptr=&arr;

    cout<<"\n"<<vptr[0]<<endl;
    cout<<"\n"<<(*vptr)[0]()<<endl;
}

live example

Note that func1() and func2() must return value, or outputting their results would cause undefined behavior

answered Sep 9, 2019 at 5:17

ikh

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robthebloke · Accepted Answer · 2019-09-09 05:31:33Z

2

typedef int (*FPTR)();
int func1(){
        cout<<"func1() being called\n";
        return 1;

}
int func2(){
        cout<<"fun2() being called\n";
        return 2;
}

FPTR arr[2] = {func1, func2}; 

// call both methods via array of pointers
cout<<"\n"<< arr[0]() <<endl;
cout<<"\n"<< arr[1]() <<endl;

FPTR (*vptr)[2] = &arr;

// call both methods via pointer to array of pointers
cout<<"\n"<< vptr[0][0]() <<endl;
cout<<"\n"<< vptr[0][1]() <<endl;

// or... 
cout<<"\n"<< (*vptr)[0]() <<endl;
cout<<"\n"<< (*vptr)[1]() <<endl;

edited Sep 9, 2019 at 5:31

answered Sep 9, 2019 at 5:16

robthebloke

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1 Comment

ikh Over a year ago

Note that func1 and func2 does not return any value so cout << (*vptr)[0]() will cause undefined behavior

n. m. could be an AI · Accepted Answer · 2019-09-09 05:23:55Z

2

A pointer to an array is not needed here. A pointer to the first array element works.

FPTR *vptr;
vptr = arr;

// vptr[0]() works

A reference to an array is also ok.

FPTR (&vptr)[2] = arr;

// vptr[0]() still works

If for some reason you need a pointer to an array, you can:

FPTR (*vptr)[2];
vptr = arr;

// (*vptr)[0]() works

To avoid confusion, prefer std::array to plain arrays.

edited Sep 9, 2019 at 5:23

answered Sep 9, 2019 at 5:18

n. m. could be an AI

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