Initializing array of structures with function pointer member in C++

Question

I am having trouble in initializing an array of structures with a function pointer as a member in it.

class Record
{
    private:
    typedef void (*display_fn_t) ();
    struct record {
        int a;
        display_fn_t disp;
    };
    static const record rec[];

    void disp1() { cout << "Display 1 with a string to display" << endl; }
    void disp2() { cout << "Display 2 with an integer to display" << endl; }

    public:
    int find() { /* logic to find the record comes here */ }
    void display() {
        for (int i = 0; i < 2; i++) {
            rec[i].disp();
        }
    }
}

const Record::record Record::rec[] = {
    { 10, disp1 },
    { 11, disp2 }
};

int main()
{
    Record r;
    if (r.find())
        r.display();
    return 0;
}

When I compile the above code, I am getting the following compilation error:

mca_record.cpp:56: error: argument of type ‘void (Record::)()’ does not match ‘void (*)()’

Pointers to member functions are not the same as pointers to non-member functions. There must be hundreds or even thousands of duplicates here on SO. The problem is that a pointer to a member function needs an actual object instance to be called on, while pointers to non-member functions doesn't. — Some programmer dude
– Some programmer dude, Commented Jun 18, 2015 at 5:41
Also, you might want to read about std::function and std::bind. — Some programmer dude
– Some programmer dude, Commented Jun 18, 2015 at 5:44

WhozCraig · Accepted Answer · 2015-06-18 06:27:05Z

1

Your syntax is wrong and isn't using the appropriate operators.

Fixing a multitude of syntax errors, and stripping out the unrelated find operation, then utilizing proper member function pointers and operator ->* gives the following (one of several ways to do this):

#include <iostream>

class Record
{
private:
    typedef void (Record::*display_memfn_t)();
    struct record
    {
        int a;
        display_memfn_t disp;
    };

    static const record rec[];

    void disp1() { std::cout << "Display 1 with a string to display" << std::endl; }
    void disp2() { std::cout << "Display 2 with an integer to display" << std::endl; }

public:
    void display();
};

const Record::record Record::rec[] =
{
    { 10, &Record::disp1 },
    { 11, &Record::disp2 }
};

void Record::display()
{
    for (size_t i=0; i<sizeof rec/sizeof*rec; ++i)
        (this->*(rec[i].disp))();
}

int main()
{
    Record r;
    r.display();
    return 0;
}

Output

Display 1 with a string to display
Display 2 with an integer to display

Compare it to your existing code, and not in particular that pointers to member functions are not simply pointers to functions. They require different handling and generally different operators to utilize. See here for different methods of member access (both variable and function).

Best of luck.

answered Jun 18, 2015 at 6:27

WhozCraig

66.5k11 gold badges82 silver badges148 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

boomerang Over a year ago

Thanks @WhozCraig. I need the find function as there goes some logic to fetch some values which are used in display functions. However, I can make use of few tweaks you mentioned. That's really cool.

marom · Accepted Answer · 2015-06-18 06:14:07Z

1

To make the call work you must invoke it like this:

        for (int i = 0; i < 2; i++) {
            (*rec[i].disp)();
        }

And initialize the table this way:

const Record::record Record::rec[] = {
    { 10, &Record::disp1 },
    { 11, &Record::disp2 }
};

answered Jun 18, 2015 at 6:14

marom

5,23813 silver badges15 bronze badges

1 Comment

boomerang Over a year ago

On looking closely, @WhozCraig has solved my problem already. But +1 for your short description attacking the problem statement directly. Thanks

Collectives™ on Stack Overflow

Initializing array of structures with function pointer member in C++

2 Answers 2

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related