How to call a function using pointer-to-member-function

Question

I have a class:

class A {
    void test_func_0(int);
    void run();

    typedef void(A::*test_func_t)(int);

    struct test_case_t{
       test_func_t test_func;
    } test_case[100];
};

Now I want to call test_func() inside run():

void A::run() 
{
    test_case[0].test_func = &test_func_0;
    test_case[0].*(test_func)(1);
}

The last line of my code, doesn't work(compile error), no matter what combination I try.

How can I avoid syntax errors when calling a member function using a pointer-to-member-function? — Robᵩ
– Robᵩ, Commented Mar 7, 2011 at 16:05

Armen Tsirunyan · Accepted Answer · 2011-03-07 11:28:58Z

5

Use this:

void A::run() 
{   
    test_case[0].test_func = &A::test_func_0;
    (this->*(test_case[0].test_func))(1);
}

Notice that you had 2 errors. The first one was how you formed the member-function-pointer. Note that the only way to do it is to use &ClassName::FuncName regardless of whether you're at class scope or not. & is mandatory too.

The second is that when you call a member via a member function pointer, you must explicitly specif y the object (of type A in your case) on which to call the member function. In this case you must specify this (and since this is a pointer we use ->* rather than .*)

HTH

edited Mar 7, 2011 at 11:28

answered Mar 7, 2011 at 11:18

Armen Tsirunyan

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Comments

Erik · Accepted Answer · 2011-03-07 11:14:22Z

1

Use:

(this->*test_case[0].test_func)(1);

answered Mar 7, 2011 at 11:14

Erik

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Comments

Sarfaraz Nawaz · Accepted Answer · 2011-03-07 11:16:09Z

-1

Member function call using pointer-to-member-function:

 test_case[0].test_func = &A::test_func_0; //note this also!
(this->*test_case[0].test_func)(1);

Demo : http://www.ideone.com/9o8C4

edited Mar 7, 2011 at 11:16

answered Mar 7, 2011 at 11:10

Sarfaraz Nawaz

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2 Comments

Erik Over a year ago

test_func is an A:: member, not a test_case_t member.

Sarfaraz Nawaz Over a year ago

@Erik: Yeah. I noticed. Actually I was testing it at ideone. So corrected and posted the link to Ideone also!

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