Can you omit the "this" keyword when using function pointer to nonstatic member function

Question

I'm delving into the topic of non-static member function pointers, and I find the syntax most disturbing. These pointers are interesting because you can do:

struct MyStruct
{
    void nonStaticMembFunc1(){};
    void nonStaticMembFunc2()
    {
         void (MyStruct::*funcPtr)() = &nonStaticMembFunc1;   // Declare local pointer

         (this->*funcPtr)();   // Use it through this instance

    }
};

So if it's the case that you can always (as far as I know) drop the use of the "this" keyword, I tried to drop it in this case, with the following attempts:

(this->*nonStaticFuncPtr)();    // Normal use, works

(->*function1)();   // Trying to drop "this" keyword            
(*function1)();     // Trying to drop the arrow, this seemed to make sense
(.*function1)();    // Trying the other way

Trying this made me confused about what scope you're in when in a nonstatic member function. If you can drop the keyword then you're already in its scope (if that's the right term), like when you can drop the use of the scope resolution :: operator if you don't need it. But I'm probably completely wrong about this.

Is your question why is there an this pointer if one is already automatically in the class scope, so this is unnecessary? — Rakete1111
– Rakete1111, Commented Nov 25, 2016 at 13:11
&nonStaticMembFunc1 is invalid and shouldn't even compile. Pointers to member functions must be fully qualified: &MyStruct::nonStaticMembFunc1. BTW MyStruct's declaration misses an ending ; — rocambille
– rocambille, Commented Nov 25, 2016 at 13:13
@Rakete No, my question is if you can drop the use of it, like you can normally. You don't need it normally because when entering the function it already has the "this" pointer if I understand right. — Zebrafish
– Zebrafish, Commented Nov 25, 2016 at 13:13
@wasthishelpful Right, but seeing as though you're already in the class's scope, you don't need the MyStruct:: , I think, at least my Visual Studio isn't showing an error. — Zebrafish
– Zebrafish, Commented Nov 25, 2016 at 13:15
@TitoneMaurice Why not simply try out? "you don't need the MyStruct:: ..." That's wrong BTW. — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Nov 25, 2016 at 13:17

πάντα ῥεῖ · Accepted Answer · 2016-11-25 13:24:09Z

1

The answer is you can't:

struct MyStruct
{
    void nonStaticMembFunc1(){};
    void nonStaticMembFunc2()
    {
         void (MyStruct::*funcPtr)() = &MyStruct::nonStaticMembFunc1;   // Declare local pointer

         (*funcPtr)(); // See error message below

    }
};

main.cpp: In member function 'void MyStruct::nonStaticMembFunc2()':
main.cpp:9:12: error: invalid use of unary '*' on pointer to member
          (*funcPtr)();   // Use it through this instance
            ^~~~~~~

Live Demo

It's a member function pointer call. This appears separated from the current instances context, hence this isn't implicit there.

edited Nov 25, 2016 at 13:24

answered Nov 25, 2016 at 13:20

πάντα ῥεῖ

88.9k13 gold badges125 silver badges199 bronze badges

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

2 Comments

Zebrafish Over a year ago

Right, but when you go (funcPtr)() , the actual operator to call the nonstatic member function isn't there, it's supposed to be -> * if called through a pointer to the object, or . if called on the object itself.

πάντα ῥεῖ Over a year ago

@TitoneMaurice I updated my answer with an explanation why this isn't (and cannot be) assumed implicit for this case.

Collectives™ on Stack Overflow

Can you omit the "this" keyword when using function pointer to nonstatic member function

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related