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Home/ Questions/Q 724121
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T06:11:09+00:00 2026-05-14T06:11:09+00:00

I’m using a library that defines an interface: template<class desttype> void connect(desttype* pclass, void

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I’m using a library that defines an interface:

template<class desttype>
void connect(desttype* pclass, void (desttype::*pmemfun)());

and I have a small hierarchy

class base {
   void foo();
};

class derived: public base { ... };

In a member function of derived, I want to call

connect(this, &derived::foo);

but it seems that &derived::foo is actually a member function pointer of base; gcc spits out

error: no matching function for call to ‘connect(derived* const&, void (base::* const&)())’

I can get around this by explicitly casting this to base *; but why can’t the compiler match the call with desttype = base (since derived * can be implicitly cast to base *)?

Also, why is &derived::foo not a member function pointer of derived?

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  1. Editorial Team
    Editorial Team
    2026-05-14T06:11:09+00:00Added an answer on May 14, 2026 at 6:11 am

    Firstly, when you do &class::member the type of the result is always based on the class that member actually declared in. That’s just how unary & works in C++.

    Secondly, the code does not compile because the template argument deduction fails. From the first argument it derives that desttype = derived, while from the second one it derives that desttype = base. This is what makes the compilation to fail. The template argument deduction rules in C++ don’t consider the fact that this can be converted to base * type. Moreover, one can argue that instead of converting this to base * type, the proper way would be to convert &derived::foo from pointer-to-base-member to pointer-to-derived-member type. Both approaches are equally viable (see below).

    Thirdly, member pointers in C++ obey the rules of contra-variance, which means that a pointer to a base class member can be implicitly converted to a pointer to a derived class member. In your case, all you need to do is to help the compiler get through template argument deduction by specifying the argument explicitly, and the code should compile

     connect<derived>(this, &derived::foo);
    

    The above should compile because of contra-variance of &derived::foo pointer, even though it is a pointer to base member. Alternatively you can do

     connect<base>(this, &derived::foo);
    

    This should also compile because of covariance of this pointer.

    You can also use explicit casts on the actual arguments (as you mention in the question) to get through the deduction ambiguity, but in my opinion in this case the explicitly specified template argument looks better.

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