Call function through pointer to class member

Question

In following code:

class foo { public: void foo_function() {}; }; class bar { public: foo foo_member; void bar_function(foo bar::*p_foo) { // what is the corrct sintax for following: this->*p_foo->foo_function(); // expression must have a pointer type?? } }; int main() { foo foo_obj; bar bar_obj; typedef foo bar::*p_foo; p_foo blah = &bar::foo_member; bar_obj.bar_function(blah); return 0; }

What would be correct syntax to make bar::bar_function work?

Precedence and type mismatch too: ((this->*p_foo).foo_function)(); — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Mar 4, 2015 at 22:13
@codekiddy p_foo blah = &bar::foo_member; In which way do you think that &bar::foo_member provides a member function pointer actually? — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Mar 4, 2015 at 22:27
@codekiddy Seems to be in eye of the reader :-P But definitely void bar_function(foo bar::*p_foo) wants a member function pointer. So how can you say we're not talking about member function pointers?? — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Mar 4, 2015 at 22:33
It wants a member pointer. foo is a class, not a function pointer typedef. — Austin Mullins
– Austin Mullins, Commented Mar 4, 2015 at 22:36

Austin Mullins · Accepted Answer · 2015-03-04 22:35:52Z

This works in ideone:

void bar_function(foo bar::*p_foo) { (this->*p_foo).foo_function(); }

It's all about having the right level of indirection. Since p_foo is a pointer to member, we need to dereference it before attempting to access it from this. At this point, you have the actual foo_member object, not a pointer to it, so you can call its foo_function by dot notation.

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