Why is the base class public method not found?

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Why is cbase::foobar() not found when called on type cderived while all members of cbase are declared public? This suggests name only is utilized for lookup and signature ignored but that is difficult to believe as one of the benefits of C++ is function overloading and selection based on argument type I would greately appreciate a reference to the language rule at cppreference.com explaining this Thank You Kindly

struct cbase
{
	void foobar() {}
};

struct cderived : public cbase
{
	void foobar(int) {}
};

int main()
{
	cderived d;
	d.foobar(); //'cderived::foobar': function does not take 0 arguments
}

What I have tried:

I tried waving a dead rubber chicken over the computer but that did not help I also tried reading cppreference.com re/ name lookup but did not find a rule explaining the compile error I also tried using cbase::foobar in cderived which solved the problem but I do not understand why it would be necessary

Posted 3-Jul-21 21:17pm

BernardIE5317

Updated 4-Jul-21 22:25pm

v2

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[no name] 4-Jul-21 3:39am

Have a look to the compiler warnings (I hope they are enabled and that the chicken not picked them :-) ).
Most probably you will find something like cderived :: foobar(int) hides cbase :: foobar()

[no name] 4-Jul-21 3:59am

Have a look here and dearch for 'hides': Standard C++[^]

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Richard MacCutchan · Answer 1 · 2021-07-04T22:25:00

Variable d is an instantiation of a cderived class. So any call to foobar requires an integer to be passed to it, as per the signature of the function in that class. In order to call the base version you need to cast d to a cbase object thus:

C++

cderived d;
static_cast<cbase>(d).foobar(); // call the base function that takes 0 arguments

BernardIE5317 · Answer 2 · 2021-07-03T22:18:00

Solution 2

My thanks to dead.rubber.chicken for directing me to the language rule which I must accept so will utilize using directive to solve problem but still do not understand why the language refuses to support what seems a logical extension to the concept of overloading

Posted 3-Jul-21 22:18pm

BernardIE5317

Comments

[no name] 4-Jul-21 4:33am

As you wrote, this seems simply to be a rule.

Especially because the compiler recognizes that there is another - better- match (and therefore the warning) I recognize no reason why not the method of base class will be taken.

Maybe somebody other can explain the 'why'

[no name] 4-Jul-21 9:47am

It won't let me go, thank you;)
There seems to be no real explanation that the c ++ compiler won't accept this. The same example (inheritance and method names) work e.g. with C # as you / I would expect.