CODEWITHSUNDEEP
c++
JavaRunner
JavaRunner

Reputation: 2545

How to call a function by a pointer?

I have some base class in which I store a pointer to a function from derived class. I need to call a function by its pointer from two places: from BaseClass and from derived class.

template < class T >
class BaseClass {
private:
    typedef void ( T::*FunctionPtr ) ();
    FunctionPtr funcPtr;
public:
    void setFunc( FunctionPtr funcPtr ) {
        this->funcPtr = funcPtr;
        ( this->*funcPtr )(); // I need to call it here also but it doesn't work
    }
};

class DerivedClass: public BaseClass < DerivedClass > {
public:
    void callMe() {
        printf( "Ok!\n" );
    }
    void mainFunc() {
        setFunc( &DerivedClass::callMe );
        ( this->*funcPtr )(); // works fine here
    }   
};

Error: left hand operand to -> * must be a pointer to class compatible with the right hand operand, but is 'BaseClass *'

Upvotes: 0

Views: 1146

Answers (1)

R Sahu
R Sahu

Reputation: 206567

( this->*funcPtr )();

is the wrong syntax to use to call funcPtr since the type of funcPtr is void T::*().

You need to use:

( (static_cast<T*>(this))->*funcPtr )();

Upvotes: 4

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