Function pointer issue: How to provide pointer from non member function to member function

Question

i need to do something like this ..., in a project of mine.

class Alpha{
    
public:
    Alpha(void* p(int, int) = nullptr);
        
    void* calculatePointer;
    
    void test();
};
    

Alpha::Alpha(void* p(int, int)) : calculatePointer(p){};

Alpha::test(){
        calculatePointer(5, 10);
    }  


  
    
void calculate(int a, int b){
    std::cout << "cum: " << a +b << "\n";  
}
    
    
int main(){
    
    Alpha A = Alpha(&calculate);
    A.test();
    
}

and it results in in these errors:

error: invalid conversion from ‘void* (*)(int, int)’ to ‘void*’ [-fpermissive]
       15 |     : calculatePointer(p),
          |                        ^
          |                        |
          |                        void* (*)(int, int)
error: cannot initialize a member subobject of type 'void *' with an lvalue of type 'void *(*)(int, int)'
     error: expression cannot be used as a function
     In constructor 'Alpha::Alpha(void* (*)(int, int))':
    error: invalid conversion from 'void (*)(int, int)' to 'void* (*)(int, int)' [-fpermissive]
note: initializing argument 1 of 'Alpha::Alpha(void* (*)(int, int))'

this is just a dummy, but that's what i gotta do.
how is it done correctly?

Answer 1

If you're confused about the exact syntax of a function pointer it's probably best to define a type alias.

If you use using, the "rhs" is <return type> (*)(<parameter types>)

class Alpha{
    
public:
    using FunctionPtr = void (*)(int, int);

    Alpha(FunctionPtr p = nullptr) : calculatePointer(p) {}
        
    FunctionPtr calculatePointer;
    
    void test()
    {
        if (calculatePointer != nullptr)
        {
            calculatePointer(5, 10);
        }
    }
};
    
void calculate(int a, int b){
    std::cout << "sum: " << (a + b) << "\n";  
}

Btw: the correct syntax without a type alias would be

Alpha(void (*p)(int, int) = nullptr);

the brackets are necessary, since the compiler treats void *p(int, int) as (void*) p(int, int).