pointer to function c++

Question

guys! I am trying to do this:

int (*filterFunc)(Medicine* criteria, Medicine*);
DynamicVector<Medicine>* filter2(Medicine* criteria, filterFunc f); //error

but I get an error: 'filterFunc' is not a type

I am trying to do this because I want a general filter so then I can do this:

int filterPrice(Pet* pet) {
    if (pet->price > 10 && pet->price < 100) {
        return 0;
    }
    return 1;
}

VectorDinamic* filter2(Pet* criteria, filterFunc f) {
    VectorDinamic* v = getAll(ctr->repo);
    VectorDinamic* rez = creazaVectorDinamic();
    int nrElems = getNrElemente(v);
    int i;
    for (i = 0; i < nrElems; i++) {
        Pet* pet = get(v, i);
        if (!f(criteria, pet)) {
            add(rez, copyPet(pet));
        }
    }
    return rez;
}

VectorDinamic* filterByPrice(float price) {
    Pet* criteria = createPet(1, "", "", price);
    VectorDinamic* rez = filter2(ctr, criteria, filterByPriceGr);
    destroyPet(criteria);
    return rez;
}

How can I solve this problem?

kennytm · Accepted Answer · 2013-04-06 10:27:46Z

9

You forgot a typedef, to declare the type. Otherwise this declaration just creates a variable of type int(*)(Medicine*,Medicine*).

  typedef int (*filterFunc)(Medicine* criteria, Medicine*);
//^^^^^^^

answered Apr 6, 2013 at 10:27

kennytm

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1 Comment

Ulrich Eckhardt Over a year ago

If you remember that any declaration can be turned into a type using typedef, it's much easier to read and write: typedef int filter_function(medicine*, medicine*) which you can then use as filter_function* f, which also doesn't blur the fact that it's a pointer.

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