Question

Edit: sorry about the stupid title; by “pointer object” I think I mean dereferenced object pointer (if that’s any clarification).

I’m new to C++, and I’ve been trying to get used to its idiosyncrasies (coming from Java).

I know it’s not usually necessary, but I wanted to try passing an object by its pointer. It works, of course, but I was expecting this to work too:

void test(Dog * d) {
*d.bark();
}

And it doesn’t. Why is this? I found out that I can do the following:

void test(Dog * d) {
Dog dawg = *d;
dawg.bark();
}

and it works flawlessly. It seems so redundant to me.

Any clarification would be appreciated. Thanks!

Answer 1

You have to use the -> operator on pointers. The . operator is for non-pointer objects.

In your first code snippet, try d->bark();. Should work fine!

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EDIT: Other answers suggest (*d).bark();. That will work as well; the (*d) dereferences the pointer (ie. turns it into a normal object) which is why the . operator works. To use the original pointer, simply d, you must use the -> operator as I described.

Hope this helps!