Calling object function through object pointers

Question

My program contains a class, here's its simplified form:

Units.h:

#define fullwidth 200
#define fullheight 200

class Units
{
public:

[...]

void MovingUp(Units* Detector[fullwidth][fullheight], Units Self);
[...]
}

It defines a function that takes two arguments as input: an array of object pointers for "Units" objects, and a special chosen "Units" object.

Here's the main part:

#include "Units.h"

[...]

int i,j;
Units* DetectorField[fullwidth][fullheight];
Units Examples[20];

for (j=0;j<fullheight;j++)
{
    for (i=0;i<fullwidth;i++)
    {
        DetectorField[i][j] = NULL;
    }
}

It creates the array of "Units" object pointers that is necessary for the function in "Units" and sets all those values to NULL at the very beginning.

After that, I try to call the function of the object through one randomly chosen pointer. Obviously, if the pointer is referenced to NULL, the call is impossible, but Visual C++ shows error even before I run the program.

DetectorField[12][12]->MovingUp(DetectorField,DetectorField[12][12]);

For this version, Visual C++ writes after compiling: cannot convert parameter 2 from 'Units *' to 'Units' - of course, since DetectorField itself is an array of pointers. Then I try to reference to the object it points, this way:

DetectorField[12][12]->MovingUp(DetectorField,&DetectorField[12][12]);

MSVC writes after compiling: cannot convert parameter 2 from 'Units **' to 'Units'

There isn't clear for me, why the second version reference to the pointer of an object pointer, not to an object. Also, how should I write the function call on the right way?

DetectorField[12][12] is a pointer. So &DetectorField[12][12] is a reference to a pointer. To get the object, you dereference the pointer using *. Try DetectorField[12][12]->MovingUp(DetectorField, *DetectorField[12][12]); (I'm just answering that point and not paying attention to questions like has memory been allocated, why not use smart pointers, vectors, etc...) — Muckle_ewe
– Muckle_ewe, Commented Jul 20, 2013 at 0:22
I'm kinda amateur at using pointers. Basic things are OK for me, but pointers and the lot of memory-handling-related operations are a bit too much for my brain yet. I hope I can learn. Thanks! — Zoltán Schmidt
– Zoltán Schmidt, Commented Jul 20, 2013 at 0:26
BTW - +1 for being very clear what you don't understand and what is your question. — Ran Eldan
– Ran Eldan, Commented Jul 20, 2013 at 0:31
Well, I was banned from asking for a very long time, so I thought it over twice before I asked. Strict expectations, quality questions. — Zoltán Schmidt
– Zoltán Schmidt, Commented Jul 20, 2013 at 0:35

greatwolf · Accepted Answer · 2013-07-20 00:29:41Z

2

For this call to be valid:

DetectorField[12][12]->MovingUp(DetectorField, DetectorField[12][12]);

The second argument has to be a Unit * type like:

class Unit
{
  // ...
  void MovingUp(Units* Detector[fullwidth][fullheight], Units *Self);
};

If you want to keep Unit::MovingUp as is then change your call to:

DetectorField[12][12]->MovingUp(DetectorField, *DetectorField[12][12]);

Note you're passing Unit in by value here. If you have other classes inheriting from Unit you risk object slicing. I suggest at least changing it to take a Unit & reference.

answered Jul 20, 2013 at 0:29

greatwolf

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1 Comment

Zoltán Schmidt Over a year ago

Indeed, "Units" already has inherited classes, thanks for mentioning it! Can you explain, what should I change? void MovingUp(Units& Detector instead of void MovingUp(Units* Detector?

Ran Eldan · Accepted Answer · 2013-07-20 00:28:57Z

1

"There isn't clear for me, why the second version reference to the pointer of an object pointer, not to an object."

When & apears this way, it mean the address of.

"Also, how should I write the function call on the right way?"

What you should do either call it that way:

DetectorField[12][12]->MovingUp(DetectorField,*DetectorField[12][12]);
                                              ^

Which means, the value pointed by DetectorField[12][12] (since it's a pointers array).

Or change the funciton decleration to:

void MovingUp(Units* Detector[fullwidth][fullheight], Units* Self);
                                                           ^

Which means this function will recive a pointer to units.

answered Jul 20, 2013 at 0:28

Ran Eldan

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1 Comment

Zoltán Schmidt Over a year ago

Well, it totally makes sense - I just didn't know that this is a valid operation, since I haven't seen using * operator this way. Thank you!

Joel · Accepted Answer · 2013-07-20 00:23:26Z

1

The type of an element in your array is Units*. You declared it that way: Units* Detector[fullwidth][fullheight] - a two-dimensional array of Units*. The address of a Units* is type Units**.

answered Jul 20, 2013 at 0:23

Joel

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