Access the elements of a multidimensional array through a pointer in another class

Question

I have a int costMap[20][20]; inside a dialog class.

Inside that class I am making an object from another class called pathFind

int costMap[20][20];
//costMap is filled with values (that represent the cost for each node -> pathFinder)
int (*firstElement)[20] = costMap;
pathFind *finder = new pathFind(*firstElement);

I would like to be able to acces the values of the multidimensional array inside the pathFind class

int (*costMap)[20];

pathFind::pathFind(int *map){
    costMap = &map;
}

However this doesn't seem to work. I get an "cannot convert int** to int(*)[20] error.

Question: How can you access the elements of a multidimensional array through a pointer in another class

Class names may not have . in them.

Lightness Races in Orbit
– Lightness Races in Orbit

2013-01-05 20:31:31 +00:00
Commented Jan 5, 2013 at 20:31 — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jan 5, 2013 at 20:31
What line gives the error?

Code-Apprentice
– Code-Apprentice

2013-01-05 20:32:05 +00:00
Commented Jan 5, 2013 at 20:32 — Code-Apprentice
– Code-Apprentice, Commented Jan 5, 2013 at 20:32
@LightnessRacesinOrbit: :p

Thomas
– Thomas

2013-01-05 20:34:39 +00:00
Commented Jan 5, 2013 at 20:34 — Thomas
– Thomas, Commented Jan 5, 2013 at 20:34

Sarfaraz Nawaz · Accepted Answer · 2013-01-05 20:38:27Z

1

You should do this:

pathFind::pathFind(int (*map)[20] ){
    costMap = map;
}

That is, match the types!

Also note that T (*)[N] and T** are not compatible types. One cannot convert to other. In your code, you're trying to do that, which is what the error message tells you.

Besides, there are other issues with your code, such as you should avoid using new and raw-pointers. Use std::vector or other containers from the Standard library, and when you need pointer, prefer using std::unique_ptr or std::shared_ptr whichever suits your need.

edited Jan 5, 2013 at 20:38

answered Jan 5, 2013 at 20:33

Sarfaraz Nawaz

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1 Comment

Lightness Races in Orbit Over a year ago

It's not really an also. It's the same thing!

K-ballo · Accepted Answer · 2013-01-05 20:31:58Z

1

You would have to write

pathFind::pathFind(int (*map)[20]){ ... }

but this being C++ you may find this better:

template< std::size_t N, std::size_t M >
pathFind::pathFind(int (&map)[N][M]){ ... }

answered Jan 5, 2013 at 20:31

K-ballo

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1 Comment

K-ballo Over a year ago

@Lightness Races in Orbit: True, thank you. That goes to show I would never do something like that

Lightness Races in Orbit · Accepted Answer · 2013-01-05 20:32:21Z

1

pathFind::pathFind(int *map){

This expects a pointer to an integer.

Elsewhere, you've already discovered the correct type to use:

pathFind::pathFind(int (*map)[20]){

Try to avoid this pointer hackery, though.

answered Jan 5, 2013 at 20:32

Lightness Races in Orbit

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