Create an array using a pointer

Question

Basically I want to create a multidimensional array using a size variable passed to my class but am not sure how to do it.

I've tried doing

//Class header
Tile * gameTile;

//Class cpp
gameTile = new Tile[size][size];

It doesnt mind new Tile[size] but not [size][size]. Why is this?

Expression must have a constant value. Ive changed the prototype and that to const but nothing :/ — Split, Jun 02 '13 at 00:34
One of the dimension must be constant, read this http://stackoverflow.com/questions/936687/how-do-i-declare-a-2d-array-in-c-using-new — Djon, Jun 02 '13 at 00:38

David G · Answer 1 · 2013-06-02T02:01:59.957

4

That fails because only the first dimension of an allocated array may have a dynamic size. To overcome this limitation you can use a 2-dimensional vector:

std::vector<std::vector<Tile> > gameTile(size, std::vector<Tile>(size));

edited Jun 02 '13 at 02:01

answered Jun 02 '13 at 00:45

David G

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How would I declare that in a class header? How would I change the size of it depending on the parameter passed to the init function? – Split Jun 02 '13 at 02:53
@Split Well vector has a [`resize()`](http://en.cppreference.com/w/cpp/container/vector/resize) member function, so you can use that. – David G Jun 02 '13 at 02:54
As for the class header - where are you intending to use this? Inside main, for a class? – David G Jun 02 '13 at 02:58
Am intending to use this in a class – Split Jun 02 '13 at 23:48

score 1 · Answer 2 · answered Jun 02 '13 at 00:42

1

Like in the link of Vlad mentions, you could use vectors, or for something more quick you could use the following code to initialize the bidimentional array:

int i;
int ** intTile;
intTile = new  int*[5];

for (i = 0; i < 5; i++) {
    intTile[i] = new int[5];
}

answered Jun 02 '13 at 00:42

Abimael Martinez

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It's not really any quicker to code than a vector of vectors and not really quicker to run than a specialized matrix class. It fails at memory management much more than both, though. – chris Jun 02 '13 at 00:43

Anthony Raimondo · Answer 3 · 2013-06-02T16:16:50.647

0

you could do

Tile *gameTile = &((new Tile[size][size])[0][0]);

this store the address for the first element in your multy dimensional array to gameTile.

edited Jun 02 '13 at 16:16

answered Jun 02 '13 at 00:48

Anthony Raimondo

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`new Tile[size][size]` is not valid C++ – kotlomoy Jun 02 '13 at 01:33
@kotlomoy it works fine for me, are you saying you cant initialize an array using 'size'? – Anthony Raimondo Jun 02 '13 at 03:34
You can, only if `size` is constant. – kotlomoy Jun 02 '13 at 09:34

score 0 · Answer 4 · answered Jun 02 '13 at 01:50

Although you want to use two dimensional arrays, I just want to point out for completeness that you can also use a one dimensional array for your purpose. If it is of dimension size by size you have:

gameTile = new Tile[size*size];
//now gameTile[y][x] becomes:
Tile t = gameTile[x + y*size];

//if array is not square
gameTile = new Tile[width*height];
//now gameTile[y][x] becomes:
Tile t = gameTile[x + y*width];

Create an array using a pointer

4 Answers4