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Array/Malloc question

Started by utwo007 Apr 22, 2002 at 3:50 PM 6 replies 1.2k views

Original Post

Apr 22, 2002 at 3:50 PM

Can I declare an array like this:

int myArray[100];

And then access the 30th element like this?:

myArray[3][10]

Also, can I do something like this:

int * myArray;
myArray = malloc(sizeof(int)*100);

And then access the 30th int like this?:

myArray[3][10];

The reason I ask is because I want to create a simply game with a map made of tiles. I will create a tile object and then create a 2D array of these tile objects to represent the map. However, different levels in the game will require different map sizes, so I want to be able to allocate the memory for the map currently in play dynamically, but also have the benifit of the 2D array. Would the following work? int * myArray = new int[x][y]; ---signature--- People get ready. I''m ready to play.

Fruny

Apr 22, 2002 at 4:29 PM

No.

Multidimensional arrays (MD arrays for short) can only be accessed in a MD fashion if they have been declared as MD arrays in the first place. Otherwise, the compiler has no way of knowing how large is each dimension.

You could create the MD array manually :

  
int** array;
array = new int*[x];
for( i=0; i<x; ++i )
  array[i] = new int[y];

array[5][5] = 3;

Then when deleting the array :

  
for( i=0; i<x;++i )
  delete[] array[i];
delete[] array;

Or, (much better), use a vector of vectors;

  
std::vector<vector<int> > array;
array.resize(x);
for(i=0; i<array.size(); ++i )
  array[i].resize(y);

array[5][5] = 3;

sjelkjd

Apr 22, 2002 at 4:36 PM

To expand on Fruny's post...
A more concise way to construct these vectors is as follows:

  
vector<int> base(y,0); //makes a vector y big, initialized with zeros

vector<vector<int> > bigvec(x,base); //makes a vector of vector<int>, all initialized with base

//now we have an x by y matrix

bigvec[5][5]=3;

Or if you wanna be really hardcore

  
vector<vector<int> > bigvec(x,vector<int>(y,0));

edit: html ate my < > s

[edited by - sjelkjd on April 22, 2002 5:36:43 PM]

eldee

Apr 22, 2002 at 4:38 PM

quote:
Original post by utwo007
Can I declare an array like this:

int myArray[100];

And then access the 30th element like this?:

myArray[3][10]

what you''re trying to do can''t be accomplished, due to the
fact taht you''re trying to access a single dimensioned
array in a multidimensional way. what you can do instead
of


myArray[3][10] to access the the 30th element, you can do this:
myArray[3*10] <= 3*10 being 30.. giving you the 30th element.

-eldee
;another space monkey;
[ Forced Evolution Studios ]

utwo007

Apr 22, 2002 at 4:54 PM

quote:

Original post by eldee
what you can do instead
of


myArray[3][10] to access the the 30th element, you can do this:
myArray[3*10] <= 3*10 being 30.. giving you the 30th element.

Very nice. I think I''ll do it this way.

---signature---
People get ready.
I''m ready to play.

Fruny

Apr 22, 2002 at 4:57 PM

quote:
Original post by sjelkjd
To expand on Fruny''s post...

/me should have remembered that.

Beer Hunter

Apr 22, 2002 at 8:19 PM

Here''s another way to access a one-dimensional array as if it were a two-dimensional array (untested):

  
int *one_dimensional_array = new int[width*height];
int **two_dimensional_array = new (int*)[height];
for(int i = 0; i < height; ++i) {
  two_dimensional_array[i] = &(one_dimensional_array[width * i]);
}
...
two_dimensional_array[row][column] = 500;
// same as one_dimensional_array[row * width + column] = 500;

...
delete[] two_dimensional_array;
delete[] one_dimensional_array;

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