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kponenation 11-23-2005 11:50 AM

Double pointers and two dimensional arrays in C
 
If a function takes a double pointer like this:

find(int **array){.....}

to call this function I have
Code:

int height = 3;
int width = 3;
int **b;
  int total,i,j;
  b=(int **)malloc(height*(sizeof(int *)));
  for (i=0; i<height; i++){
    b[i]=(int *)malloc(width*sizeof(int));
    b[3][3] = {{1,1,0},{0,0,1},{1,0,1}}; // I GET AN ERROR here
    find(b);
  };

when I compile this i get an error at this line "b[3][3] = {{1,1,0},{0,0,1},{1,0,1}}; "
How can I fix this?

I'm little confused about the two dimenstional array and double pointers.

Thank you,
James

tuxdev 11-23-2005 11:57 AM

try making your array like
Code:

int **array=new (int *)[width]
for(int i=0;i<width;i++)array[i]=new int[height];


kponenation 11-23-2005 12:03 PM

thanks but i get an error at this line

int **array=new (int *)[width];

tuxdev 11-23-2005 12:19 PM

try making width and height const. What the errors are is more important than just the line number. Post them!

dmail 11-23-2005 12:24 PM

tuxdev please look at the title for this thread. C doesn't have the new operator, this was introduced in C++.

kponenation 11-23-2005 12:27 PM

so i also tried int **array=(int *)[width]; but it still doesn't work.

kponenation 11-23-2005 12:29 PM

and the error i get is " syntax error before '[' token"

dmail 11-23-2005 12:35 PM

Quote:

Originally posted by kponenation
so i also tried int **array=(int *)[width]; but it still doesn't work.
if you want to dynamicly create it C you need to use malloc,calloc etc.
I havn't post the answer yet,because im trying to figure it myself, its a long time since i used C.

Dark_Helmet 11-23-2005 12:37 PM

I haven't tried to compile this, but please correct me if I'm wrong. Your assignment is not doing what you think.
Code:

b[3][3] = {{1,1,0},{0,0,1},{1,0,1}};
First of all, index '3' is out of bounds. For a height and width of 3, the row and column indices range in value from 0 to 2.

Second, "b[3][3]" is a reference to a single int. "{{1,1,0},{0,0,1},{1,0,1}}" is a literal 3x3 integer array. You can't store an array in an int.

You'll need to reexamine how you're approaching your data. There looks like some confusion exists on how to deal with pointers, arrays, or both.

dmail 11-23-2005 12:55 PM

that quite true Dark_Helmet
@kponenation there is nothing wrong with your creation of the array
Code:

        int height = 3;
        int width = 3;
        int i;
        int**b;
        b=(int **)malloc( height * ( sizeof(int*) ) );

        for (i=0; i<height; i++)
                { b[i] = (int*)malloc( width * sizeof(int) ); }

if you hadn't of already created the array, you could create and fill on the fly like so:
Code:

int b[3][3] = { {1,1,0}, {0,0,1}, {1,0,1} };

kponenation 11-23-2005 12:57 PM

i'm pretty sure b[3][3] = {{1,1,0},{0,0,1},{1,0,1}};
is correct for

1 1 0
0 0 1
1 0 1

dmail 11-23-2005 01:04 PM

Quote:

Originally posted by kponenation
i'm pretty sure b[3][3] = {{1,1,0},{0,0,1},{1,0,1}};
is correct for

1 1 0
0 0 1
1 0 1

Im afraid its not. you have already created the array like i said, if you hadn't then you could do it by declaring and initialising in the same line like the above post.
what you are trying to do here Dark_Helmet has pointed(haha) out is incorrect.

kponenation 11-23-2005 01:05 PM

then how do i assign values to the 2 dimensional array?

and thanks for your help

dmail 11-23-2005 01:12 PM

it depends on where the values come from, but in my experience:
b[0][0] = 1;
b[0][1] = 1;
b[0][2] = 0;
if anybody know of another way(not including loops etc) i would like to hear it.

tuxdev 11-23-2005 01:13 PM

Code:

b[0][0]=1;
b[0][1]=1;
b[0][2]=0;
b[1][0]=0;
...

I don't know of any elegant way of doing it.

kponenation 11-23-2005 01:16 PM

i have a 20x20 array that i need to assign values (can't be random values)
that's why i want to know if i could do a[] []={{........}}
but i guess I need to type in 400 values by typing...
Thank you dmail and tuxdev.
If you have any suggestion please reply me.

James

dmail 11-23-2005 01:20 PM

well as an alternative create and intailise the array in a header file and make it global, you then dont need to pass the address of the array from func to func and neither do you have to create the array using memory funcs.

kalebris 11-24-2005 02:50 AM

As far as I know the array[i][j] means *(*array+i)+j correct me if i'm wrong and this is a simple int value. As they told you cannot store a 2 dimensional array in an int. But the
Code:

{
    int array[3][3]={{ , , ,},{ , , , },....};
    ...
    ...
}

is working fine at the beginning of the block, but not in the middle and unfortunately i and j have to be calculable in compile time.

vivekr 11-24-2005 03:15 AM

May be like this
Code:

int m[][3]={{1,2,3},{5,6,7},{8,9,10}};
And I think to you Mr.kpone... I gave a long string comparison program. You can very well refer to it on how to dynamically allocate two dimensional arrays

kalebris 11-24-2005 03:53 AM

sorry my bad. Could you explain shortly why is it working? This pointer stuff is not totaly clear for me.

bigearsbilly 11-24-2005 05:37 AM

Code:

b[3][3] = {{1,1,0},{0,0,1},{1,0,1}};
you can only use this syntax for initialisation.
I.e. when you declare the variable.
Not at runtime.

sylmarsh 03-06-2011 12:49 PM

2-D Array Initializing in C Programming
 
One can initialize a 2-D integer Array in C as below :

int b[3][3] = {1,1,0,0,0,1,1,0,1};

Pls. note that in all there are 9 elements (3 rows and 3 cols).

I know this post is quite old but just in case somebody still needs it.

ta0kira 03-06-2011 06:45 PM

You do realize the original poster could have completed both a bachelor's and a master's in the time that's passed since 2005...
Kevin Barry

Peverel 03-16-2011 11:43 AM

The key to this problem is that C does not have multidimensional arrays, only one dimensional ones. The declaration

int **b;

creates a pointer to an array, each element of which is a pointer to a one dimensional array. This structure could be set up by

b = (int **)malloc(3*sizeof(int *));
for (i=0;i<3;i++) b[i] = (int *)malloc(3*sizeof(int));

Now

b[3][3]=1;

sets the last element of the array. Note that

b[3][3] = {{1,1,0},{0,0,1},{1,0,1}};

was wrong on two counts: b[3][3] would be an integer variable if it had existed, which it did not; in any case it would be out of range, since arrays are zero-indexed.

Gnu C has an extension: a declaration like

int b[3][3];

sets up a local 3 by 3 array, which may be initialised by

int b[3][3] = {{1,1,0},{0,0,1},{1,0,1}};

but observe that this is an initialisation, performed at compile time, not an assignment. Notice also that both dimensions must be given explicitly. I suspect that b in this case is really an array of 9 integers which is treated as two dimensional, but I cannot be sure. It may be possible to assign (in either method) something like

b[1]={0,0,1};

I cannot say without trying it, not possible on a borrowed MAC without Linux.

There is a fundamental difference between the original C version and the Gnu extension. Malloc creates space on the heap, which is global, whereas the extension allocates local space, scoped to the function in which b is declared.


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