Thread: Help with multi-dimentional arrays in C

Surely you've been here long enough to stop using void main
FAQ > main() / void main() / int main() / int main(void) / int main(int argc, char *argv[]) - Cprogramming.com

> I know when you pass a array as an argument it gets decomposed into a pointer, but with a multi-dimensional array this is not the case.
It's a pointer all-right, but it's type is just not what you expect.

Some examples

Code:

#include<stdio.h>
void display1(int array)
{
    printf("%d\n",array);
}
void display2(int array[],int size)
{
    for(int j=0;j<size;j++)
    {
        printf("%d\n",array[j]);
    }
}
void display3(int *array,int size)
{
    for(int j=0;j<size;j++)
    {
        printf("%d\n",array[j]);
    }
}
void display4(int array[][10],int size)
{
  for ( int i = 0 ; i < size ; i++ )
    for(int j=0;j<10;j++)
    {
        printf("%d\n",array[i][j]);
    }
}
void display5(int (*array)[10],int size)
{
  for ( int i = 0 ; i < size ; i++ )
    for(int j=0;j<10;j++)
    {
        printf("%d\n",array[i][j]);
    }
}
int main()
{
    int array[2][10]={ {1,2,3,4,5,6,7,8,9,10},
                    {11,12,13,14,15,16,17,18,19,20}
                   };
    display1(array[0][0]);
    display2(&array[0][4],5);
    display3(&array[0][4],5);
    display2(array[0],5);
    display3(array[0],5);
    display4(array,2);
    display5(array,2);
    return 0;
}

A function taking an array parameter can always be written in one of two ways, using either array notation or pointer notation.
As far as the compiler is concerned, they are identical.

> note: expected 'int (*)[10]' but argument is of type 'int *'|
See how this matches the parameter type of display5() ?

Thanks for the info about void main() and int main(). It'll change the way I think about main() function.

Anyways the code provided was of real help. I managed to write something similar to access the given number of elements from a 2D array with a mentioned starting point.

Here I want to read the five elements from the 2D array, starting from the element 15.

Code:

#include<stdio.h>
void display(int *array,int number)
{
    int i;
    for(i=0;i<number;i++)
    {
        printf("%d\n",*(array+i));
    }
}
int main(void)
{
int array[2][10]={
                    {1,2,3,4,5,6,7,8,9,10},
                    {11,12,13,14,15,16,17,18,19,20}
                  };
int  *pointer;
int i,j;
for(i=0;i<2;i++)
 {
for(j=0;j<10;j++)
    {
        if (array[i][j] ==15)
        {
            pointer = &array[i][j];
            display(pointer,5);
            break;
        }
    }
 }
return 0;
}

Is this approach correct? Given that the pointer is passed any changed made inside the function will also be reflected in the main too.

Update : I realized just now that the function display() I've written now is exactly the same as display3() in the example provided

Just because your parameter is int* array, doesn't mean you can't do array[i] to index it inside your loop - as opposed to saying the cumbersome *(array+i)

pointer = &array[i][j];
display(pointer,5);
Nor does this buy anything over saying
display(&array[i][j],5);

Unless you have some other cause to use pointer for something else as well.