5 replies to this topic

[wiwbiz]

How do you pass a 2D array with initialized dimensions to a function ?

[hbk]

you pass it like this

in your function call

int main()
{
.................
....................

function(array[][SIZE],....);
}

the second definition of 2d array must be explicitly defined.
as how i have size defined

[wiwbiz]

That's what I needed to know.What if you don't know value of SIZE ?

[bobdark]

pass it as a value of some pointer type, and pass the dimensions as parameters - that is if you are the one who implements the function that receive the array.
For example:

void foo(int** arr, int rows, int cols){
    int i=0, j=0;
    for (i=0; i < rows; i++)
        for (j=0; j<cols; j++) {
            /*do something*/
        }
    return;
}

int main(){
    /*whatever you do here, lets say eventually, 'arr' is a two dimensional int
    array of m rows on n columns*/
    ...
    foo(arr, m, n);
    ...
}

[wiwbiz]

That won't work as arr is pointer to pointer, but without number of pointer pointed to.

[bobdark]

Correct but I assumed that if you don't know the size, you will have to use dynamic memory allocation, something like that:

int** arr = (int**) malloc (sizeof(int*)*rows);
for (i==0; i< rows; i++)
    arr[i] = (int*) malloc (sizeof(int)*cols);

And then passing this guy to the function.