passing a 2d array to a function

change

void myFunc(int **array)

to

void myFunc(int** &array)

I tried it, but I get this error:

parse error before '&' token

I don't beleive that is correct syntax...thanks, though. Any other thoughts?

[edited by - regularkid on July 14, 2003 2:09:13 AM]

int** is a pointer to a pointer, and can be used for a multi-dim dynam array,

int array[4][4] can be accessed via a pointer to the first element (or any other really) since the elements are contiguous in memory

i.e. void f(int* array)
{
for (int i = 0; i < 4; i++)
for (int j = 0; j < 4; j++)
array[j] = 0;
}

for functions like f that may operate on unknown array-dims, just pass the array-dims as params.

PS myarray is an int*
HTH

Wizza Wuzza?

[edited by - chris hare on July 14, 2003 2:25:54 AM]

[edited by - Chris Hare on July 14, 2003 2:27:26 AM]

[edited by - Chris Hare on July 14, 2003 2:28:41 AM]

In f(...) it should be array[ i ][ j ], I'm bad at this post-HTML stuff

Wizza Wuzza?

[edited by - Chris Hare on July 14, 2003 2:30:42 AM]

your options are

void myFunc( int array[4][4] )
{
    int i;
    int j;

    for(i = 0; i < 4; i++)
    {
        for(j = 0; j < 4; j++)
        {
            cout << array[i][j] << endl;
        }
    }
}

or

void myFunc( int array[][4] )
{
    int i;
    int j;

    for(i = 0; i < 4; i++)
    {
        for(j = 0; j < 4; j++)
        {
            cout << array[i][j] << endl;
        }
    }
}

or

void myFunc( int* array )
{
    int i;
    int j;

    for(i = 0; i < 4; i++)
    {
        for(j = 0; j < 4; j++)
        {
            cout << array[i*4+j] << endl;
        }
    }
}

And nothing else.

A bidimensional array is not an int**.

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if I were to use this option:

void myFunc(int array[][4])
{
...
}

what is the syntax for passing the array into this function? Thanks

quote:

---

what is the syntax for passing the array into this function? Thanks

---

For that option (which I think is the most recommended) use:
myFunc( myArray )

Since myArray is of type int*[4] (or int[][4]), it''s fully compatible.

You would just call

myFunc(array);



I don''t believe you can use myFunc(int *array). array is not an int*, it''s an int(*)[4]. By the way, if you''re just using a static 4x4 array, you could make it a single dimensional array and save yourself some funky syntax.

Cool! That worked great...just one more question (i promise!). Ok, the array that I have (i forgot to mention) is a constant array, ie:

const int array[][4] =
{
    ...
}

How, would I go about passing this in? I tried casting it, but I don''t know how to cast to the type [][4]? Thanks a lot!

RegularKid : declare myFunc as
myFunc( const int array[][4] ); or
myFunc( const int array[4][4] ); or
myFunc( const int* array );

Then just pass the array identifier to the function : myFunc( myArray );

quote:

---

Original post by kdogg
I don't believe you can use myFunc(int *array). array is not an int*, it's an int(*)[4].

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C++ multidimensional arrays are row-major, continuous blocks of memory. Accessing them through a pointer has the effect of flattening them and providing access as if it were a single-dimensional array (with array[j][k] becoming array[j*width+k]).

See Stroustrup's "The C++ Programming Language", section C.7 in which the problem ofmultidimensional arrays as parameters is directly addressed.

See also the C++ Standard section 8.3.4 on arrays.

As a final (loosely related) note, in C++, dynamic array is spelled std::vector<sometype>, with multidimensional dynamic arrays being vectors of vectors.

---

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[edited by - Fruny on July 14, 2003 5:30:17 AM]