Passing a variable length 2D array to a function

Question

If I dynamically allocate a 2D array(malloc it),

int r,c;
scanf("%d%d",&r,&c);
int **arr = (int **)malloc(r * sizeof(int *));

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

and then later on pass it to a function whose prototype is:

fun(int **arr,int r,int c)

There is no issue. But when I declare a 2D array like VLA i.e.

int r,c;
scanf("%d%d",&r,&c);
int arr2[r][c];

It gives me an error when I try to pass it to the same function. Why is it so? Is there any way by which we can pass the 2nd 2D array(arr2) to a function?

I Know there are lots of similar questions but I didn't find one which address the same issue.

ok. I tried the code in c++. It was not working. Probably it won't work in c either. — Raman
– Raman, Commented Aug 7, 2015 at 5:29

R Sahu · Accepted Answer · 2016-10-25 20:40:19Z

7

A 1D array decays to a pointer. However, a 2D array does not decay to a pointer to a pointer.

When you have

int arr2[r][c];

and you want to use it to call a function, the function needs to be declared as:

void fun(int r, int c, int arr[][c]);

and call it with

fun(r, c, arr2);

edited Oct 25, 2016 at 20:40

answered Aug 7, 2015 at 5:29

R Sahu

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2 Comments

Raman Over a year ago

probably you are saying arr is of type int** but arr2 is of type int *[c], right?

R Sahu Over a year ago

arr, the argument in fun, is of type int (*)[c], not int**.

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