How to pass a two-dimensional array to a function in C? - c

Question

My function prototype is
int** rotate(int **arr, int row, int col, int fl);
where arr is the two dimensional array, row and col is the number of row and columns of the 2D array respectively, fl is a flag variable. If the value of fl is 0 then the array will be rotated right, if fl is 1 then the array will be rotated left.
I have called the function as follows:
int **res= rotate(arr, row, col, fl);
But I got one warning and one note
[Warning] passing argument 1 of 'rotate' from incompatible pointer type.
[Note] expected 'int **' but argument is of type 'int (*)[20]'

Answer 1

A pointer to a pointer is different from a pointer to an array. Array-to-pointer decaying can only happen on the left-most side (e.g. int [3][20] to int (*)[20]).
Change your function declaration to
int** rotate(int (*arr)[20], int row, int col, int fl);
or more obviously,
int** rotate(int arr[][20], int row, int col, int fl);
Note you have to fix the size at compile-time.

Answer 2

If your compiler supports variable length arrays then the function declaration can look the following way
void rotate( size_t row, size_t col, int arr[][col], int fl);
or
void rotate( size_t row, size_t col, int arr[][col], _Bool fl);
In this case you can use arrays with different sizes.
Here is a demonstrative program
#include <stdio.h>
void rotate( size_t row, size_t col, int a[][col], _Bool fl )
{
for ( size_t i = 0; i < ( fl ? row : col ); i++ )
{
for ( size_t j = 0; j < ( fl ? col : row ); j++ )
{
printf( "%d ", a[fl ? i : j][fl ? j : i] );
}
putchar( '\n' );
}
}
#define N1 3
int main(void)
{
int a[][3] =
{
{ 1, 2, 3 },
{ 4, 5, 6 }
};
rotate( sizeof( a ) / sizeof( *a ), N1, a, 0 );
putchar( '\n' );
rotate( sizeof( a ) / sizeof( *a ), N1, a, 1 );
putchar( '\n' );
return 0;
}
Its output is
1 4
2 5
3 6
1 2 3
4 5 6
Otherwise if within the function you are going to create new arrays then the function can look as it is shown in the following demonstrative program.
#include <stdio.h>
#include <stdlib.h>
int ** rotate( size_t, size_t, int a[][*], _Bool fl );
int ** rotate( size_t row, size_t col, int a[][col], _Bool fl )
{
int **p = malloc( col * sizeof( int * ) );
for ( size_t i = 0; i < col; i++ )
{
p[i] = ( int * )malloc( row * sizeof( int ) );
}
if ( fl )
{
for ( size_t i = 0; i < row; i++ )
{
for ( size_t j = 0; j < col; j++ )
{
p[col - j - 1][i] = a[i][j];
}
}
}
else
{
for ( size_t i = 0; i < row; i++ )
{
for ( size_t j = 0; j < col; j++ )
{
p[j][i] = a[row - i - 1][j];
}
}
}
return p;
}
#define M 2
#define N 3
int main(void)
{
int a[M][N] =
{
{ 1, 2, 3 },
{ 4, 5, 6 }
};
int **p = rotate( M, N, a, 0 );
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < M; j++ )
{
printf( "%d ", p[i][j] );
}
putchar( '\n' );
}
putchar( '\n' );
for ( size_t i = 0; i < N; i++ )
{
free( p[i] );
}
free( p );
p = rotate( M, N, a, 1 );
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < M; j++ )
{
printf( "%d ", p[i][j] );
}
putchar( '\n' );
}
putchar( '\n' );
for ( size_t i = 0; i < N; i++ )
{
free( p[i] );
}
free( p );
return 0;
}
Its output is
4 1
5 2
6 3
3 6
2 5
1 4

Answer 3

From the example, I assume that you are using a static definition of arr.
The Pointer to pointer is not the same as a 2D array.
Change
int** rotate(int **arr, int row, int col, int fl);
to
int** rotate(int arr[][20], int row, int col, int fl);
Note that the no of columns will have to be defined before compilation.