Reputation: 8947
Below program (a toy program to pass around arrays to a function) doesn't compile. Please explain me, why is the compiler unable to compile(either because of technical reason or because of standard reason?)
I will also look at some book explaining pointers/multi dimensional arrays(as I am shaky on these), but any off-the-shelf pointers here should be useful.
void print2(int ** array,int n, int m);
main()
{
int array[][4]={{1,2,3,4},{5,6,7,8}};
int array2[][2]={{1,2},{3,4},{5,6},{7,8}};
print2(array,2,4);
}
void print2(int ** array,int n,int m)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
printf("%d ",array[i][j]);
printf("\n");
}
}
Upvotes: 9
Views: 27270
Reputation: 1194
In your code the double pointer is not suitable to access a 2D array because it does not know its hop size i.e. number of columns. A 2D array is a contiguous allotted memory.
The following typecast and 2D array pointer would solve the problem.
# define COLUMN_SIZE 4
void print2(int ** array,int n,int m)
{
// Create a pointer to a 2D array
int (*ptr)[COLUMN_SIZE];
ptr = int(*)[COLUMN_SIZE]array;
int i,j;
for(i = 0; i < n; i++)
{
for(j = 0; j < m; j++)
printf("%d ", ptr[i][j]);
printf("\n");
}
}
Upvotes: 0
Reputation: 79003
In C99, as a simple rule for functions that receive "variable length arrays" declare the bounds first:
void print2(int n, int m, int array[n][m]);
and then your function should just work as you'd expect.
Edit: Generally you should have a look into the order in which the dimension are specified. (and me to :)
Upvotes: 5
Reputation: 182794
This (as usual) is explained in the c faq. In a nutshell, an array decays to a pointer only once (after it decayed, the resulting pointer won't further decay).
An array of arrays (i.e. a two-dimensional array in C) decays into a pointer to an array, not a pointer to a pointer.
Easiest way to solve it:
int **array; /* and then malloc */
Upvotes: 12