Reputation: 703
My background is Java therefore I'm not used to pointers, the following code throws error and I can't see way:
#include <stdio.h>
#define DIM 2
void sort_intervals(int** intervals, int n);
int main()
{
int a[3][DIM] = {
{1, 6} ,
{4, 9} ,
{3,17} };
sort_intervals(a, 3);
return 0;
}
void sort_intervals(int** intervals, int n)
{
printf("%d ", intervals[0][0]);//<--- error here
}
Error: Access violation reading location
I'm not allowed to change the function signiture
Upvotes: 0
Views: 72
Reputation: 141881
Then you need an array of pointers that point to arrays.
int a_1[DIM] = {1, 6};
int a_2[DIM] = { ... };
int a_3[DIM] = { ... };
int *a[3] = { a_1, a_2, a_3, }; // array of 3 pointers
sort_intervals(a, 3);
or exactly equivalent using compound literals you can:
int *a[3] = { (int[DIM]){1, 6}, (int[DIM]){2, 7}, (int[DIM]){3, 17}};
sort_intervals(a, 3);
even:
sort_intervals((int*[3]){
(int[DIM]){1, 6},
(int[DIM]){2, 7},
(int[DIM]){3, 17},
}, 3);
Upvotes: 2
Reputation: 16876
If you can't change the signature as you explained in your comment, then you can get it done using a jagged array:
#define DIM 2
void sort_intervals(int** intervals, int n);
int main()
{
int **a = malloc(3 * sizeof(int*));
for (int i = 0; i < 3; i++) {
a[i] = malloc(DIM * sizeof(int));
}
a[0][0] = 1; a[0][1] = 6;
a[1][0] = 4; a[1][1] = 9;
a[2][0] = 3; a[2][1] = 17;
sort_intervals(a, 3);
for (int i = 0; i < 3; i++) {
free(a[i]);
}
free(a);
return 0;
}
void sort_intervals(int** intervals, int n)
{
printf("%d ", intervals[2][0]);//<--- error here
}
Upvotes: 1
Reputation: 1417
I am assuming you can change main function.
You could also initialize a
as double pointer such as:
int **a = (int**)calloc(3, sizeof(int*)); //setting a as array of 3 pointers
for (int i = 0; i < 3; i++)
*a = (int*) calloc(DIM, sizeof(int)); //assigning pointer to a[0], a[1] and a[2]
Here a is array of 3 integer pointers. a[0],a[1] and a[2] are int pointers. You can modify a as a[0][0] = 5;
Upvotes: 1