Reputation: 413
A function dynamically creates an int
array whose elements are predetermined to be int[2]
. Is there any way to have a function assign values to that array and then return it to the caller.
The below code accomplishes this task but throws warning: initialization from incompatible pointer type [enabled by default]
#include <stdlib.h>
#include <stdio.h>
int *get_values()
{
int (*x)[2] = malloc(sizeof(int[2])*3);
x[0][0] = 1;
x[0][1] = 2;
x[1][0] = 11;
x[1][1] = 12;
x[2][0] = 21;
x[2][1] = 22;
return x;
}
int main()
{
int (*x)[2] = get_values();
int i;
for (i=0; i!=3; i++)
printf("x[%d] = { %d, %d }\n", i, x[i][0], x[i][1]);
}
I'm aware of the alternative where you dynamically allocate both dimensions, but this is something that I am curious about.
Upvotes: 0
Views: 91
Reputation: 212939
Rather than keep repeating the same clunky syntax it can be helpful to define a typedef in cases like this. This makes it easier to declare the correct return type for get_values
:
#include <stdlib.h>
#include <stdio.h>
typedef int I2[2];
I2 * get_values(void)
{
I2 * x = malloc(sizeof(I2) * 3);
x[0][0] = 1;
x[0][1] = 2;
x[1][0] = 11;
x[1][1] = 12;
x[2][0] = 21;
x[2][1] = 22;
return x;
}
int main()
{
I2 * x = get_values();
int i;
for (i=0; i!=3; i++)
printf("x[%d] = { %d, %d }\n", i, x[i][0], x[i][1]);
free(x);
}
Recommended reading: Don't repeat yourself (DRY).
Upvotes: 7
Reputation: 25752
And this is how it looks without a typedef:
int (*get_values(void))[2]
{
return NULL;
}
Pretty unreadable.
Notice in that function definition, if you replace get_values(void)
with x you get: int (*x)[2]
, which is exactly what the pointer definition looks like.
Upvotes: 4