Changing variables from outside - c

Question

In main() function I initialize a couple of variables (int and int* array). Then I print something out and read them from the console scanf.
I want to place this functionality into some external function so that the main will look like this:
int main()
{
int n = 0, x = 0;
int *arr = NULL;
load(&n, &x, &arr);
}
After load() function call I want the variables to be exactly as they were set inside of the load() function. How can I do this?
And second question, just out of curiosity:
/**
* Description of the function
*
* #param int n Foo
* #param int x Bar
* #param int *arr Does something
*/
void load(int n, int x, int *arr)
{
// something
}
Is this documentation useful in C coding, and is it a good practice?

Answer 1

You are passing address of two int and one pointer(third argument), you should receive first two arguments in pointer(one *) to int and third argument in pointer to pointer(two **) of int:
void load(int* n, int* x, int **arr){
// ^ ^ one* ^ two **
*n = 10;
*x = 9;
}
In load function you can assign values to *n and *x because both points to valid memory addresses but you can't do **arr = 10 simply because arr doesn't points to any memory (points to NULL) so first you have to first allocate memory for *arr, do like:
void load(int* n, int* x, int **arr){
*n = 10;
*x = 9;
*arr = malloc(sizeof(int));
**arr = 10;
}
Is this documentation useful in C coding, and is it a good practice?
Yes
but Sometimes I documents my function arguments like in following ways:
void load(int n, // is a Foo
int x, // is a Bar
int **arr){ // do some thing
// something
}
A reference: for document practice
Edit As you are commenting, do like below I am writing, it will not give any error/because of malloc().
#include<stdio.h>
#include<stdlib.h>
void load(int* n, int* x, int **arr){
*n = 10;
*x = 9;
*arr = malloc(sizeof(int));
**arr = 10;
printf("\n Enter Three numbers: ");
scanf("%d%d%d",n,x,*arr);
}
int main(){
int n = 0, x = 0;
int *arr = NULL;
load(&n, &x, &arr);
printf("%d %d %d\n", n, x, *arr);
free(arr);
return EXIT_SUCCESS;
}
Compile and run like:
~$ gcc ss.c -Wall
:~$ ./a.out
Enter Three numbers: 12 13 -3
12 13 -3
As Commented by OP:
"Invalid convertion from void* to int*" when I change this to arr = malloc(sizeof(int)(*n));
syntax of malloc():
void *malloc(size_t size);
malloc() returns void* and *arr type is int* that is the reason compiler messages because of different types : "Invalid convertion from void* to int*"
But I avoid casting when malloc(), since: Do I cast the result of malloc? (read Unwind's answer)