Changing variables from outside

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)