Pointing to arrays using void function - c

Question

Sorry for that title. I really didn't know how to define this problem.
I was needed to declare integer array of N numbers and to fill it with random nums in void function. Then that array needs to be printed in main. The thing is that i am not allowed to use printf in void function so only way to print in main is to use pointers I guess. My knowledge is limited as I am beginner at pointers. Thx in advance and sorry for bad english.
Here is my code so far. When I compile it marks segmentation error.
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
void form();
int main()
{
int N, a[100];
printf("Input index: \n");
scanf("%d", &N);
form(N, &a);
printf("Array: \n");
for (int i = 0; i < N; i++) {
printf("a[%d] = %d", i, a[i]);
}
}
void form(int N, int *ptr[100])
{
srand(time(NULL));
for (int i = 0; i < N; i++) {
*ptr[i] = rand() % 46;
}

Answer 1

There are several issues in your code.
1) Your array decalaration form() is obsolete. Use proper prototype.
2) For declaring a VLA, declare it after reading N instead of using a fixed size array.
3) An array gets converted into a pointer to its first element when passed to a function. See: What is array decaying?
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
void form(int, int*); /* see (1) */
int main(void) /* Standard complaint prototype for main.
If you need to pass arguments you can use argc, and argv */
{
int N;
printf("Input size: \n");
scanf("%d", &N);
int a[N]; /* see (2) */
form(N, a); /* see (3) */
printf("Array: \n");
for (int i = 0; i < N; i++) {
printf("a[%d] = %d", i, a[i]);
}
}
void form(int N, int *ptr) { /* Modified to match the prototype
srand(time(NULL));
for (int i = 0; i < N; i++) {
ptr[i] = rand() % 46;
}
}

Answer 2

So a couple things:
void form();
As Olaf was alluding to, this declaration is incorrect - you are missing the applicable parameters. Instead, it should be
void form(int N, int ptr[100]);
The main reason your program is crashing is because of the following line:
*ptr[i] = rand() % 46;
You are dereferencing the pointer at i, which is actaully giving you a number - what you want is to assign the value of the pointer at i the new random value:
ptr[i] = rand() % 46;
As related reading, see this question about passing an array in as a function parameter (basically, int ptr[] is the same thing as int * ptr)

Answer 3

Small modifications on your code:
1) Correction and simplification of parameter handling at function call. Just hand over "a", it's an array, so it is an address, you can use int *ptr, or int ptr[], or int ptr[100] in the formal parameter list for it. So you can use simply ptr[i] in your function.
2) Make a prototype for function from old-style declaration providing parameter list.
3) int i; declaration before the for loop - not mandatory, depends on your compiler standard
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
void form(int N, int *ptr);
int main()
{
int N, a[100];
printf("Input index: \n");
scanf("%d", &N);
form(N, a);
printf("Array: \n");
int i;
for (i = 0; i < N; i++) {
printf("a[%d] = %d", i, a[i]);
}
}
void form(int N, int *ptr)
{
srand(time(NULL));
int i;
for (i = 0; i < N; i++) {
ptr[i] = rand() % 46;
}
}