How to change the value of a variable passed as an argument and also use the value for an array

This is my example code:

#include <stdio.h>
void Func(int a[], int b) {
    a[b] = 1;
    b += 5;
}
int main(void) {
    int a[10];
    int b = 0;
    printf("%d\n", b);
    Func(a, b);
    printf("%d\n", a[0]);
    printf("%d\n", b);
}

And I want the program to print:

0
1
5

I've tried changing the function by making b a pointer:

void Func(int a[], int *b)

... and by writing *b += 5. Also, I've changed the call of the function to:

int a[10];
int *b = 0;
// ...
Func(a, &b);

Is it possible to manipulate b and to use it?

Note: I've tried following this post

> c
> void Func(int a[], int b){
> a[b] = 1; // OK
> b += 5; // useless
> }
>
The problem with this code is that Func is assigning a local copy of b, and this has no effect outside of the function.

To get the expected output:

#include <stdio.h>
void Func(int a[], int *b) {
   a[*b] = 1;
   *b += 5;
}
int main(void) {
    int a[10];
    int b = 0;
    printf("%d\n", b);    // print value of b
    Func(a, &b);          // pass address of b to Func
    printf("%d\n", a[0]);
    printf("%d\n", b);    // print value of b again
}

Func still has a local copy named b, but this copy is a pointer which may point to the b object in main. This allows us to modify the pointed-to object with *b += 5.

For that you want to use a and b as a pointer, because a is an array and b you want to modify the value inside the function.

But you can't send directly b inside the function is why you to send as &b to give the address where the variable is stored in that way you don't copy the value of b but directly change it.

void foo(int *a, int *b) {
    a[*b] = 1;
    *b += 5;
}
int main(void) {
    int a[10] = {0};
    int b = 0;
    printf("b = [%i]\n", b);
    foo(a, &b);
    printf("a[0] = [%i]\n", a[b]);
    printf("b = [%i]\n", b);
}