I the function read I need to access values of integer a and integer b from the main function without declaring it in the prototype of the function read, using pointers. Pointer x should point to integer a, and pointer y should point to integer b.

#include <stdio.h>
void read(int zzz[], int n) {
    int *arr = zzz, *x=a,*y=b;
}
int main() {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    return 0;
}

How this could be implemented?

CodePudding user response：

There are two ways that the read function can read the values or addresses of a and b:

• Pass them as parameters
• Make a and b global

So if you don't want to make them parameters, you need to move them outside of the main function and before the read function.

Also, read is the name of a system function, so you should name it something else so you don't conflict with it.

CodePudding user response：

You have already received the more conventional answer. It is the most logical choice for your given restriction.

However ... Silly artificial restrictions sometimes deserve a silly contrived solution in kind.

A less conventional way would be to create an understanding with the function that the values will be passed in with the array pointer in some non-standard way.

For example, you can make two extra array members to represent a and b.

int array_for_my_read[array_size   2];
/* instead of a and b, you use those extra array members */
...
int *x = zzz   n, *y = x   1;

Alternatively, you could create a special structure that holds the array, and the pointers. Then the function recovers the pointer to the structure from the array pointer.

struct extra_parameters {
    int *a;
    int *b;
    int zzz[zzz_size];
};
...
struct extra_parameters x;
int a, b;
x.a = &a;
x.b = &b;
read(x.zzz, zzz_size);
...
void *p = (char *)zzz - offsetof(struct extra_parameters, zzz);
struct extra_parameters *xp = p;
int *a = xp->a, *b = xp->b;