size of array in C without sizeof and not in main function

Question

I want to understand how to know the size of array without using sizeof and not in main function. and i really want to understand why my code not working when i do it in other function.(in main it works same as sizeof). >>> The rasult i got are some trash numbers

*with char array is understood but i dont know how to do it with other data specifiers.

#include <stdio.h>

void arrSize(int a[],int b[]){
    int size_a = *(&a+1) - a;
    int size_b = *(&b+1) - b;
    printf("%d,%d",size_a,size_b);
}

int main(){
    int arr[] = {2,12,14,16};
    int brr[] = {8,53,2,4,16};
    
    arrSize(arr,brr);

    return 0;
}

Answer 1

This function declaration

void arrSize(int a[],int b[]){
    int size_a = *(&a+1) - a;
    int size_b = *(&b+1) - b;
    printf("%d,%d",size_a,size_b);
}

is adjusted by the compiler to the following declaration

void arrSize(int *a,int *b){
    int size_a = *(&a+1) - a;
    int size_b = *(&b+1) - b;
    printf("%d,%d",size_a,size_b);
}

That is parameters having array types are adjusted by the compiler to pointers to array element types.

On the other hand, in this call

arrSize(arr,brr);

the arrays are implicitly converted to pointers to their first elements.

Having a pointer to the first element of an array you are unable to determine the array size pointed to by the pointer.

the initializers in these declarations

    int size_a = *(&a+1) - a;
    int size_b = *(&b+1) - b;

invoke undefined behavior because you are dereferencing pointers that do not point to valid object.