当 arrays 通过函数作为 arguments 传递时，它们是否被零索引？ C编程

Question

Are array zero indexed when they are passed throu an function as argument? Or did C just copy the array A's memory into two different arrays?

I never need to "reset" the orginal position of an array when I pass the same array throu another function?

static const int A[3] = {1, 5, 8};

void fun(const int B[]){
    printf("val = %d\n", *B);
    B++;
    printf("val = %d\n", *B);
}

int main() {

    fun(A);
    fun(A);

    return 0;
}

Output:

val = 1
val = 5
val = 1
val = 5

For example when I don't use const , I can see that they share the same memory, but that'only when I don't use const .

static int A[3] = {1, 5, 8};

void fun(int B[]){
    printf("val = %d\n", *B);
    *B = 10;
    printf("val = %d\n", *B);
}

int main() {

    fun(A);
    fun(A);

    return 0;
}

Output:

val = 1
val = 10
val = 10
val = 10

Answer 1

You modify local pointer. Why pointer? Because array are passed as pointers. To modify the original pointer you need to pass pointer to pointer

int A[] = {1, 5, 8, 10};

void fun(const int **B){
    printf("val = %d\n", **B);
    (*B)++;
    printf("val = %d\n", **B);
}

int main() 
{
    int *C = A;
    fun(&C);
    fun(&C);

    return 0;
}