How to pass pointer of arrays to functions

Question

With my programm I try to change the order of numbers in the int array. To the first function, I just passed both arrays and printed the array called arraytemp with the changed order. After that I printed in the main function the same array, just to see if the array was filled too. I havented used any pointers in the first function - how did the array got filled? Does the arrays adress get passed to functions anyway?

Then I wanted to pass arrays with the same content to the second function, but this time I used pointers. I have no clue, how to get the same result printed, because I get a stack smashing error. I am kinda comfused with '*' and '&'. So, how should I pass these arrays when using pointers?

#include <stdio.h>

void switchnum (int arraytemp[6], int array[], int laenge) {
    printf("\n\nAfter (in function 1):\n");
    for(int i = 0 ; i<laenge ; i++) {
        arraytemp[i] = array[laenge-1-i];
        printf("%d ", arraytemp[i]);
    }
    return 0;
}

void switchnum2 (int *arraytemp2[6], int array2[], int laenge2) {
    printf("\nAfter (in function2):\n");
    for(int j = 0 ; j<laenge2 ; j++) {
        arraytemp2[j] = array2[laenge2-1-j];
        printf("%d ", arraytemp2[j]);
    }
    return 0;
}
int main() {
    int array[] = {4,8,1,3,0,9};
    int arraytemp[6];
    printf("Before (main):\n");
    for(int i = 0 ; i<6 ; i++) {
        printf("%d ", array[i]);
    }  
    
    switchnum(arraytemp, array, 6);

    printf("\nAfter (in main):\n");
    for(int i = 0 ; i<6 ; i++) {
        printf("%d ", arraytemp[i]);
    }   
    
    
    int array2[] = {4,8,1,3,0,9};
    int arraytemp2[6];
    
    switchnum2(arraytemp2, array2, 6);    
    
    return 0;
}

Answer 1

The compiler adjusts a parameter having an array type to pointer to the array element type.

So this function declaration

void switchnum (int arraytemp[6], int array[], int laenge);

is equivalent to the following declaration

void switchnum (int arraytemp[], int array[], int laenge);

and the same way is equivalent to the following declaration

void switchnum (int *arraytemp, int *array, int laenge);

As for this function declaration

void switchnum2 (int *arraytemp2[6], int array2[], int laenge2);

then it is adjusted by the compiler to the declaration

void switchnum2 (int **arraytemp2, int *array2, int laenge2);

So the used argument expression and the function parameter have incompatible pointer types.

Pay attention to that in this call

switchnum(arraytemp, array, 6);

the both arrays are converted to pointers to their first elements of the type int *.

In fact this call is equivalent to

switchnum( &arraytemp[0], &array[0], 6);