Shifting elements in an array in C — pointer-based

So I'm trying to learn C right now, and I have two functions: one that shifts elements in an array:

void shift(int a[], int n) {
   int i;
   for(i = 0; i != n-1; i++){
      a[i] = a[i+1];
   }
}

and a version of the same function, except pointer-based:

void pointer_shift(int *a[], int n) {
   int i;
   for (i = 0; i != n - 1; i++) {
      *a[i] = *a[i + 1];
   }
}

I don't know whether the pointer-based version is correct or not, but I guess my most important question is how I'm supposed to actually test that both/either work. Besides the definition of these two functions, I have:

#include <stdio.h>

void shift(int a[], int n);
void pointer_shift(int *a[], int n);

int main(void) {
    printf("Begin execution of testing Problem 1\n");
    int a1[] = {100, 101, 102};
    int i;
    for(i = 0; i<3;i++)
        printf("Before Shift: " "%d\n", a1[i]);
    //shift(a1, 3);
    pointer_shift(&a1, 3);
    for(i = 0; i<3;i++)
        printf("After Shift In Main: " "%d\n", a1[i]);
    return 0;
}

shift(a1, 3)

works fine, but I, for the life of me, can't figure out how to correctly test pointer_shift.

I get two errors; one is that in the line

pointer_shift(&a1, 3)

I am passing argument 1 from an incompatible pointer type. The other error is indecipherable, but I was hoping the problem would be obvious enough that someone would be able to help me. So... how to test my two functions in my main?

Thanks!

Answers


Change your pointer based function like this:

void pointer_shift(int *a, int n) {
   int i;
   for (i = 0; i != n - 1; i++) {
      *(a+i) = *(a+i+1);
   }
}

What you are receiving is an array of pointers whereas you pass an array from main. Since an array decays into pointer, calling shift(a1, 3) is sufficient here.


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