Insertion Sort Doing Nothing - c

Question

I wrote the following insertion sort yesterday (I started learning C 3 days ago). For some reason, the sort does not modify the array AT ALL.
#include <stdio.h>
int *insert(int arr[], int index, int item);
int *isort(int arr[]);
int main() {
int a[17] = {1, 2, 9, 5, 3, 2, 1, 6, 5, 9, 0, 1, 3, 4, 2, 3, 4};
int *b = isort(a);
for (int i = 0; i < 17; i += 1) {
printf("%d ", b[i]);
}
return 0;
}
int *insert(int arr[], int index, int item) {
--index;
while (index >= 0 && item < arr[index]) {
arr[index + 1] = arr[index];
--index;
}
arr[index + 1] = item;
return arr;
}
int *isort(int arr[]) {
for (int i = 1; i < sizeof(arr) - 1; i++) {
arr = insert(arr, i, arr[i]);
}
return arr;
}
I'm thinking it could be my compiler, as I'm running a compiler that is on a non unix machine: lcc-win, but I'm not sure. Is there some fundamental thing I'm missing here?

Answer 1

int *isort(int arr[]) {
for (int i = 1; i < sizeof(arr) - 1; i++) {
arr = insert(arr, i, arr[i]);
}
return arr;
}
In this function sizeof(arr) actually returns the size of the pointer and not the size of the array.
In C a special rule says an array parameter is actually adjusted to a parameter of the corresponding pointer type.
That is:
int *isort(int arr[]) { /* ... */ }
is equivalent to this:
int *isort(int *arr) { /* ... */ }
To fix this, add a new parameter in your function that takes the size of the array:
int *isort(int arr[], size_t size) { /* ... */ }

Answer 2

The first problem, as has been pointed out, is that the isort function uses the sizeof operator on a pointer. The way C treats arrays is a little strange at first glance. The name of an array is a pointer to its first element. So when you call isort like this:
int *b = isort(a);
you are simply pushing a pointer to the array onto the stack. In the definition of isort,
int *isort(int arr[])
declares arr to be a pointer to int just like
int *isort(int *arr)
C is even more confusing in this respect: if you had said:
int *isort(int arr[17])
the arr variable is still just a pointer to int ... the "17" here is discarded! Even with this syntax, sizeof(arr) will still be the size of a pointer to int.
On a 32-bit system (ILP32), sizeof(arr) will always be 4, however big the array is.
Therefore, you need to pass the size of the array to isort. A good general way to do this is to define a macro like this:
#define NITEMS(arr) (sizeof(arr)/sizeof(arr[0]))
This will calculate the number of elements in an array of any type.
Your next problem is more a style one than an actual error:
arr = insert(arr, i, arr[i]);
This calls the insert function with a reference to "arr". The array is modified through this reference, and then a pointer to that array is returned. It will always be the same pointer as you sent in the first place, so this assignment actually does nothing, harmlessly. Like I say, a style problem, not a code error.
The final issue is your isort function stops one short (after you correct the sizeof problem), since you went from 1 to sizeof-1. Here is a fixed version:
#include <stdio.h>
#define NITEMS(arr) (sizeof(arr)/sizeof(arr[0]))
int *insert(int arr[], int index, int item);
int *isort(int arr[], size_t nitems);
int main() {
int a[17] = {1, 2, 9, 5, 3, 2, 1, 6, 5, 9, 0, 1, 3, 4, 2, 3, 4};
int *b = isort(a, NITEMS(a));
for (int i = 0; i < NITEMS(a); i += 1) {
printf("%d ", b[i]);
}
printf("\n");
return 0;
}
int *insert(int arr[], int index, int item) {
--index;
while (index >= 0 && item < arr[index]) {
arr[index + 1] = arr[index];
--index;
}
arr[index + 1] = item;
return arr;
}
int *isort(int arr[], size_t nitems) {
for (int i = 1; i < nitems; i++) {
insert(arr, i, arr[i]);
}
return arr;
}