Sorting Algorithm Not Passing

Question

I'm trying to take a file and run it through a sorting algorithm, insertion sort in this case. After I pass it through the algorithm and print the array, it doesn't look sorted. My thinking is because I'm trying to pass an array to the function, and view the information through a pointer, but I think I'm messing it up.

void insertionSort(int arr[]) {
    int key, j;
    for(int i = 1; i < (sizeof(arr) / sizeof(int)); i++) {
        int key = arr[i];
        int j = i-1;
        while(j >= 0 && arr[j] > key) {
            arr[j+1] = arr[j];
            j = j - 1;
        }
        arr[j+1] = key;
    }
}

I have the function set to void because I thought I'd be able to just access the array since it is a pointer. Would it be better if I returned the array? My logic is obviously wrong, so any pointers (no pun intended) would be great.

Here is the block of code from my main()

case 2:
    printf("Filling array...\n");

    fp = fopen("Random.txt", "r");
    if (fp == NULL) {
        printf("\nError opening Random.txt\n");
        exit(1);
    }

    for (int i = 0; i < (sizeof(arr) / sizeof(int)); i++) {
        fscanf(fp, "%d", &arr[i]);
    }

    insertionSort(arr);

    for (int i = 0; i < (sizeof(arr) / sizeof(int)); i++) {
        printf("%d\n", arr[i]);
    }

    fclose(fp);
    break;
}

Hint 1: void insertionSort(int arr[]) { is exactly the same as void insertionSort(int *arr) { (yes it's a pointer even when you write []). Hint 2: what's sizeof(int*)/sizeof(int)? — Stack Exchange Broke The Law
– Stack Exchange Broke The Law, Commented Jun 22, 2018 at 1:08
sizeof(arr) in the insertionSort routine will be sizeof(int*) which is 8 byte, so what you are actually doing is 8 / 4 , which is 2 — 0.sh
– 0.sh, Commented Jun 22, 2018 at 1:09
@immibis I'm using sizeof to iterate through the entire array. I'm under the impression that I can't use .length() for example because C is not OOP? @0.sh Read above comment. Using sizeof to iterate through array. If there's a better method, I'm open to it. — InfoSecNick
– InfoSecNick, Commented Jun 22, 2018 at 1:14

Stephen Docy · Accepted Answer · 2018-06-22 01:14:02Z

1

void insertionSort(int arr[]) {

Arrays decay to pointers when passed to functions, so this is basically

void insertionSort(int *arr) {

which means this will not give you the number of elements in the array

for(int i = 1; i < (sizeof(arr) / sizeof(int)); i++) {

You need to explicitly pass the size of the array to the function.

void insertionSort(int arr[], int size) {
    for (int i = 1; i < size; i++) {
        int key = arr[i];
        int j = i - 1;
        while (j >= 0 && arr[j] > key) {
            arr[j + 1] = arr[j];
            j = j - 1;
        }
        arr[j + 1] = key;
    }
}

To call the function:

insertionSort(arr, sizeof(arr) / sizeof(int));

In main(), where the array is created, you can use sizeof(A) / sizeof(int), because main() knows the size of the array.

answered Jun 22, 2018 at 1:14

Stephen Docy

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2 Comments

David C. Rankin Over a year ago

In other words, sizeof (array) only reports the size of the array in the scope in which array is declared. Otherwise, it's just sizeof (a pointer).

InfoSecNick Over a year ago

Ah, I didn't know that an array decays to a pointer. Is there any reason to this? Or is this just C? Thanks again.

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