Insertion sort algorithm in C is not working

Question

I am trying to sort an array of integers for finding the median of the same array. The code I am using sorts only the first two elements of the 10 element array. I have cross checked the swapping of variables in the loops, but everything seems okay.

void sort(int *arr) {
    //get the size of this array
    int size = sizeof(arr) / sizeof(arr[0]);
    for (int i = 0; i < size; i++) {
        for (int j = i + 1; j < size; j++) {
            if (arr[i] > arr[j]) {
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
}

Application

#include <stdlib.h>
#include <stdio.h>

int main() {
    int numbers[10];
    numbers[0] = 10;
    numbers[1] = 9;
    numbers[2] = 8;
    numbers[3] = 7;
    numbers[4] = 6;
    numbers[5] = 5;
    numbers[6] = 4;
    numbers[7] = 3;
    numbers[8] = 2;
    numbers[9] = 1;

    sort(numbers);
    //code sorts the element in the first and second index, the rest are unsorted please help

    return 0;
}

What am I doing wrong?

Answer 1

The function parameter has a pointer type

void sort(int* arr){

Even if you will rewrite the function declaration like

void sort( int arr[10] ){

nevertheless the compiler will adjust the function parameter having the array type to pointer to the element type as it is written in your original function declaration

void sort(int* arr){

And in this call of the function

sort(numbers);

the array designator is implicitly converted to a pointer to its first element. That is the call above in fact is the same as

sort( &numbers[0] );

So using the operator sizeof with a pointer expression yields the size of a pointer. That is this line

//get the size of this array
int size=sizeof(arr)/sizeof(arr[0]);

is equivalent to

//get the size of this array
int size=sizeof( int * )/sizeof( int );

and yields either 2 or 1 depending on the used system.

You need to declare the function like

void sort( int *arr, size_t n );

and pass to the function the number of elements in the array explicitly.

Bear in mind that in general the user can use the function for a dynamically allocated array.

Pay attention to that the used by you algorithm is not the insertion sort algorithm. It is the selection sort algorithm with redundant swaps.

A function that implements the insertion sort algorithm can look for example the following way as it is shown in the demonstration program below.

#include <stdio.h>

void insertion_sort( int a[], size_t n )
{
    for ( size_t i = 1; i < n; i++ )
    {
        int current = a[i];
        size_t j = i;
        for ( ; j != 0 && current < a[j - 1]; j-- )
        {
            a[j] = a[j - 1];
        }

        if ( j != i ) a[j] = current;
    }
}

int main()
{
    int a[] = { 9,8, 7, 6, 5, 4, 3, 2, 1, 0 };
    const size_t N = sizeof( a ) / sizeof( *a );

    for (size_t i = 0; i < N; i++)
    {
        printf( "%d ", a[i] );
    }
    putchar( '\n' );

    insertion_sort( a, N );

    for (size_t i = 0; i < N; i++)
    {
        printf( "%d ", a[i] );
    }
    putchar( '\n' );
}

The program output is

9 8 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 9

As for the function that implements the selection sort algorithm then without redundant swaps it can look the following way as it is shown in the next demonstration program.

#include <stdio.h>

void selection_sort( int a[], size_t n )
{
    for ( size_t i = 0; i < n; i++ )
    {
        size_t min = i;

        for ( size_t j = i + 1; j < n; j++ )
        {
            if (a[j] < a[min]) min = j;
        }

        if ( min != i )
        {
            int tmp = a[i];
            a[i] = a[min];
            a[min] = tmp;
        }
    }
}

int main()
{
    int a[] = { 9,8, 7, 6, 5, 4, 3, 2, 1, 0 };
    const size_t N = sizeof( a ) / sizeof( *a );

    for (size_t i = 0; i < N; i++)
    {
        printf( "%d ", a[i] );
    }
    putchar( '\n' );

    selection_sort( a, N );

    for (size_t i = 0; i < N; i++)
    {
        printf( "%d ", a[i] );
    }
    putchar( '\n' );
}

The program output is the same as shown above

9 8 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 9

Answer 2

This is because int size = sizeof(arr) / sizeof(arr[0]); will return 2 , because sizeof(arr) will give the sizeof a pointer to int , which in your case is 8 bytes and then sizeof(arr[0]) will give a sizeof int which in your case is 4 bytes .

So,

8 / 4 = 2

Fix:

• Add another parameter for the length of the array.
• type-cast numbers to (int *) for sort() function
• Your sort() function does not implement insertion sort algorithm, instead it is selection sort algorithm. READ MORE
• There's no need of stdlib.h header file in your code

Like this: TRY IT ONLINE

#include <stdio.h>

void sort(int *arr, size_t len)
{
    for (int i = 0; i < len; i++)
    {
        for (int j = i + 1; j < len; j++)
        {
            if (arr[i] > arr[j])
            {
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
}

// using `const` keyword, because `arr` isn't modifying inside `print()` function
void print(const int *arr, size_t len){
    for (size_t i = 0; i < len; i++)
    {
        printf("%d\n", arr[i]);
    }
}

int main(void)
{
    int numbers[10];
    numbers[0] = 10;
    numbers[1] = 9;
    numbers[2] = 8;
    numbers[3] = 7;
    numbers[4] = 6;
    numbers[5] = 5;
    numbers[6] = 4;
    numbers[7] = 3;
    numbers[8] = 2;
    numbers[9] = 1;

    size_t len = sizeof(numbers) / sizeof(numbers[0]);
    sort((int *)numbers, len);

    print(numbers, len);
    return 0;
}