CODEWITHSUNDEEP
carrayssortingwhile-loopbubble-sort
W.N
W.N

Reputation: 29

Sorting array only with while and if

I get a message when I try to run the program. Why?

Segmentation fault

my code:

#include <stdio.h>

void sort_array(int *arr, int s);

int main() {
    int arrx[] = { 6, 3, 6, 8, 4, 2, 5, 7 };

    sort_array(arrx, 8);
    for (int r = 0; r < 8; r++) {
        printf("index[%d] = %d\n", r, arrx[r]);
    }
    return(0);
}

sort_array(int *arr, int s) {
    int i, x, temp_x, temp;
    x = 0;
    i = s-1;
    while (x < s) {
        temp_x = x;
        while (i >= 0) {
            if (arr[x] > arr[i]) {
                temp = arr[x];
                arr[x] = arr[i];
                arr[i] = temp;
                x++;
            }
            i++;
        }
        x = temp_x + 1;
        i = x;
    }
}

I think that the problem is in the if statement. What do you think? Why does it happen? I think that I use in positive way with the pointer to the array.

Upvotes: 0

Views: 4354

Answers (3)

Adam Shaikh
Adam Shaikh

Reputation: 1

**Simplest Way**

  function sort(arr) {
  for (let i = 0; i < arr.length; i++) {
  for (let j = 0; j < arr.length; j++) {
   if (arr[j] < arr[j + 1]) {
    [arr[j], arr[j + 1]] = [arr[j + 1], arr[j]];
   }
  }
 }
  return arr;
   }
 console.log("2 sort array descending", sort(arr));

Upvotes: -4

Vlad from Moscow
Vlad from Moscow

Reputation: 311028

This loop in your program

    while (i >= 0) {
        //...
        i++;
    }

does not make sense because i is increased unconditionly.

The program can look the following way

#include <stdio.h>

void bubble_sort( int a[], size_t n )
{
    while ( !( n < 2 ) )
    {
        size_t i = 0, last = 1;

        while ( ++i < n )
        {
            if ( a[i] < a[i-1] )
            {
                int tmp = a[i]; 
                a[i] = a[i-1];
                a[i-1] = tmp;
                last = i;
            }
        }

        n = last;
    }
}   

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

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

    bubble_sort( a, N );

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

    return 0;
}

The program output is

6 3 6 8 4 2 5 7 
2 3 4 5 6 6 7 8

If you want that the sorting function had only one while loop then you can implement it the following way

void bubble_sort( int a[], size_t n )
{
    size_t i = 0;

    while ( ++i < n )
    {
        if ( a[i] < a[i-1] )
        {
            int tmp = a[i]; 
            a[i] = a[i-1];
            a[i-1] = tmp;
            i = 0;
        }
    }
}

Upvotes: 4

chqrlie
chqrlie

Reputation: 144780

In your inner loop, you increment i beyond the size of the array. Your algorithm should require you to decrement i instead, but I am not sure this would be enough to fix the sorting algorithm.

You should first try to implement Bubble sort with a single while loop where you compare adjacent items and step back whenever you swap them.

Upvotes: 2

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