Sorting numbers in C

Question

I have created a small C program which sorts odd and even numbers in descending order, with the user inputting whether they wish to sort the odd or the even numbers. So in order to try and make it more complex, I was wondering if there was a way in which I could have my program sort only the even numbers while leaving the odd numbers in their current place and vice versa so for instance:

Input:

odd

1 7 3 5 2 4 20

Output:

1 3 5 7 2 4 20

There is probably multiple ways to go about this, I'm wanting to include it in

There should be a way, but it's a very broad question for this site. — Iharob Al Asimi, Mar 10 '17 at 00:24
I couldn't find anything on the web in relation to sorting only specific numbers and leaving the rest unaltered, so any information is good. — John Anderson, Mar 10 '17 at 00:34
Try extracting only the even numbers to a new array, sort that, and then loop to insert into the old array only when the value in the old array at the given index is even. — AntonH, Mar 10 '17 at 00:40

score 0 · Answer 1 · answered Mar 10 '17 at 00:31

0

Maybe something like, if the number is divisible by 2(remainder 0) it is even, so do nothing. If the number + 1 is divisible by 2(and so is odd), then sort.

answered Mar 10 '17 at 00:31

takintoolong

140
9

Vlad from Moscow · Accepted Answer · 2017-03-10T01:05:13.717

0

First of all you should place the sort algorithm in a separate function. Secondly you can add one more parameter that will specify the predicate for elements of the array that need to be sorted.

For example

#include <stdio.h>

void sort( int a[], size_t n, int predicate( int ) )
{
    for ( size_t i = 0; i < n; i++ )
    {
        if ( predicate( a[i] ) )
        {
            for ( size_t j = i + 1; j < n; j++ )
            {
                if ( predicate( a[j] ) && a[i] < a[j] )
                {
                    int tmp = a[i];
                    a[i] = a[j];
                    a[j] = tmp;
                }
            }
        }
    }
}   

int even( int x ) { return ( x & 1 ) == 0; }
int odd( int x ) { return ( x & 1 ); }

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

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

    sort( a, N, even );

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

    sort( a, N, odd );

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

    return 0;
}

The program output is

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

edited Mar 10 '17 at 01:05

answered Mar 10 '17 at 00:59

Vlad from Moscow

301,070
26
186
335

A very portable "even" test is `( x %2 ) == 0` as `( x & 1 ) == 0` fails with rare 1's complement platforms. A 2's complement platform with an optimizing compiler will emit efficient code just like `x&1 == 0` for original code `( x %2 ) == 0`. [ref](http://stackoverflow.com/q/160930/2410359) – chux - Reinstate Monica Mar 10 '17 at 03:56
@chux This does not make sense because positive and negative zeroes compared with zero yield true. – Vlad from Moscow Mar 10 '17 at 13:22
The comment is unclear to me. How would `(x%2) == 0` not work for you? – chux - Reinstate Monica Mar 10 '17 at 15:17
@chux I did not say that it won't work. – Vlad from Moscow Mar 10 '17 at 15:33
What specifically is your concern then with [here](http://stackoverflow.com/questions/42708001/sorting-numbers-in-c/42708303?noredirect=1#comment72558646_42708303)? – chux - Reinstate Monica Mar 10 '17 at 15:45
@chux I pointed out that you can compare negative and positive zeroes with zero and the result will be true independing on the sign of the zero. – Vlad from Moscow Mar 10 '17 at 15:50
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/137773/discussion-between-chux-and-vlad-from-moscow). – chux - Reinstate Monica Mar 10 '17 at 16:13
@chux I am sorry I do not take part in chats. – Vlad from Moscow Mar 10 '17 at 16:24

score 0 · Answer 3 · answered Mar 10 '17 at 03:46

... sort only the even numbers while leaving the odd numbers in their current place and vice versa

When encountering a value that matches SkipType, skip that array element.

// Return true when odd, else false
bool IsOdd(int x) {
  return x%2;  // x%2 --> -1,0,1
}

...
// Select skipping odd or even
bool SkipType = IsOdd(1);

for (i = 0; i < ArraySize; ++i) {
  if (IsOdd(number[i]) == SkipType) continue;
  for (j = i + 1; j < ArraySize /*size*/; ++j) {
    if (IsOdd(number[j]) == SkipType) continue;
    if (number[i] > number[j]) {  // I think OP want > here, not <
      a = number[i];
      number[i] = number[j];
      number[j] = a;
    }
  }
}

Sorting numbers in C

3 Answers3