Remove even numbers from array in c

Question

Hello i\'m trying for about 2 hours to create a program which will remove even numbers from a dinamyc allocated array(with malloc)in c.Can somebody help me with some tips or cre

Answer 1

Let's assume that you already allocated dynamically an array of n elements and initialized it.

In this case the function that removes elements with even values can look the following way

size_t remove_even( int *a, size_t n )
{
    size_t m = 0;

    for ( size_t i = 0; i < n; i++ )
    {
        if ( a[i] % 2 != 0 )
        {
            if ( i != m ) a[m] = a[i];
            ++m;
        }
    }

    return m;
}

It can be called the following way

size_t m = remove_even( p, n );

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

where p is the pointer to your dynamically allocated array of n elements.

The function actually removes nothing. It simply moves odd elements to the beginning of the array.

You can then use standard C function realloc to delete physically the removed elements.

For example

int *tmp = realloc( p, m * sizeof( int ) );

if ( tmp != NULL ) p = tmp;

Here is a demonstrative program

#include 
#include 

size_t remove_even( int a[], size_t n )
{
    size_t m = 0;

    for ( size_t i = 0; i < n; i++ )
    {
        if ( a[i] % 2 != 0 )
        {
            if ( i != m ) a[m] = a[i];
            ++m;
        }
    }

    return m;
}

#define N   10

int main( void )
{
    int *a = malloc( N * sizeof( int ) );

    for ( size_t i = 0; i < N; i++ ) a[i] = i;

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

    size_t m = remove_even( a, N );

    int *tmp = realloc( a, m * sizeof( int ) );

    if ( tmp != NULL ) a = tmp;

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

    free( a );
}

Its output is

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