diamond star pattern with number of rows v1 .c

/*
    Corrected code.
 diamond star pattern with number of rows v1 .c

 If number_of_rows is even number, we have to add 1 to be odd number.
 This is because diamond must be simetrical.

----*-
---*-*-
--*-*-*-
-*-*-*-*-
*-*-*-*-*-
-*-*-*-*-
--*-*-*-
---*-*-
----*-

row  spaces     *

 0.     4       1
 1.     3       2
 2.     2       3
 3.     1       4
 4.     0       5

 5.     1       4
 6.     2       3
 7.     3       2
 8.     4       1


    Upper half of diamond

For number of spaces in a row we have to find relation with r and nr, (nr/2 - r).

For number of "*-" in a row we have to find relation with r, (r+1).

    Lower half of diamond

For number of spaces in a row we have to find relation with r and nr, (r - nr/2).

For number of "*-" in a row we have to find relation with r and nr, (nr-r).


If number_of_rows nr is even number, we increment it for 1, to be odd number,
because diamond must be simetrical.

*/

#include "stdio.h"

int main(void)
{
/*
    r   row,
    nr  number_of_rows,
    ns  number_of_spaces,
    na  number_of_asterisks
*/
    int i, r, nr, ns, na;

    printf("\n Diamond of how much rows, nr = ");
    scanf("%d", &nr);           // try 8, 9 or 50

    if(nr%2==0)                 // If number_of_rows nr is even number,
        nr++;                   // we increment it for 1, to be odd number,
                                // because diamond must be simetrical.
    printf("\n");               // print new row

    for (r=0; r<nr; r++)            // print rows one by one row
    {
        if(r<nr/2+1)                // Upper half of diamond
        {
            ns = nr/2-r;            // calculate ns and na
            na = r+1;
        }
        else                        // Lower half of diamond
        {
            ns = r-nr/2;            // calculate ns and na
            na = nr-r;
        }

        for (i=0; i<ns; i++)        // print ns spaces "-"
            printf("-");            // replace "-" with " "

        for (i=0; i<na; i++)        // print na "*-"
            printf("*-");           // replace "*-" with "* "

        printf("\n");               // print new row
    }

    return 0;                       // This is the simplest and most comprehensive code.
}