#define PR1(z)  printf("i=%03d  ", z );
#define PRn     printf("\n");
#define PR2(x, n, y)  for (i=1; i<n; i++) printf(x, y[i]);
int rek2(int), przes2(int), sort2(int), a[9]={0,63,35,15,48,15,27,17,63};
main()
{  int n=9, (*fptr)(int);
   fptr = rek2;    printf("Wynik dzialania rek2 = %d\n", (*fptr)(n));
   fptr = przes2;  printf(" => Wynik = %d\n", (*fptr)( 2*n+2>>2 ));
   fptr = sort2;
   printf("Co to za obiekt: int (*fptr)(float *, char)?\n"); PRn;
   printf("W wydruku zestawu 2 ma byc %d+1 linii", (*fptr)(n+1>>1)&0xa);
}
int rek2(int n)
{  printf ("Wywolanie rek2 z argumentem (%d)\n", n);
   if ( n<=0 ) return n&15;
   else return ( rek2 ( n - 5 >> 1 ) + n | 7);    }
int przes2 (int m)
{  int *ptr, i, n = 1;
   ptr=(int *) malloc(m*sizeof(int)<<1); *(ptr+n++) = m;
   {int i, j, m;
   for ( j=7, i=5, m=i&j; j>i; j--) { *(ptr + n++) = m + 1 >> 1;
            { int i, m; for ( i=4, m=j^i; i<j-1; i++) *(ptr+n++) = --m;}
        ptr[n++] = m^3; }  ptr[n++] = m/2-i; }  ptr[n++] = m<<1;
       PR2("%d", 2*m+1, ptr); return ( 4 - m >> 2 );
 }
int sort2(int n)
{  int i, j, k, l, p, x;
   PR1(1); PR2("%3o ",n,a); PRn; l = 2; p = k = n - 1;
   do { for(j=p; j>=l; j--)
           if ( a[j-1] > a[j] ) { x=a[j-1]; a[j-1]=a[j]; a[j]=x; k=j;}
        l=k+1; PR1(l); PR2("%3x ", n, a);  PRn;
        for (j=l; j<=p; j++)
	   if ( a[j-1] > a[j] )   { x=a[j-1]; a[j-1]=a[j]; a[j]=x; k=j;}
        p=k-1; PR1(p); PR2("%3d ", n, a);  PRn;
      } while (l<p); return(sizeof(a)/2|7);        }