不用循环和条件判断打印11000

//z 不用循环和条件判断打印1-1000
//z 2011-05-24 19:16:07@is2120

#include  <iostream>
template <int  N>
struct  NumberGeneration{
  static  void  out(std::ostream& os)
  {
    NumberGeneration<N-1 >::out(os);
    os << N << std::endl;
  }
};
template <>
struct  NumberGeneration<1 >{
  static  void  out(std::ostream& os)
  {
    os << 1  << std::endl;
  }
};
int  main(){
   NumberGeneration<1000 >::out(std::cout);
}

/*
———————————————————————————————————————————— */
/*
@PP, that's quite lengthy to explain, but basically, j is initially 1 because
it's actually argc, which is 1 if the program is called without arguments. Then,
j/1000 is 0 until j becomes 1000, after which it's 1. (exit - main) is, of
course, the difference between the addresses of exit() and main(). That means
(main + (exit - main)*(j/1000)) is main() until j becomes 1000, after which it
becomes exit(). The end result is that main() is called when the program starts,
then calls itself recursively 999 times while incrementing j, then calls exit().
Whew :)
*/
#include  <stdio.h>
#include  <stdlib.h>

void  main(int  j) {
  printf(" %d /n " , j);
  (&main + (&exit - &main)*(j/1000 ))(j+1 );
}

#include  <stdio.h>
#include  <stdlib.h>

void  f(int  j)
{
    static  void  (*const  ft[2 ])(int ) = { f, exit };

    printf(" %d /n " , j);
    ft[j/1000 ](j + 1 );
}

int  main(int  argc, char  *argv[])
{
    f(1 );
}
/* ———————————————————————————————————————————— */

/*
I'm surprised nobody seems to have posted this -- I thought it was the most
obvious way. 1000 = 5*5*5*8.
*/

#include  <stdio.h>
int  i = 0 ;
p()    { printf(" %d /n " , ++i); }
a()    { p();p();p();p();p(); }
b()    { a();a();a();a();a(); }
c()    { b();b();b();b();b(); }
main() { c();c();c();c();c();c();c();c(); return  0 ; }
/* ———————————————————————————————————————————— */
[ Edit: (1 ) and  (4 ) can be used for  compile time constants only, (2 ) and  (3 ) can
be used for  runtime expressions too — end edit. ]

// compile time recursion
template <int  N> void  f1()
{
    f1<N-1 >();
    cout << N << '/n' ;
}

template <> void  f1<1 >()
{
    cout << 1  << '/n' ;
}

// short circuiting (not a conditional statement)
void  f2(int  N)
{
    N && (f2(N-1 ), cout << N << '/n' );
}

// constructors!
struct  A {
    A() {
        static  int  N = 1 ;
        cout << N++ << '/n' ;
    }
};

int  main()
{
    f1<1000 >();
    f2(1000 );
    delete [] new  A[1000 ]; // (3)
    A data[1000 ]; // (4) added by Martin York
}

/* ———————————————————————————————————————————— */
#include  <stdio.h>
#define MAX  1000
int  boom;
int  foo(n) {
    boom = 1  / (MAX-n+1 );
    printf(" %d /n " , n);
    foo(n+1 );
}
int  main() {
    foo(1 );
}

//z 2011-05-24 19:16:11@is2120