不用循环和条件判断打印1-1000

//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