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 usedfor compile time constants only, (2 )and (3 ) can
be usedfor 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