我觉得这个答案会很简单，很容易理解。

int print1000(int num=1)
{
    printf("%d\n", num);

    // it will check first the num is less than 1000. 
    // If yes then call recursive function to print
    return num<1000 && print1000(++num); 
}

int main()
{
    print1000();
    return 0;        
}

template <int To, int From = 1>
struct printer {
    static void print() {
        cout << From << endl; 
        printer<To, From + 1>::print();
    }
};    

template <int Done>
struct printer<Done, Done> {
     static void print() {
          cout << Done << endl;
     }
};

int main() 
{
     printer<1000>::print();
}

适合c++爱好者

int main() {
  std::stringstream iss;
  iss << std::bitset<32>(0x12345678);
  std::copy(std::istream_iterator< std::bitset<4> >(iss), 
            std::istream_iterator< std::bitset<4> >(),
            std::ostream_iterator< std::bitset<4> >(std::cout, "\n")); 
}

#include <iostream>
#include <iterator>
using namespace std;

int num() { static int i = 1; return i++; }
int main() { generate_n(ostream_iterator<int>(cout, "\n"), 1000, num); }

这是标准C:

#include <stdio.h>

int main(int argc, char **argv)
{
    printf("%d ", argc);
    (void) (argc <= 1000 && main(argc+1, 0));

    return 0;
}

如果不带参数调用它，它将输出从1到1000的数字。注意，&&操作符不是一个“条件语句”，尽管它具有相同的目的。

Javascript是为了好玩。包括自动停止在1000:

var max = 1000;
var b = ["break"];
function increment(i) {
    var j = Math.abs(i - max);
    console.log(j);           
    b[(i/i) - 1].toString();
    i--;
    increment(i);    
}
increment(max);