被接受答案的c++变体:

void print(vector<int> &v, int ind)
{
    v.at(ind);
    std::cout << ++ind << std::endl;
    try
    {
        print(v, ind);
    }
    catch(std::out_of_range &e)
    {
    }
}

int main()
{
    vector<int> v(1000);
    print(v, 0);
}

丑陋的C答案(每10的幂只展开一个堆栈帧):

#define f5(i) f(i);f(i+j);f(i+j*2);f(i+j*3);f(i+j*4)
void f10(void(*f)(int), int i, int j){f5(i);f5(i+j*5);}
void p1(int i){printf("%d,",i);}
#define px(x) void p##x##0(int i){f10(p##x, i, x);}
px(1); px(10); px(100);

void main()
{
  p1000(1);
}

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();
}

如果POSIX解决方案被接受:

#include <stdio.h>
#include <signal.h>
#include <stdlib.h>
#include <sys/time.h>
#include <pthread.h>

static void die(int sig) {
    exit(0);
}

static void wakeup(int sig) {
    static int counter = 1;
    struct itimerval timer;
    float i = 1000 / (1000 - counter);

    printf("%d\n", counter++);

    timer.it_interval.tv_sec = 0;
    timer.it_interval.tv_usec = 0;
    timer.it_value.tv_sec = 0;
    timer.it_value.tv_usec = i; /* Avoid code elimination */
    setitimer(ITIMER_REAL, &timer, 0);
}

int main() {
    pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
    signal(SIGFPE, die);
    signal(SIGALRM, wakeup);
    wakeup(0);
    pthread_mutex_lock(&mutex);
    pthread_mutex_lock(&mutex); /* Deadlock, YAY! */
    return 0;
}

#include <boost/mpl/range_c.hpp>
#include <boost/mpl/for_each.hpp>
#include <boost/lambda/lambda.hpp>
#include <iostream>

int main()
{
  boost::mpl::for_each<boost::mpl::range_c<unsigned, 1, 1001> >(std::cout << boost::lambda::_1 << '\n');
  return(0);
}

应该在任何不喜欢0 / 0的机器上工作。如果需要，可以用空指针引用替换它。程序可以在打印1到1000后失败，对吧?

#include <stdio.h>

void print_1000(int i);
void print_1000(int i) {
    int j;
    printf("%d\n", i);
    j = 1000 - i;
    j = j / j;
    i++;
    print_1000(i);
}

int main() {
    print_1000(1);
}