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;
}

Manglesky的解决方案很棒，但还不够模糊。: -):

#include <stdio.h>
#define TEN(S) S S S S S S S S S S
int main() { int i = 1; TEN(TEN(TEN(printf("%d\n", i++);))) return 0; }

与宏!

#include<stdio.h>
#define x001(a) a
#define x002(a) x001(a);x001(a)
#define x004(a) x002(a);x002(a)
#define x008(a) x004(a);x004(a)
#define x010(a) x008(a);x008(a)
#define x020(a) x010(a);x010(a)
#define x040(a) x020(a);x020(a)
#define x080(a) x040(a);x040(a)
#define x100(a) x080(a);x080(a)
#define x200(a) x100(a);x100(a)
#define x3e8(a) x200(a);x100(a);x080(a);x040(a);x020(a);x008(a)
int main(int argc, char **argv)
{
  int i = 0;
  x3e8(printf("%d\n", ++i));
  return 0;
}

这只使用O(log N)堆栈，并使用麦卡锡评估http://en.wikipedia.org/wiki/Short-circuit_evaluation作为其递归条件。

#include <stdio.h>

int printN(int n) {
  printf("%d\n", n);
  return 1;
}

int print_range(int low, int high) {
  return ((low+1==high) && (printN(low)) ||
      (print_range(low,(low+high)/2) && print_range((low+high)/2, high)));
}

int main() {
  print_range(1,1001);
}

该任务从未指定程序必须在1000之后终止。

void f(int n){
   printf("%d\n",n);
   f(n+1);
}

int main(){
   f(1);
}

(如果你run ./a，可以缩写为this。没有额外的参数)

void main(int n) {
   printf("%d\n", n);
   main(n+1);
}

#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; }

我很惊讶似乎没有人张贴这个——我认为这是最明显的方式。1000 = 5*5*5*8。