这是我的两个解。第一个是c#，第二个是C语言:

C#:

const int limit = 1000;

Action<int>[] actions = new Action<int>[2];
actions[0] = (n) => { Console.WriteLine(n); };
actions[1] = (n) => { Console.WriteLine(n);  actions[Math.Sign(limit - n-1)](n + 1); };

actions[1](0);

C:

#define sign(x) (( x >> 31 ) | ( (unsigned int)( -x ) >> 31 ))

void (*actions[3])(int);

void Action0(int n)
{
    printf("%d", n);
}

void Action1(int n)
{
    int index;
    printf("%d\n", n);
    index = sign(998-n)+1;
    actions[index](++n);
}

void main()
{
    actions[0] = &Action0;
    actions[1] = 0; //Not used
    actions[2] = &Action1;

    actions[2](0);
}

令人惊讶的是，如果你放弃了“必须是C或c++”的要求，事情会变得多么简单:

Unix shell:

echo {1..1000} | tr ' ' '\n'

or

yes | nl | awk '{print $1}' | head -1000

如果你在一个没有yes命令的Unix变体上运行，使用其他进程至少生成1000行:

find / 2> /dev/null | nl | awk '{print $1}' | head -1000

or

cat /dev/zero | uuencode - | nl | awk '{print $1}' | head -1000

or

head -1000 /etc/termcap | nl -s: | cut -d: -f1

易如反掌!: P

#include <iostream>

static int current = 1;

struct print
{
    print() { std::cout << current++ << std::endl; }
};

int main()
{
    print numbers [1000];
}

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;        
}

如果你的C编译器支持block，那么你可以这样写:

#include <stdio.h>

void ten(void (^b)(void)) { b();b();b();b();b();b();b();b();b();b(); }

int main() {
    __block int i = 0;
    ten(^{
        ten(^{
            ten(^{
                printf("%d\n", ++i);
            });
        });
    });
    return 0;
}

c++利用RAII

#include <iostream>
using namespace std;

static int i = 1;
struct a
{
    a(){cout<<i++<<endl;}
    ~a(){cout<<i++<<endl;}
}obj[500];

int main(){return 0;}

C语言开发宏

#include <stdio.h>

#define c1000(x) c5(c5(c5(c4(c2(x))))) 
#define c5(x) c4(x) c1(x) //or x x x x x
#define c4(x) c2(c2(x))   //or x x x x
#define c2(x) c1(x) c1(x) //or x x
#define c1(x) x

int main(int i){c1000(printf("%d\n",i++);)return 0;}

编辑:还有一个，这个很简单

#include <stdio.h>
#define p10(x) x x x x x x x x x x
int main(int i){p10(p10(p10(printf("%d\n",i++);)))return 0;} 

解析:选C

编辑:此c代码包含<=和?:操作符

#include <stdio.h>

int main(int i){return (i<=1000)?main(printf("%d\n",i++)*0 + i):0;}