很难看透所有已经提出的解决方案，所以这可能是一个重复。

我想要一些相对简单的东西，只有纯C，而不是c++。它使用递归，但与我看到的其他解相反，它只做对数深度的递归。通过查找表可以避免使用条件。

typedef void (*func)(unsigned, unsigned);
void printLeaf(unsigned, unsigned);
void printRecurse(unsigned, unsigned);


func call[2] = { printRecurse, printLeaf };

/* All array members that are not initialized 
   explicitly are implicitly initialized to 0 
   according to the standard. */
unsigned strat[1000] = { 0, 1 };


void printLeaf(unsigned start, unsigned len) {
  printf("%u\n", start);
}

void printRecurse(unsigned start, unsigned len) {
  unsigned half0 = len / 2;
  unsigned half1 = len - half0;
  call[strat[half0]](start, half0);
  call[strat[half1]](start + half0, half1);
}

int main (int argc, char* argv[]) {
  printRecurse(0, 1000);
}

这甚至可以通过使用一个指针动态地完成。相关的变化:

unsigned* strat = 0;

int main (int argc, char* argv[]) {
  strat = calloc(N, sizeof(*strat));
  strat[1] = 1;
  printRecurse(0, N);
}

很难看透所有已经提出的解决方案，所以这可能是一个重复。

我想要一些相对简单的东西，只有纯C，而不是c++。它使用递归，但与我看到的其他解相反，它只做对数深度的递归。通过查找表可以避免使用条件。

typedef void (*func)(unsigned, unsigned);
void printLeaf(unsigned, unsigned);
void printRecurse(unsigned, unsigned);


func call[2] = { printRecurse, printLeaf };

/* All array members that are not initialized 
   explicitly are implicitly initialized to 0 
   according to the standard. */
unsigned strat[1000] = { 0, 1 };


void printLeaf(unsigned start, unsigned len) {
  printf("%u\n", start);
}

void printRecurse(unsigned start, unsigned len) {
  unsigned half0 = len / 2;
  unsigned half1 = len - half0;
  call[strat[half0]](start, half0);
  call[strat[half1]](start + half0, half1);
}

int main (int argc, char* argv[]) {
  printRecurse(0, 1000);
}

这甚至可以通过使用一个指针动态地完成。相关的变化:

unsigned* strat = 0;

int main (int argc, char* argv[]) {
  strat = calloc(N, sizeof(*strat));
  strat[1] = 1;
  printRecurse(0, N);
}

未经测试，但应该是香草标准C:

void yesprint(int i);
void noprint(int i);

typedef void(*fnPtr)(int);
fnPtr dispatch[] = { noprint, yesprint };

void yesprint(int i) {
    printf("%d\n", i);
    dispatch[i < 1000](i + 1);
}

void noprint(int i) { /* do nothing. */ }

int main() {
    yesprint(1);
}

        #include <stdio.h>
        #include <stdlib.h>
        #include <string.h>

        typedef void(*word)(int);

        word words[1024];

        void print(int i) {
                printf("%d\n", i);
                words[i+1](i+1);
        }

        void bye(int i) {
                exit(0);
        }

        int main(int argc, char *argv[]) {
                words[0] = print;
                words[1] = print;
                memcpy(&words[2], &words[0], sizeof(word) * 2); // 0-3
                memcpy(&words[4], &words[0], sizeof(word) * 4); // 0-7
                memcpy(&words[8], &words[0], sizeof(word) * 8); // 0-15
                memcpy(&words[16], &words[0], sizeof(word) * 16); // 0-31
                memcpy(&words[32], &words[0], sizeof(word) * 32); // 0-63
                memcpy(&words[64], &words[0], sizeof(word) * 64); // 0-127
                memcpy(&words[128], &words[0], sizeof(word) * 128); // 0-255
                memcpy(&words[256], &words[0], sizeof(word) * 256); // 0-511
                memcpy(&words[512], &words[0], sizeof(word) * 512); // 0-1023
                words[1001] = bye;
                words[1](1);
        }

正如它是回答:)

在不使用任何循环或条件语句的情况下，从1到1000的数字。不要只写printf()或cout语句1000次。”);

#include <stdio.h>
void main(int i){printf("%d\n",i)&&i++<1000&&(*((int*)&i-1)-=5);} 

另一个:

#include <stdio.h>
int main(int i){return i<=1000&&printf("%d\n",i)&&main(++i);}