How Can You Print Numbers 1 to 1000 Without Loops or Conditional Statements?

Printing Numbers from 1 to 1000 Without Loops or Conditionals

A challenge posed to programmers is to print numbers from 1 to 1000 without employing any loop structures or conditional statements. This task requires a creative approach to avoid the typical methods for iterating through a range of numbers.

One solution in C or C exploits the recursive nature of function calls. The following code bypasses loops and conditionals:

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

void main(int j) {
  printf("%d\n", j);
  (&&main + (&exit - &main)*(j/1000))(j+1);
}</stdlib.h></stdio.h>

Here, the magic lies in the use of function pointers. The &main expression represents the address of the main function, while &exit - &main calculates the size of the function in memory. By multiplying (j/1000) with this value, the function recursively calls itself, shifting its location in memory by the appropriate amount. This allows for incrementing j by 1 and continuing the printing process without any explicit looping mechanism.

Since the original code had issues with pointer arithmetic, an improved version in standard C is provided below:

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

void f(int j)
{
    static void (*const ft[2])(int) = { f, exit };

    printf("%d\n", j);
    ft[j/1000](j + 1);
}

int main(int argc, char *argv[])
{
    f(1);
}</stdlib.h></stdio.h>

In this version, a static array of function pointers is utilized to avoid the pointer arithmetic concerns. The main function initializes the array with two elements: f itself for continuing the recursion and exit for when the final j value is reached, signaling the end of the process.

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