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

Printing 1 to 1000: Beyond Loops and Conditionals

In the realm of programming, we often rely on loops and conditional statements to iterate through elements. But what if we seek a unique approach, one that defies the conventional wisdom of loops and conditionals? This article presents an enigmatic technique for printing the numbers from 1 to 1000 without using these constructs.

The solution leverages a clever interplay of function recursion and pointer arithmetic. Consider the following code in C:

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

void main(int j) {
  printf("%d\n", j);
  (&&main + (&&exit - &amp;main)*(j/1000))(j+1);
}

Initially, it may seem like we're attempting to compute an address, but the true magic lies in the function pointer and pointer arithmetic. The snippet essentially creates a recursive function call mechanism, where the next function to be called is determined by pointer arithmetic based on the value of j.

When j reaches 1000, the pointer arithmetic evaluates to (&&exit - &main)*1000, which effectively points to the exit function. Calling exit terminates the program, ending the recursive calls and printing the desired sequence from 1 to 1000.

To ensure compatibility with standard C, the code can be modified as follows:

#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);
}

This solution demonstrates the power of recursion and pointer arithmetic, providing an elegant and unconventional method for printing numbers without relying on loops or conditionals. It serves as an intriguing reminder that creative thinking can lead to innovative solutions in the world of programming.

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