Printing 1 to 1000 without loop or conditionals in C/C++

Here we will see how to print 1 to 1000 without loop or any conditional statements. As the loops cannot be used, so we can try recursions, but here another constraint that, we cannot use the conditions also. So the base case of the recursion will not be used.

Here we are solving this problem using static members. At first we are initializing the static member with 1, then in the constructor we are printing the value and increase its value. Now create an array of 1000 objects of that class, so 1000 different objects are created, so the constructor is called 1000 times. Thus we can print 1 to 1000.

Example

#include<iostream>
using namespace std;
class PrintN {
   public:
      static int value;
      PrintN() {
         cout<< value++ <<", ";
      }
};
int PrintN::value = 1; //initialize to 1
main() {
   int N = 1000;
   PrintN obj[N]; //create an array of size 1000 to create 1000 objects
}

Output

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, .... 990, 991, 992, 993, 994, 995, 996, 997, 998, 999, 1000,

Anvi Jain

Updated on 30-Jul-2019 22:30:25

• Related Questions & Answers
• Printing all subsets of {1,2,3,…n} without using array or loop in C program
• Print 1 to 100 in C++, without loop and recursion
• Printing Items in 0/1 Knapsack in C++
• Program to print numbers from 1 to 100 without using loop
• How will you print numbers from 1 to 100 without using loop in C?
• How to print all the Armstrong Numbers from 1 to 1000 using C#?
• Print a character n times without using loop, recursion or goto in C++
• Print a pattern without using any loop in C++
• C program to print number series without using any loop
• Printing Pyramid in C++
• Execute MySQL query from the terminal without printing results?
• String Without AAA or BBB in C++
• Printing Heart Pattern in C
• Printing Interesting pattern in C++
• Fleury’s Algorithm for printing Eulerian Path or Circuit in C++