Write a program to print numbers from 1 to 100?

Below is the program to print the Numbers from 1 to 100 in C++, C, Python3, C#, PHP and JavaScript.

C++

// C++ program to How will you print
//  numbers from 1 to 100 without using loop?
#include <iostream>
using namespace std;
 
class gfg
{
  
// Prints numbers from 1 to n
public:
void printNos(unsigned int n)
{
 if(n > 0)
 {
 printNos(n - 1);
 cout << n << " ";
 }
 return;
}
};
 
// Driver code
int main()
{
 gfg g;
 g.printNos(100);
 return 0;
} 

C

#include <stdio.h>
 
void printNos(unsigned int n)
{
 if(n > 0)
 {
 printNos(n - 1);
 printf("%d ", n);
 }
 return;
}
 
// Driver code
int main()
{
 printNos(100);
 getchar();
 return 0;
}

C#

using System;
 
class GFG
{
  
 // Prints numbers from 1 to n
 static void printNos(int n)
 {
 if(n > 0)
 {
 printNos(n - 1);
 Console.Write(n + " ");
 }
 return;
 }
 
// Driver Code
public static void Main()
{
 printNos(100);
}
}

Java

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
 
class GFG
{
 // Prints numbers from 1 to n
 static void printNos(int n)
 {
 if(n > 0)
 {
 printNos(n - 1);
 System.out.print(n + " ");
 }
 return;
 }
 
 // Driver Code
 public static void main(String[] args)
 {
 printNos(100);
 }
}

PHP

<?php
// PHP program print numbers
// from 1 to 100 without
// using loop   
 
// Prints numbers from 1 to n
function printNos($n)
{
 if($n > 0)
 {
 printNos($n - 1);
 echo $n, " ";
 }
 return;
}
 
// Driver code
printNos(100);
 
?>

Python 3

def printNos(n):
 if n > 0:
 printNos(n - 1)
 print(n, end = ' ')
 
# Driver code
printNos(100)

JavaScript

<script>
  
 function printNos(n)
 {
 if(n > 0)
 {
 printNos(n - 1);
 document.write(n + " ");
 }
 return;
 }
  
 printNos(100);
 
</script>

Output:

Source: geeksofgeeks

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