How will you print numbers from 1 to 100 without using loop?

If we take a look at this problem carefully, we can see that the idea of “loop” is to track some counter value e.g. “i=0” till “i <= 100". So if we aren't allowed to use loop, how else can be track something in C language!

Well, one possibility is the use of ‘recursion’ provided we use the terminating condition carefully. Here is a solution that prints numbers using recursion.

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; 
} 
  
  
// This code is contributed by SoM15242 

C

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

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); 
    } 
} 
  
// This code is contributed by Manish_100 

Python3

# Python3 program to Print 
# numbers from 1 to n  
  
def printNos(n): 
    if n > 0: 
        printNos(n - 1) 
        print(n, end = ' ') 
  
# Driver code  
printNos(100) 
  
# This code is contributed by Smitha Dinesh Semwal 

C#

// C# code for print numbers from  
// 1 to 100 without using loop 
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); 
} 
} 
  
// This code is contributed by Ajit 

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); 
  
// This code is contributed by vt_m 
?> 

Output :

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 
16 17 18 19 20 21 22 23 24 25 26 27 
28 29 30 31 32 33 34 35 36 37 38 39
40 41 42 43 44 45 46 47 48 49 50 51
52 53 54 55 56 57 58 59 60 61 62 63
64 65 66 67 68 69 70 71 72 73 74 75
76 77 78 79 80 81 82 83 84 85 86 87
88 89 90 91 92 93 94 95 96 97 98 99 
100

Time Complexity : O(n)

Now try writing a program that does the same but without any “if” construct.
Hint — use some operator which can be used instead of “if”.

Please note that recursion technique is good but every call to the function creates one “stack-frame” in program stack. So if there’s constraint to the limited memory and we need to print large set of numbers, “recursion” might not be a good idea. So what could be the other alternative?

Another alternative is “goto” statement. Though use of “goto” is not suggestible as a general programming practice as “goto” statement changes the normal program execution sequence yet in some cases, use of “goto” is the best working solution.

So please give a try printing numbers from 1 to 100 with “goto” statement. You can use GfG IDE!

Print 1 to 100 in C++, without loop and recursion

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