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How will you print numbers from 1 to 100 without using loop? | Set-2

  • Difficulty Level : Basic
  • Last Updated : 30 Jun, 2021

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!
It can be done in many ways to print numbers using any looping conditions such as for(), while(), do-while(). But the same can be done without using loops (using recursive functions, goto statement).

Printing numbers from 1 to 100 using recursive functions has already been discussed in Set-1 . In this post, other two methods have been discussed: 

1. Using goto statement: 

C++




#include <iostream>
using namespace std;
 
int main()
{
    int i = 0;
     
begin:
    i = i + 1;
    cout << i << " ";
 
    if (i < 100)
    {
        goto begin;
    }
    return 0;
}
 
// This code is contributed by ShubhamCoder

C




#include <stdio.h>
 
int main()
{
    int i = 0;
begin:
    i = i + 1;
    printf("%d ", i);
 
    if (i < 100)
        goto begin;
    return 0;
}

C#




using System;
 
class GFG{
 
static public void Main ()
{
    int i = 0;
    begin:
        i = i + 1;
        Console.Write(" " + i + " ");
 
        if (i < 100)
        {
            goto begin;
        }
}
}
 
// This code is contributed by ShubhamCoder
Output: 
1 2 3 4 . . . 97 98 99 100

 

2. Using recursive main function: 

C++




// C++ program to count all pairs from both the
// linked lists whose product is equal to
// a given value
#include <iostream>
using namespace std;
 
int main()
{
    static int i = 1;
     
    if (i <= 100)
    {
        cout << i++ << " ";
        main();
    }
    return 0;
}
 
// This code is contributed by ShubhamCoder

C




// C program to count all pairs from both the
// linked lists whose product is equal to
// a given value
#include <stdio.h>
 
int main()
{
    static int i = 1;
    if (i <= 100) {
        printf("%d ", i++);
        main();
    }
    return 0;
}

Java




// Java program to count all pairs from both the
// linked lists whose product is equal to
// a given value
class GFG
{
    static int i = 1;
 
    public static void main(String[] args)
    {
 
        if (i <= 100)
        {
            System.out.printf("%d ", i++);
            main(null);
        }
    }
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to count all pairs from both
# the linked lists whose product is equal to
# a given value
def main(i):
     
    if (i <= 100):
        print(i, end = " ")
        i = i + 1
        main(i)
         
i = 1
main(i)
 
# This code is contributed by SoumikMondal

C#




// C# program to count all pairs from both the
// linked lists whose product is equal to
// a given value
using System;
 
class GFG
{
    static int i = 1;
 
    public static void Main(String[] args)
    {
        if (i <= 100)
        {
            Console.Write("{0} ", i++);
            Main(null);
        }
    }
}
 
// This code is contributed by Rajput-Ji
Output: 
1 2 3 4 . . . 97 98 99 100

 

 

Want to learn from the best curated videos and practice problems, check out the C Foundation Course for Basic to Advanced C.



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