Print 1 to 100 without loop using Goto and Recursive-main

Our task is to print all numbers from 1 to 100 without using a loop. There are many ways to print numbers from 1 to 100 without using a loop. Two of them are the goto statement and the recursive main.

Print numbers from 1 to 100 Using Goto statement 

Follow the steps mentioned below to implement the goto statement:

• declare variable i of value 0
• declare the jump statement named begin, which increases the value of i by 1 and prints it.
• write an if statement with condition i < 100, inside which call the jump statement begin

Below is the implementation of the above approach:

#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

#include <stdio.h>

int main()
{
    int i = 0;
begin:
    i = i + 1;
    printf("%d ", i);

    if (i < 100)
        goto begin;
    return 0;
}

public class Main {
    public static void main(String[] args) {
        int i = 0;

        while (i < 100) {
            i = i + 1;
            System.out.print(i + " ");
        }
    }
}

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

i = 0

while i < 100:
    i = i + 1
    print(i, end=' ')

// Javascript
let i = 0;

while (i < 100) {  // loop will iterate 100 times
    i += 1;
    console.log(i, end=' '); // print i and add a space
}

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(1)
Auxiliary Space: O(1)

Print numbers from 1 to 100 Using recursive-main

Follow the steps mentioned below to implement the recursive main:

• declare variable i of value 1.
• keep calling the main function till i < 100.

Below is the implementation of the above approach:

// 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;
}

// 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

// 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 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# 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

// Program to count all pairs from both
//the linked lists whose product is equal to
// a given value
function main(i) {
    if (i <= 100) {
        console.log(i + " ");
        i = i + 1;
        main(i);
    }
}

let i = 1;
main(i);

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(1)
Auxiliary Space: O(1)