How will you print numbers from 1 to 100 without using loop? | Set-2
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 valueclass 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 valuedef main(i): if (i <= 100): print(i, end = " ") i = i + 1 main(i) i = 1main(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 valueusing 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
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