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# Practice Questions for Recursion | Set 3

• Difficulty Level : Easy
• Last Updated : 09 Apr, 2021

Explain the functionality of below recursive functions.

Question 1

## C++

 void fun1(int n){   int i = 0;     if (n > 1)     fun1(n - 1);   for (i = 0; i < n; i++)     cout << " * ";} // This code is contributed by shubhamsingh10

## C

 void fun1(int n){   int i = 0;    if (n > 1)     fun1(n-1);   for (i = 0; i < n; i++)     printf(" * ");}

## Java

 static void fun1(int n){   int i = 0;     if (n > 1)     fun1(n - 1);   for (i = 0; i < n; i++)     System.out.print(" * ");}  // This code is contributed by shubhamsingh10

## Python3

 def  fun1(n):    i = 0    if (n > 1):        fun1(n - 1)    for i in range(n):        print(" * ",end="") # This code is contributed by shubhamsingh10

## C#

 static void fun1(int n){    int i = 0;    if (n > 1)        fun1(n-1);    for (i = 0; i < n; i++)        Console.Write(" * ");} // This code is contributed by shubhamsingh10

## Javascript



Answer: Total numbers of stars printed is equal to 1 + 2 + …. (n-2) + (n-1) + n, which is n(n+1)/2.

Question 2

## C++

 #define LIMIT 1000void fun2(int n){  if (n <= 0)     return;  if (n > LIMIT)    return;  cout << n <<" ";  fun2(2*n);  cout << n <<" ";} // This code is contributed by shubhamsingh10

## C

 #define LIMIT 1000void fun2(int n){  if (n <= 0)     return;  if (n > LIMIT)    return;  printf("%d ", n);  fun2(2*n);  printf("%d ", n);}

## Java

 int LIMIT = 1000;void fun2(int n){    if (n <= 0) return;    if (n > LIMIT) return;     System.out.print(String.format("%d ", n));    fun2(2 * n);    System.out.print(String.format("%d ", n));}

## Python3

 LIMIT = 1000def fun2(n):    if (n <= 0):        return    if (n > LIMIT):        return    print(n, end=" ")    fun2(2 * n)    print(n, end=" ") # This code is contributed by shubhamsingh10

## C#

 int LIMIT = 1000void fun2(int n){    if (n <= 0)        return;    if (n > LIMIT)        return;    Console.Write(n+" ");    fun2(2*n);    Console.Write(n+" ");} // This code is contributed by Shubhamsingh10

## Javascript



Answer: For a positive n, fun2(n) prints the values of n, 2n, 4n, 8n … while the value is smaller than LIMIT. After printing values in increasing order, it prints same numbers again in reverse order. For example fun2(100) prints 100, 200, 400, 800, 800, 400, 200, 100.
If n is negative, the function is returned immediately.