# Practice Questions for Recursion | Set 2

Explain the functionality of following functions.

**Question 1**

## C

`/* Assume that n is greater than or equal to 1 */` `int` `fun1(` `int` `n)` `{` ` ` `if` `(n == 1)` ` ` `return` `0;` ` ` `else` ` ` `return` `1 + fun1(n / 2);` `}` |

## Java

`/* Assume that n is greater than or equal to 1 */` `static` `int` `fun1(` `int` `n)` `{` ` ` `if` `(n == ` `1` `)` ` ` `return` `0` `;` ` ` `else` ` ` `return` `1` `+ fun1(n / ` `2` `);` `}` `// This code is contributed by shubhamsingh10` |

## Python3

`# Assume that n is greater than or equal to 1 */` `def` `fun1(n):` ` ` `if` `(n ` `=` `=` `1` `):` ` ` `return` `0` ` ` `else` `:` ` ` `return` `1` `+` `fun1(n` `/` `/` `2` `)` `# This code is contributed by shubhamsingh10` |

## C#

`/* Assume that n is greater than or equal to 1 */` `static` `int` `fun1(` `int` `n)` `{` ` ` `if` `(n == 1)` ` ` `return` `0;` ` ` `else` ` ` `return` `1 + fun1(n / 2);` `}` `// This code is contributed by shubhamsingh10` |

## Javascript

`<script>` `/* Assume that n is greater than or equal to 1 */` `function` `fun1(n)` `{` ` ` `if` `(n == 1)` ` ` `return` `0` ` ` `else` ` ` `return` `1 + fun1(n / 2)` `}` `// This code is contributed by gottumukkalabobby` `</script>` |

## C++

`/* Assume that n is greater than or equal to 1 */` `int` `fun1(` `int` `n)` `{` ` ` `if` `(n == 1)` ` ` `return` `0;` ` ` `else` ` ` `return` `1 + fun1(n / 2);` `}` `// This code is contributed by shubhamsingh10` `// Improved by Adwitiya Mourya` |

Answer: The function calculates and returns

. For example, if n is between 8 and 15 then fun1() returns 3. If n is between 16 to 31 then fun1() returns 4.**Question 2**

## C++

`/* Assume that n is greater than or equal to 0 */` `void` `fun2(` `int` `n)` `{` `if` `(n == 0)` ` ` `return` `;` `fun2(n/2);` `cout << n%2 << endl;` `}` `//This code is contributed by shubhamsingh10` |

## C

`/* Assume that n is greater than or equal to 0 */` `void` `fun2(` `int` `n)` `{` ` ` `if` `(n == 0)` ` ` `return` `;` ` ` `fun2(n/2);` ` ` `printf` `(` `"%d"` `, n%2);` `} ` |

## Java

`/* Assume that n is greater than or equal to 1 */` `static` `void` `fun2(` `int` `n)` `{` `if` `(n == ` `0` `)` ` ` `return` `;` ` ` `fun2(n/` `2` `);` `System.out.println(n%` `2` `);` `}` `// This code is contributed by Shubhamsingh10` |

## Python3

`# Assume that n is greater than or equal to 0 */` `def` `fun2(n):` ` ` `if` `(n ` `=` `=` `0` `):` ` ` `return` ` ` ` ` `fun2(n ` `/` `2` `)` ` ` `print` `(n ` `%` `2` `, end` `=` `"")` `# This code is contributed by shubhamsingh10` |

## C#

`void` `fun2(` `int` `n)` `{` `if` `(n == 0)` ` ` `return` `;` ` ` `fun2(n/2);` `Console.Write(n%2);` `}` `// This code is contributed by shubhamsingh10` |

## Javascript

`<script>` `// Assume that n is greater than or equal to 1` `function` `fun2(n)` `{` ` ` `if` `(n == 0)` ` ` `return` `;` ` ` ` ` `fun2(n / 2);` ` ` `document.write(n % 2 + ` `" "` `)` `}` `// This code is contributed by gottumukkalabobby` `</script>` |

Answer: The function fun2() prints binary equivalent of n. For example, if n is 21 then fun2() prints 10101.

Note that above functions are just for practicing recursion, they are not the ideal implementation of the functionality they provide.

Please write comments if you find any of the answers/codes incorrect.

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