# 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);  ` `}  ` ` `  `// This code is contributed by shubhamsingh10 `

## 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 `

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 `

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.

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Improved By : SHUBHAMSINGH10

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