Lisp is often used in educational contexts, where students learn to understand and implement recursive algorithms.

Production code written in Common Lisp or portable code has several issues with recursion: They do not make use of implementation-specific features like *tail call optimization*, often making it necessary to avoid recursion altogether. In these cases, implementations:

- Usually have a
*recursion depth limit*due to limits in stack sizes. Thus recursive algorithms will only work for data of limited size. - Do not always provide optimization of tail calls, especially in combination with dynamically scoped operations.
- Only provide optimization of tail calls at certain optimization levels.
- Do not usually provide
*tail call optimization*. - Usually do not provide
*tail call optimization*on certain platforms. For example, implementations on JVM may not do so, since the JVM itself does not support*tail call optimization*.

Replacing tail calls with jumps usually makes debugging more difficult; Adding jumps will cause stack frames to become unavailable in a debugger. As alternatives Common Lisp provides:

- Iteration constructs, like
`DO`

,`DOTIMES`

,`LOOP`

, and others - Higher-order functions, like
`MAP`

,`REDUCE`

, and others - Various control structures, including low-level
`go to`

```
(defun fn (x)
(cond (test-condition1 the-value1)
(test-condition2 the-value2)
...
...
...
(t (fn reduced-argument-x))))
CL-USER 2788 > (defun my-fib (n)
(cond ((= n 1) 1)
((= n 2) 1)
(t (+
(my-fib (- n 1))
(my-fib (- n 2))))))
MY-FIB
CL-USER 2789 > (my-fib 1)
1
CL-USER 2790 > (my-fib 2)
1
CL-USER 2791 > (my-fib 3)
2
CL-USER 2792 > (my-fib 4)
3
CL-USER 2793 > (my-fib 5)
5
CL-USER 2794 > (my-fib 6)
8
CL-USER 2795 > (my-fib 7)
13
```

```
(defun fn (x)
(cond (test-condition the-value)
(t (fn reduced-argument-x))))
```

```
;;Find the nth Fibonacci number for any n > 0.
;; Precondition: n > 0, n is an integer. Behavior undefined otherwise.
(defun fibonacci (n)
(cond
( ;; Base case.
;; The first two Fibonacci numbers (indices 1 and 2) are 1 by definition.
(<= n 2) ;; If n <= 2
1 ;; then return 1.
)
(t ;; else
(+ ;; return the sum of
;; the results of calling
(fibonacci (- n 1)) ;; fibonacci(n-1) and
(fibonacci (- n 2)) ;; fibonacci(n-2).
;; This is the recursive case.
)
)
)
)
```

```
;;Recursively print the elements of a list
(defun print-list (elements)
(cond
((null elements) '()) ;; Base case: There are no elements that have yet to be printed. Don't do anything and return a null list.
(t
;; Recursive case
;; Print the next element.
(write-line (write-to-string (car elements)))
;; Recurse on the rest of the list.
(print-list (cdr elements))
)
)
)
```

To test this, run:

```
(setq test-list '(1 2 3 4))
(print-list test-list)
```

The result will be:

```
1
2
3
4
```

One easy algorithm to implement as a recursive function is factorial.

```
;;Compute the factorial for any n >= 0. Precondition: n >= 0, n is an integer.
(defun factorial (n)
(cond
((= n 0) 1) ;; Special case, 0! = 1
((= n 1) 1) ;; Base case, 1! = 1
(t
;; Recursive case
;; Multiply n by the factorial of n - 1.
(* n (factorial (- n 1)))
)
)
)
```

This modified text is an extract of the original Stack Overflow Documentation created by following contributors and released under CC BY-SA 3.0

This website is not affiliated with Stack Overflow

Email: [email protected]