Edit detail for LispInAxiom revision 1 of 1

It is possible to enter Lisp expressions directly into Axiom (including function definitions) using the )lisp command, but this is rather awkward for long Lisp programs.

)lisp (defun foo1 (x y) (+ x y))

Value = FOO1

)lisp (foo1 2 3)

Value = 5

On the Axiom wiki we can enter Lisp code in the same way that we enter Axiom code. For example:

  \begin{lisp}
  (defun FOO2 (x y)
    (+ x y))
  \end{lisp}

The text between the \begin{lisp} ... \end{lisp} is saved in a temporary file (e.g. tempfile.lisp). It is then compiled and loaded into Lisp via the command:

  )list (load (compile-file "tempfile.lisp"))

;;; Just a silly example
(defun FOO2 (x y)
  (+ x y))

Value = 304

Click +lisp to check the Lisp output for errors.

We can call the function from Axiom like this:

FOO2(3,4)$Lisp

(1)

Or we can call the function from Lisp like this:

(format t "(FOO2 1 2) = ~a~%" (FOO2 1 2) )

(FOO2 1 2) = 3
Value = 88

Notice that the Value = 40 is the result of the (load ...) function call, not the value of the FOO2 function.

In the case of an error:

(defunny FOO3 (x y)
  (+ x y))

 
   >> System error:
   The function DEFUNNY is undefined.

+lisp shows the error message.

This is a lisp expression that generates some output.

(format t "hello, world")

hello, world
Value = 88

Lisp Tutorial 1

http://www.cs.sfu.ca/CC/310/pwfong/Lisp/1/tutorial1.html

Common LISP reads in an expression, evaluates it, and then prints out the result. For example, if you want to compute the value of (2 * cos(0) * (4 + 6)), you type in:

)lisp (* 2 (cos 0) (+ 4 6))

Value = 20.0

Here is a more complex recursive function.

(defun fibonacci (N)
  "Compute the N'th Fibonacci number."
  (if (or (zerop N) (= N 1))
      1
    (+ (fibonacci (- N 1)) (fibonacci (- N 2)))))

Value = 600

We can trace its execution.

)lisp (trace fibonacci)

Value = (FIBONACCI)

)lisp (fibonacci 5)

  1> (FIBONACCI 5)
    2> (FIBONACCI 4)
      3> (FIBONACCI 3)
        4> (FIBONACCI 2)
          5> (FIBONACCI 1)
          <5 (FIBONACCI 1)
          5> (FIBONACCI 0)
          <5 (FIBONACCI 1)
        <4 (FIBONACCI 2)
        4> (FIBONACCI 1)
        <4 (FIBONACCI 1)
      <3 (FIBONACCI 3)
      3> (FIBONACCI 2)
        4> (FIBONACCI 1)
        <4 (FIBONACCI 1)
        4> (FIBONACCI 0)
        <4 (FIBONACCI 1)
      <3 (FIBONACCI 2)
    <2 (FIBONACCI 5)
    2> (FIBONACCI 3)
      3> (FIBONACCI 2)
        4> (FIBONACCI 1)
        <4 (FIBONACCI 1)
        4> (FIBONACCI 0)
        <4 (FIBONACCI 1)
      <3 (FIBONACCI 2)
      3> (FIBONACCI 1)
      <3 (FIBONACCI 1)
    <2 (FIBONACCI 3)
  <1 (FIBONACCI 8)
Value = 8

Lisp Links & Notes (advanced)

http://www.cs.ucla.edu/classes/winter04/cs161/l1/lispLinks.html

by Michael G. Dyer http://www.cs.ucla.edu/classes/winter04/cs161/l1

lambda, closures, apply, funcall, and mapcar

One thing to note here is that the word "thunk" is often used as a synonym for "closure".

A closure is defined as a double whose two elements are a pointer to code and a pointer to an environment. In the adder example below, the reason we need the "environment pointer" is that we need a way to remember what n was bound to at the time that (adder 3) was created.

This particular "adder" example comes from the CL Cookbook

(defun adder (n)
   (function (lambda (x) (+ x n)))
)

Value = 536

)lisp (adder 3)

Value = #<compiled-closure 0000000001efbea0>

)lisp (apply (adder 3) '(4))

Value = 7

)lisp (funcall (adder 3) 4)

Value = 7

)lisp (mapcar (adder 3) '(1 2 3 4))

Value = (4 5 6 7)

There are two slightly different ways of using mapcar. One is by passing it a lambda block, which is created by simply prepending the #' in front of the function name. The other is to pass it a closure:

(defun increment (i)
   (+ i 1)
)

Value = 304

)lisp #'increment

Value = #<compiled-function INCREMENT>

)lisp (mapcar #'increment '(1 2 3))

Value = (2 3 4)

)lisp #'(lambda (x) (increment x))

Value = (LAMBDA-CLOSURE () () () (X) (INCREMENT X))

)lisp (mapcar #'(lambda (x) (increment x)) '(1 2 3))

Value = (2 3 4)

compose

Let's take a look at a "higher-order" function, which means that it takes functions as arguments. The compose function is the same as in mathematics. f(g(x)) = compose(f,g)(x)

(defun increment (i)
  (+ i 1)
)

(defun square (z)
  (* z z)
)

(defun compose (f g)
   #'(lambda (x) (funcall f (funcall g x)))
)

Value = 1048

)lisp (funcall (compose #'increment #'square) 3)

Value = 10

curry and uncurry

(defun printthree (arg1 arg2 arg3)
  (format *query-io* "arg1 = ~A, arg2 = ~A, arg3 = ~A~%" arg1 arg2 arg3)
)

Value = 320

)lisp (printthree 3 4 4)

arg1 = 3, arg2 = 4, arg3 = 4
Value = NIL

(defun curry (function &rest args)
  (function (lambda (&rest more-args)
   (apply function (append args more-args)))))

Value = 744

)lisp (curry #'printthree 'a)

Value = #<compiled-closure 0000000001f25240>

)lisp (apply (curry #'printthree 'a) '(b c))

arg1 = A, arg2 = B, arg3 = C
Value = NIL

)lisp (curry #'printthree 'a 'b)

Value = #<compiled-closure 0000000001f25120>

)lisp (apply (curry #'printthree 'a 'b) '(c))

arg1 = A, arg2 = B, arg3 = C
Value = NIL

(defun uncurry (f)
   #'(lambda (x y)  (funcall (funcall f x) y))
)

(defun uncurry (f)
   #'(lambda (x y)  (funcall (funcall f x) y))
)

Warning: UNCURRY has a duplicate definition in this file
Value = 1088

)lisp (uncurry (curry #'+ 1))

Value = #<compiled-closure 0000000001f347e0>