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  • Math operation abstraction

    From Daniel Cerqueira@21:1/5 to All on Mon May 27 23:29:56 2024
    Hi.

    I have this function

    ```
    (defun muo (func a b)
    (cond
    ((eq b 1) a)
    (t (funcall func a (muo func a (1- b))))))
    ```

    Which I can use for obtaining the multiplication operation using the
    addition. As an example:

    * (muo #'+ 5 2)
    10

    and "muo #'+" means "use the multiplication".

    Now if I want the exponentiation I have to do:

    * (muo #'(lambda (a b) (muo #'+ a b)) 5 2)
    25

    This creates the exponentiation.

    I would like to obtain the exponentiation using "(muo #'+ 5 2 2)"
    instead, being the last argument ("2") the number of times the operation
    is applied. In the case of exponentiation, it applies the multiplication
    2 times, resulting in the exponentiation math operation.

    I am hoping I am explaining this concept well enough.

    Basically, "muo #'+" creates the addition twice, which is the
    multiplication, and "muo #'+ a b 2" should create the multiplication
    twice, which is the exponentiation. If I use "muo #'+ a b 3" it should
    create the exponentiation twice, which is the exponentiation of the exponentiation.

    I need help in making this code modification. Seeing code is what I
    want. If you can write the code and explain it, better.

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  • From B. Pym@21:1/5 to Daniel Cerqueira on Mon May 27 23:54:43 2024
    On 5/27/2024, Daniel Cerqueira wrote:

    * (muo #'(lambda (a b) (muo #'+ a b)) 5 2)

    Pascal Bourguignon wrote:

    #'(lambda (letter)
    (add-counter letters letter))
    word))
    (read-words infile)))

    (function (lambda ...)) is a pleonasm. lambda is a macro that
    already expands to (function (lambda ...)).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)