newbie questions: why doesn't ((lambda (x) (x 2)) (lambda (n) (+ n 3))) work in elisp?

newbie questions: why doesn't ((lambda (x) (x 2)) (lambda (n) (+ n 3))) work in elisp?

[Tahsin Alam]

[-- Attachment #1: Type: text/plain, Size: 483 bytes --]


As the subject says - why does the following expression give error when I try
to evaluate it in the *scratch* buffer?

((lambda (x) (x 2)) (lambda (n) (+ n 3)))


---

This e-mail may contain confidential and/or privileged information. If you
are not the intended recipient (or have received this e-mail in error)
please notify the sender immediately and destroy this e-mail. Any
unauthorized copying, disclosure or distribution of the material in this
e-mail is strictly forbidden.

[-- Attachment #2: Type: text/html, Size: 595 bytes --]

Re: newbie questions: why doesn't ((lambda (x) (x 2)) (lambda (n) (+ n 3))) work in elisp?

[Joost Diepenmaat]

Tahsin Alam <tahsin.alam@db.com> writes:

> As the subject says - why does the following expression give error when I try to evaluate it in the *scratch* buffer?
>
> ((lambda (x) (x 2)) (lambda (n) (+ n 3)))

Because ((some call evaluating to a function)) isn't valid in emacs
lisp, which also means

(lambda (x) (x 2))

has no real meaning in elisp (it does in scheme, but not in elisp or
common lisp, because elisp and cl have separate namespaces for
functions and variables). You want either

(funcall (lambda (x) (funcall x 2)) (lambda (n) (+ n 3)))

or possibly

(list (lambda (x) (list x 2)) (lambda (n) (+ n 3)))

or some permutation of both.

-- 
Joost Diepenmaat | blog: http://joost.zeekat.nl/ | work: http://zeekat.nl/

Re: newbie questions: why doesn't ((lambda (x) (x 2)) (lambda (n) (+ n 3))) work in elisp?

[Joost Diepenmaat]

Joost Diepenmaat <joost@zeekat.nl> writes:
> Tahsin Alam <tahsin.alam@db.com> writes:
>
>> As the subject says - why does the following expression give error when I try to evaluate it in the *scratch* buffer?
>>
>> ((lambda (x) (x 2)) (lambda (n) (+ n 3)))
>
> Because ((some call evaluating to a function)) isn't valid in emacs
> lisp, which also means
>
> (lambda (x) (x 2))

unless, of course, you've got a (defun x (y) ...) somewhere...

> has no real meaning in elisp (it does in scheme, but not in elisp or
> common lisp, because elisp and cl have separate namespaces for
> functions and variables). You want either
>
> (funcall (lambda (x) (funcall x 2)) (lambda (n) (+ n 3)))
>
> or possibly
>
> (list (lambda (x) (list x 2)) (lambda (n) (+ n 3)))
>
> or some permutation of both.

-- 
Joost Diepenmaat | blog: http://joost.zeekat.nl/ | work: http://zeekat.nl/

Re: newbie questions: why doesn't ((lambda (x) (x 2)) (lambda (n) (+ n 3))) work in elisp?

[Thien-Thi Nguyen]

⎛⎞ Tahsin Alam <tahsin.alam@db.com>
⎝⎠ Fri, 30 May 2008 13:03:49 -0400

   As the subject says - why does the following expression give
   error when I try to evaluate it in the *scratch* buffer?

   ((lambda (x) (x 2)) (lambda (n) (+ n 3)))

One reason is because Emacs Lisp is not Scheme (sigh).
You can try this, instead:

(funcall (lambda (x) (funcall x 2)) (lambda (n) (+ n 3)))

thi

Re: newbie questions: why doesn't ((lambda (x) (x 2)) (lambda (n) (+ n 3))) work in elisp?

[Johan Bockgård]

Thien-Thi Nguyen <ttn@gnuvola.org> writes:

> ⎛⎞ Tahsin Alam <tahsin.alam@db.com>
> ⎝⎠ Fri, 30 May 2008 13:03:49 -0400
>
>    As the subject says - why does the following expression give
>    error when I try to evaluate it in the *scratch* buffer?
>
>    ((lambda (x) (x 2)) (lambda (n) (+ n 3)))
>
> One reason is because Emacs Lisp is not Scheme (sigh).
> You can try this, instead:
>
> (funcall (lambda (x) (funcall x 2)) (lambda (n) (+ n 3)))

The first funcall is unnecessary.

-- 
Johan Bockgård