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]

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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)))


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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

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/