Cannot use function call as first argument in s-exp [duplicate]

Answer 1

Common Lisp has several different namespaces, and in function forms, the functional value of the operator is used. Your lambda example works, because lambda forms are a separate branch in the evaluation rules. You can just google for "Common Lisp namespaces" "Lisp-1 vs. Lisp-2" for more details, and there are quite a few questions and answers here on SO that cover those topics.

But, to answer your particular question, use funcall (you can also take a look at apply):

(funcall (makefun 1) 2)

Answer 2

If answered a similar question already on Stackoverflow. It's a duplicate. Would need to find it.

Anyway.

(defun makefun(x) ;1
  (lambda (z)
    (+ x z)))

((makefun 1) 2)

Common Lisp does not allow evaluation in the function position. It requires you to put a symbol or an actual lambda expression there.

Remember: Common Lisp has a separate function namespace. The value namespace is different. Here MAKEFUN returns a value and this value is not available in the function namespace.

There are only two syntactic ways to call a function:

(foo 1 2)

and

((lambda (a b) (list a b)) 1 2)

See CLHS (Conses as Forms). Here we have a Lambda Form.

Adding the capability to write ((some-function-returning-function) 1 2) would make function calling in Common Lisp more confusing:

• (foo 1 2) would use the function namespace
• ((some-function-returning-function) 1 2) would use the value namespace