### New Copies Discovered.  ; Let us define f and L on the following way:

(set 'f (lambda(x)(eval x)))
(set 'L '(0))

; what is the result of the evaluation of (f 'L)?

(println (f 'L)) ; output: (0)

; It appears that that particular function call evaluates to the
; value of L. However, just like in my last post - it is not actual
; value of the L, but its copy. We can test it on the following way:

(print "Attempt to push 1. ")
(push 1 L)
(println "L=" L) ; output: L=(1 0), attempt succeeded

; However,

(print "Attempt to push 2. ")
(push 2 (f 'L))
(println "L=" L) ; output: L=(1 0), attempt failed

; The value of (f 'L) is EQUAL to the value of L, but not the SAME.
; In some purely functional language, it wouldn't be the issue,
; however, Newlisp is not purely functional language - and in my
; opinion, it is good that it is not. Primitive function push is
; mutator - its result is not important, its side effect on one
; particular instance of the value is.

; Why f didn't returned exactly the value of L, but its copy?
; Who made this copy? Since f is (lambda(x)(eval x)), there are
; two possibilities. The first is that eval is responsible for a
; copy, and the second is that function does not return the result
; of the evaluation of the last experssion, but the COPY of that result.

; We can test whether it is eval who did the copy easily. Let
; us evaluate:

(print "Attempt to push 3. ")
(push 3 (eval 'L))
(println "L=" L) ; output: L=(3 1 0), attempt succeeded

; So, the function is responsible. It returns the copy of the result
; of the evaluation of the last expression in its body.

; I tested several other syntactical constructs of Newlisp and
; observed that many of do the same. The most notable, macros
; do.

(print "Attempt to push 4. ")
(set 'f (lambda-macro(x)(eval x)))
(push 4 (f L))
(println "L=" L) ; output: L=(3 1 0), attempt failed

; And also:

(print "Attempt to push 5. ")
(push 5 (let()L))
(println "L=" L) ; output: L=(3 1 0), attempt failed

(print "Attempt to push 6. ")
(push 6 (begin L))
(println "L=" L) ; output: L=(3 1 0), attempt failed

; Programmer cannot write the function (or even macro) that behaves
; exactly as primitive function eval on his own. It is impossible
; even by using actual eval in the function definition - because
; whatever one does, function will never return actual value of
; some variable, say L - but always some copy of that value - and
; as we've seen (eval L) returns actual value of L.

; The solution is to return the symbol L, not its value. Such a
; symbol can be evaluated in the environment of the caller, and
; result will be exactly the value of L.

(print "Attempt to push 7. ")
(set 'f '(lambda(x)x))

(push 7 (eval (f 'L)))
(println "L=" L)  ; output: L=(7 3 1 0), attempt succeeded
(exit)

; Semantically, difference is not big. Syntactically, however,
; extra eval in the caller environment is annoyance, more from
; eaesthetical than practical point of vew. But, aesthetic is the
; pride of Lisp, and it could be worth to research the possibility
; of the syntactical constructs equivalent to lambda, lambda macro,
; and possibly begin and let expressions - differing only by
; returning actual values of the last subexpression.