Unquoting without eval - macros

Question

I store some macros in quoted form (because in fact they produce lambdas with tricky lexical environment and I prefer store and serialize them as lists). So now I'm trying:
(defun play (s)
(funcall (macroexpand s)))
Macroexpand evaluates quoted lambda, so funcall can't run it. How to unquote result of macroexpand without eval? Because in my case it would cause indefensible security hole.
MORE INFO:
What I get look like this (in simplest case):
FUNCALL: #1=#'(LAMBDA (#:G6008) (SYMBOL-MACROLET NIL T)) is not a function name; try using a symbol instead
and symbol-macrolet is what actually builds up "tricky lexical environment" inside lambda.

Answer 1

Macroexpand evaluates quoted lambda, so funcall can't run it. How to
unquote result of macroexpand without eval? Because in my case it
would cause indefensible security hole.
I think that Sylwester's comment about the XY problem is probably right here; it sounds like you're trying to do something that might be done better in a different way. That said, if you have a list that's a lambda expression, you can use coerce to get a function object instead of using eval. That is, you can do this:
CL-USER> (funcall '(lambda () 42))
; Error, like you've been having
CL-USER> (funcall (coerce '(lambda () 42) 'function))
42 ; turned the list (lambda () 42) into a function and called it
This is described in the documentation for coerce; when the "output" type is function, this is what happens with the object argument:
If the result-type is function, and object is any function name that
is fbound but that is globally defined neither as a macro name nor as
a special operator, then the result is the functional value of object.
If the result-type is function, and object is a lambda expression,
then the result is a closure of object in the null lexical
environment.
Thus, if you have a function that returns list of the form (lambda ...), you can use coerce and funcall with its result. This includes macroexpansions, although you may want to use macroexpand-1 rather than macroexpand, because lambda is already a macro, so if you expand too far, (lambda () ...) turns into (function (lambda () ...)).
CL-USER> (defmacro my-constantly (value)
`(lambda () ,value))
MY-CONSTANTLY
CL-USER> (macroexpand-1 '(my-constantly 36))
(LAMBDA () 36)
T
CL-USER> (funcall (coerce (macroexpand-1 '(my-constantly 36)) 'function))
36
If you try that with the plain macroexpand, though, there's a problem. Consider yourself warned:
CL-USER> (macroexpand '(my-constantly 36))
#'(LAMBDA () 36) ; not a list, but a function
T
CL-USER> (funcall (coerce (macroexpand '(my-constantly 36)) 'function))
; Error. :(

Answer 2

I think this is a case where you will find that the REPL is your friend. To get you started:
cl-user> (defmacro foo () #'(lambda () :hi))
foo
cl-user> (foo)
#<Compiled-function (:internal foo) (Non-Global) #x3020014F82FF>
cl-user> (funcall *)
:hi
cl-user> (macroexpand '(foo))
#<Compiled-function (:internal foo) (Non-Global) #x3020014F82FF>
t
cl-user> (funcall *)
:hi
cl-user>
I'll note in passing that the lambda form that appears in your example takes an argument, while your funcall doesn't provide one.

Answer 3

There are several issues with this. First I think this is a XY problem so if you show more of your problem I guess we might find a solution for it.
It has nothing with unquoting since when evaluation a quoted expression it's not longer quoted, but it's turn into data representation of the original quoted expression. You must eval data if you want to run it.
With eval you won't get the current lexical scope. So since you mention eval and lexical environment in the same post makes me thing you won't get what you want.
Now it's no problem making a list of functions, like this:
(list (lambda (x) (+ x x))
This is not quoted since I use list and teh lambda is evaluated to a closure. If it was assigned to variable x you could call it with (funcall (car x) 10)) ; ==> 20
EDIT
I actually made the same error message with this code:
(defmacro test () '#'(lambda (x) x))
(macroexpand '(test)) ; ==> #'(lambda (x) x) ; t
(funcall (macroexpand '(test)) 5) ; ==>
*** - funcall: #'(lambda (x) x) is not a function name; try using a symbol
instead
It doesn't work since you cannot call a lambda unevaluated. You need to call the closure (function) which is the result of the evaluation of the lambda form. If you would instead not quote it you'll have that evaluation in the macro:
(defmacro test () #'(lambda (x) x))
(macroexpand '(test)) ; ==> #<function :lambda (x) x> ;
(funcall (macroexpand '(test)) 5) ; ==> 5
Actually I cannot see why you would need to make this a macro. Lets make it a function instead.
(defun test () #'(lambda (x) x))
(funcall (test) 5 ) ; ==> 5
Even if this was more comples in most cases you would do with a closure.