> Eg no modern Lisp stores a function as an S-Expr of its syntax Many Common Lisp implementations do that, especially those with an interpreter: SBCL (yes, SBC

> Eg no modern Lisp stores a function as an S-Expr of its syntax Many Common Lisp implementations do that, especially those with an interpreter: SBCL (yes, SBCL has an interpreter, too -> one can tell SBCL to use an interpreter), LispWorks, a Lisp Machine, Allegro CL, CLISP, ... LispWorks: CL-USER 18 > (defun foo (a b) (* a (let ((a (* a a))) (break) (+ b (expt a b))))) FOO CL-USER 19 > (foo 1 2) Break. 1 (continue) Return from break. 2 (abort) Return to top loop level 0. Type :b for backtrace or :c <option number> to proceed. Type :bug-form "<subject>" for a bug report template or :? for other options. CL-USER 20 : 1 > :n Call to INVOKE-DEBUGGER CL-USER 21 : 1 > :n Call to BREAK CL-USER 22 : 1 > :n Interpreted call to FOO You'll see that LispWorks says that FOO is running using the s-expression interpreter. Next we'll have a look at this s-expression: CL-USER 23 : 1 > :lambda (LAMBDA (A B) (DECLARE (SYSTEM::SOURCE-LEVEL #<EQ Hash Table{0} 82200D6A93>)) (DECLARE (LAMBDA-NAME FOO)) (* A (LET (#) (BREAK) (+ B #)))) The list is a value, which we can pretty print: CL-USER 24 : 1 > (pprint *) (LAMBDA (A B) (DECLARE (SYSTEM::SOURCE-LEVEL #<EQ Hash Table{0} 82200D6A93>)) (DECLARE (LAMBDA-NAME FOO)) (* A (LET (#) (BREAK) (+ B #)))) If you would modify this s-expression in the debugger, the program is directly changed.