Re: How does letf work?

[Florian Beck]

Thank you, I think I got it now.

If I understand correctly, it works like this:

{1} means pointer to cons cell 1; <...> means a shadowed cons cell

Simplified example.

(letf*                ; x undefined
    ((x '(A B))       ; x={1} 1=[A|{2}]  2=[B|()]
     ((cdr x) '(C)))  ;                 <2=[C|()]>
  x                   ; x={1} 1=[A|{2}] <2=[C|()]>
  )                   ; returns {1}, i.e. pointer to cell {1}
                      ; which still point to {2}, but {2}'s
                      ; original value has been restored

Same with car:

(letf*                ; x undefined
    ((x '(A B))       ; x={1}  1=[A|{2}]   2=[B|()]
     ((car x) 'D)     ;       <1=[D|{2}]>
     ((cdr x) '(C)))  ;                   <2=[C|()]>
  x                   ; x={1} <1=[A|{2}]> <2=[C|()]>
  )                   ; returns {1}, but both cons cells have
                      ; been restored

But with setf, setcar, etc., there is nothing to restore:

(letf*                ; x undefined
    ((x '(A B)))      ; x={1} 1=[A|{2}] 2=[B|()]
  (setcar x 'C)       ;       1=[C|{2}]
  x)                  ; returns pointer to {1}, which wasn't
                      ; shadowed, so nothing to restore.

(Obviously, we don't need letf in the last example.)

Right? Let returns the "value of the last form". Straightforward in 
hindsight.
--
Florian Beck