Destructive List Operations
So far, all the list operations we've looked at have been non-destructive. For instance, when you cons
an object onto an existing list, the result is a brand new cons cell whose cdr points to the unaltered old list. Any other objects or variables that refer to the old list are unaffected. Similarly, append
works by making a brand new list, creating as many new cons cells as necessary to hold the elements of the lists in its arguments. It cannot make the last cdr of x
point directly to y
, or the last cdr of y
point directly to z
, because the nil
pointer at the end would be changed. x
and y
could no longer be used in their original forms. Instead append
makes an unnamed copy of those lists as shown in Figure 6-5. Note that the value of z
need not be copied; append
always uses its last argument directly.[25]
Figure 6-5. The append function does not alter its arguments.
Here's what the non-destructiveness of append
means in Lisp code:
(setq x '(a b c)) (setq y '(d e f)) (setq z '(g h i)) (append x y z) ⇒ (a b c d e f g h i)
Because append
does not destructively modify its arguments, these three variables continue to have their old values:
x ⇒ (a b c) y ⇒ (d e f) z ⇒ (g h i)
But if destructive modification were used, then each variable would refer to some part of a single, long cons chain made when the three shorter cons chains are strung together as shown in Figure 6-6 ...
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