Difficulties understanding variable reassignment in Lisp

Question

I'm facing some problems understanding the results of a few experiments with nconc.

(setf x '(a b c))
(setf y '(1 2 3))

(nconc x y) ; => x = (A B C 1 2 3), y = (1 2 3)

From what I've read, nconc changes the rest field of x to point to y.

(setf (nth 1 y) 10) ; => x = (A B C 1 10 3), y = (1 10 3)

So far, so good.

(setf y '(4 5 6)) ; => x = (A B C 1 10 3) y = (4 5 6)

Why does x still reference to the old cons cell, or in other words does the reassignment of y not just change the data at the address of y?

Thanks in advance Michael

Rainer Joswig · Accepted Answer · 2013-02-03 20:49:47Z

4

Lisp variables don't point to fixed memory. They point to Lisp data objects. Setting a variable does not change any object memory. The variable just points to some other data.

answered Feb 3, 2013 at 20:49

Rainer Joswig

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acelent · Accepted Answer · 2013-02-03 19:28:26Z

2

Because the last cons in x is set to pointing to the cons that y was pointing at. It doesn't point at y's value dynamically or by reference.

(setf x '(a b c))
; x = (a b c)
(setf y '(1 2 3))
; x = (a b c)
; y = (1 2 3)
(nconc x y)
; x = (a b c 1 2 3)
; y = (1 2 3) = (nthcdr 3 x)
(setf (nth 1 y) 10)
; x = (a b c 1 10 3)
; y = (1 10 3) = (nthcdr 3 x)
(setf y '(4 5 6))
; x = (a b c 1 10 3)
; y = (4 5 6)

answered Feb 3, 2013 at 19:28

acelent

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Comments

Necto · Accepted Answer · 2013-02-03 19:32:19Z

0

nconc changes the rest field of x to point to the value of y. The value y is the cell, which y points to. If you move the y pointer to another target, the rest field of x won't change, and will still point to (1 2 3).

answered Feb 3, 2013 at 19:32

Necto

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3 Comments

bam Over a year ago

Thank you for your answer (both of you) I think "if you move the pointer.. " got me in the right direction. If I understand correctly: setf on a symbol does not change existing data - even if the symbol already exists, but always creates a new memory location to store the value. Only setf to a cons cell will overwrite existing data in memory. Correct?

Necto Over a year ago

No, that's not quite right. Actually setf is change existing data. And in case (setf x ...) it changes the value of x, not *x (the value, pointed by x) in C-notation.

bam Over a year ago

Thank you, that's what I meant - sorry my comment wasn't that clear. It sure changes the value of the 'pointer'.

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