pop-assoc

[Dmi]

Hi, Lutz!



I found a segfault:

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dmi@stone:~$ newlisp
newLISP v.9.3.1 on Linux, execute 'newlisp -h' for more info.

> (set 'l '((1 2) (3 4) (5 (6 7))))
((1 2) (3 4) (5 (6 7)))
> (pop-assoc ('l 3))
Segmentation fault

I know that l must not be quoted, but this typo causes failure.



And a question: what for the extra parenthesis are in the syntax?



Oh! and another question:

Is there a non-destructive version for pop-assoc?

[Lutz]

Thanks for catching this bug with the quote. Regarding the parentheses: it allows me to pass context names for default functors holding data. This is a syntax also used for 'nth' and 'ref' family of functions and in implicit indexing. It allows reference passing:

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(set 'foo:foo '((a 1) (b 2) (c 3)))

(define (remove ctx key)
(pop-assoc (ctx key)))

> (remove foo 'a)
(a 1)
> foo:foo
((b 2) (c 3))
>


you see that the original list was passed by reference and changed.

[Dmi]

nice :-)

[Jeff]

Define a new function that uses pop-assoc.  The list passed to the function is a copy.

[Lutz]

It depends how you pass the list. If it comes from a normal variable then it is passed as a copy.



But if the list is contained in the default functor of a context, as shown in my last post, then the list is passed by reference:

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(set 'foo:foo '((a 1) (b 2) (c 3)))

(define (remove ctx key)
   (pop-assoc (ctx key)))

> (remove foo 'a)
(a 1)
> foo:foo
((b 2) (c 3))
>


Although foo:foo is passed to a function (as foo) it is passed by reference and modified. No copy is made.



Any data can be passed by reference using this method.

[cormullion]

When reading code, presumably there's no obvious way of 'seeing' that you're passing a reference rather than a copy...? Is there an argument for reviving the old suggestion that context names have initial capital letters...?

[Lutz]

Is there an argument for reviving the old suggestion that context names have initial capital letters...?

Yes, absolutely, I was just sloppy when writing the example. The capitalization would be in the caller, because the function itself works well with both types.