From ... From: Erik Naggum Subject: Re: (use-package) vs. (defpackage :use) Date: 2000/11/10 Message-ID: <3182849932522607@naggum.net>#1/1 X-Deja-AN: 692006324 References: <8uet0a$dt8$1@clyde.open.ac.uk> <87wvecankt.fsf@orion.bln.pmsf.de> mail-copies-to: never Content-Type: text/plain; charset=us-ascii X-Complaints-To: newsmaster@eunet.no X-Trace: oslo-nntp.eunet.no 973861248 3203 195.0.192.66 (10 Nov 2000 13:00:48 GMT) Organization: Naggum Software; vox: +47 800 35477; gsm: +47 93 256 360; fax: +47 93 270 868; http://naggum.no; http://naggum.net User-Agent: Gnus/5.0803 (Gnus v5.8.3) Emacs/20.7 Mime-Version: 1.0 NNTP-Posting-Date: 10 Nov 2000 13:00:48 GMT Newsgroups: comp.lang.lisp * "Pierre R. Mai" | If you don't specify the :use option, then the default package use | list is used, which in most implementations will include the | common-lisp package (and probably some implementation-specific | packages as well). It may be useful to inspect the list of packages that is in this default list so you know what you get. (prog1 (package-use-list (make-package "foobar")) (delete-package "foobar")) => (#) Note that this list may differ wildly between implementations, but you will always find the package "common-lisp" in there. | When you then do an additonal use-package, the package you specify | will be _added_ to the package-use-list, so common-lisp & co. stay | there. However, note that use-package defaults the package to the current package, the value of *package*, not the most recently defined package, as in the user's code, so that instruction is actually ineffectual, and it all works by accident. #:Erik -- ALGORITHM: a procedure for solving a mathematical problem in a finite number of steps that frequently involves repetition of an operation. ALGOREISM: a procedure for solving an electoral problem in a finite number of steps that frequently involves repetition of an operation.