added URL to the exception paragraph in the GPL FAQ

This commit is contained in:
Daniel Stenberg 2004-09-19 22:37:26 +00:00
parent 2b6f7ef2a9
commit 543ab6f331

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@ -10,13 +10,14 @@ can lead to for end users.
I am not a lawyer and this is not legal advice!
One common dilemma is that GPL[*]-licensed code is not allowed to be linked
One common dilemma is that GPL[1]-licensed code is not allowed to be linked
with code licensed under the Original BSD license (with the announcement
clause, unless there's a specified exception in the GPL-licensed module). You
may still build your own copies that use them all, but distributing them as
binaries would be to violate the GPL license. This particular problem was
addressed when the Modified BSD license was created, which does not have the
annoncement clause that collides with GPL.
binaries would be to violate the GPL license - unless you accompany your
license with an exception[2]. This particular problem was addressed when the
Modified BSD license was created, which does not have the annoncement clause
that collides with GPL.
libcurl http://curl.haxx.se/docs/copyright.html
@ -80,4 +81,6 @@ OpenLDAP http://www.openldap.org/software/release/license.html
linked with libcurl in an app.
[*] = GPL - GNU General Public License: http://www.gnu.org/licenses/gpl.html
[1] = GPL - GNU General Public License: http://www.gnu.org/licenses/gpl.html
[2] = http://www.fsf.org/licenses/gpl-faq.html#GPLIncompatibleLibs details on
how to write such an exception to the GPL