added URL to the exception paragraph in the GPL FAQ

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

        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