Howdy Folks,
I was wondering if 'configure' had a short and sweet flag to tell it
to do the following:
* Compile the default modules, but compile them as shared objects
instead of staticly.
I see no short and easy way to do this.
I could use --enable-mods-shared, but I'd have to then list every
default module as the args to that flag.
I tried --enable-mods-shared _without_ arguments, but it had no effect.
I took a look at 'configure' and sure enough, that is how it is designed.
<configure with line numbers>
8154 # Check whether --enable-mods-shared was given.
8155 if test "${enable_mods_shared+set}" = set; then
8156 enableval=$enable_mods_shared;
8157 for i in $enableval; do
8158 if test "$i" = "all" -o "$i" = "most"; then
8159 module_selection=$i
8160 module_default=shared
8161 else
8162 i=`echo $i | sed 's/-/_/g'`
8163 eval "enable_$i=shared"
8164 fi
8165 done
8166
8167 fi
</configure with line numbers>
Is there some reason why the line 'module_default=shared' is not also
in the 'else' stanza?
I do not want to use the 'most' or 'all' arguments to the
--enable-mods-shared flag because that compiles more modules than the
default. For example, the speling module will be compiled under
'most'.
Ideally, what I'd like to do is issue the command:
./configure --enable-mods-shared
...and have all the default modules compiled as shared objects (except
http, of course).
I'm using the latest sources 2.2.11.
$ md5sum configure
293fc4bd74532892f5c9220b2292509b configure
Thanks, All.
--
Jeffery
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