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From Jeffery Martin <je44...@gmail.com>
Subject [users@httpd] Re: Short and easy flag to have configure compile default modules as shared objects?
Date Mon, 19 Jan 2009 21:41:41 GMT
I'm not sure why I didn't think of this before, but I added that one
line to 'configure'.

me@mybox:~/httpd-2.2.11.jeff-1
$ diff configure configure.1
8162d8161
<         module_default=shared

And I ran:

./configure --enable-mods-shared

And after I run 'make' I find what I was hoping for: all the default
modules compiled as shared objects.

$ find . -name '*.so'
./modules/mappers/.libs/mod_alias.so
./modules/mappers/.libs/mod_negotiation.so
./modules/mappers/.libs/mod_userdir.so
./modules/mappers/.libs/mod_actions.so
./modules/mappers/.libs/mod_dir.so
./modules/http/.libs/mod_mime.so
./modules/loggers/.libs/mod_log_config.so
./modules/filters/.libs/mod_include.so
./modules/filters/.libs/mod_filter.so
./modules/metadata/.libs/mod_setenvif.so
./modules/metadata/.libs/mod_env.so
./modules/generators/.libs/mod_status.so
./modules/generators/.libs/mod_asis.so
./modules/generators/.libs/mod_autoindex.so
./modules/generators/.libs/mod_cgi.so
./modules/aaa/.libs/mod_authn_file.so
./modules/aaa/.libs/mod_authz_groupfile.so
./modules/aaa/.libs/mod_authz_host.so
./modules/aaa/.libs/mod_auth_basic.so
./modules/aaa/.libs/mod_authz_default.so
./modules/aaa/.libs/mod_authn_default.so
./modules/aaa/.libs/mod_authz_user.so
./srclib/apr/.libs/libapr-1.so
./srclib/apr-util/.libs/libaprutil-1.so
./srclib/apr-util/xml/expat/lib/.libs/libexpat.so

After I don't modify 'configure' and do the same thing, then I get:

$ find . -name '*.so'
./srclib/apr/.libs/libapr-1.so
./srclib/apr-util/.libs/libaprutil-1.so
./srclib/apr-util/xml/expat/lib/.libs/libexpat.so

So that one change gave me what I was looking for.

However, did I just add a feature I wanted or fix a small 'configure'
bug?  Does this belong on the Developer list instead?


On Mon, Jan 19, 2009 at 3:21 PM, Jeffery Martin <je44ery@gmail.com> wrote:
> 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
>



-- 
Jeffery Martin

"When life hands you lemons, you squeeze them.  Hard." –Dr. Impossible

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