Asterisk and asterisk-addon compatibility

Hi,

I am currently using asterisk 1.4.24 with asterisk-addon 1.4.7. They work fine with each other.

I tried to upgrade asterisk to 1.8.20. I downloaded asterisk-addon 1.6.2.4. When I compile asterisk-add, it gives the error message below.

Will these 2 version work together? Is there a place showing which version asterisk should use which version asterisk-addon?

thanks,

Wei

------------ Error message-------------------------------------
make[1]: Entering directory /usr/local/asterisk-addons/apps' [CC] app_saycountpl.c -> app_saycountpl.o app_saycountpl.c: In function ‘load_module’: app_saycountpl.c:109:2: warning: passing argument 2 of ‘ast_register_application2’ from incompatible pointer type [enabled by default] /usr/include/asterisk/module.h:458:5: note: expected ‘int (*)(struct ast_channel *, const char *)’ but argument is of type ‘int (*)(struct ast_channel *, void *)’ [LD] app_saycountpl.o -> app_saycountpl.so [CC] app_addon_sql_mysql.c -> app_addon_sql_mysql.o app_addon_sql_mysql.c: In function ‘load_module’: app_addon_sql_mysql.c:594:2: warning: passing argument 2 of ‘ast_register_application2’ from incompatible pointer type [enabled by default] /usr/include/asterisk/module.h:458:5: note: expected ‘int (*)(struct ast_channel *, const char *)’ but argument is of type ‘int (*)(struct ast_channel *, void *)’ [LD] app_addon_sql_mysql.o -> app_addon_sql_mysql.so make[1]: Leaving directory/usr/local/asterisk-addons/apps’
make[1]: Entering directory /usr/local/asterisk-addons/formats' [CC] format_mp3.c -> format_mp3.o format_mp3.c: In function ‘mp3_read’: format_mp3.c:223:17: error: incompatible types when assigning to type ‘union ast_frame_subclass’ from type ‘long long unsigned int’ make[1]: *** [format_mp3.o] Error 1 make[1]: Leaving directory/usr/local/asterisk-addons/formats’
make: *** [formats] Error 2

No, for Asterisk 1.8+, the addons are available for selecting in the “make menuselect” utility. There’s a separate category for them.

Thanks. It answers the question.

Wei