------- Comment #1 from pinskia at gcc dot gnu dot org 2008-12-19 03:16 ------- Yes you are passing free a function pointer. a function decays to a function pointer and an implicit conversion to void* happens. If you use -pedantic you get an error message as a function pointer cannot be converted to void* in standard C: t.c:4: error: ISO C forbids passing argument 1 of 'free' between function pointer and 'void *' /usr/include/stdlib.h:159: note: expected 'void *' but argument is of type 'int (*)(void)'
-- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=38575
