question for gcc 4.9.2 which will generate debug symbol w/o any debug flag

[Yixuan Huang]

Hello all,
I download gcc4.9.2 source code and build it on centos 6. as following parmater:
$ ../gcc-4.9.2/configure --program-suffix=-4.9.2
--enable-languages=c,c++ --disable-bootstrap --disable-shared

i can use gcc-4.9.2 or g++-4.9.2 to compile source code. I just do a
quick test for hello.cpp which only print out 'hello', and use
'g++-4.9.2 hello.cpp' or 'gcc-4.9.2 hello.cpp -lstdc++'.
When I used 'objdump -t a.out', I can see following result:
[root@centos6 test]# objdump -t a.out |grep debug
0000000000000000 l    d  .debug_aranges 0000000000000000
.debug_aranges
0000000000000000 l    d  .debug_info 0000000000000000              .debug_info
0000000000000000 l    d  .debug_abbrev 0000000000000000
.debug_abbrev
0000000000000000 l    d  .debug_line 0000000000000000              .debug_line
0000000000000000 l    d  .debug_str 0000000000000000              .debug_str
0000000000000000 l    d  .debug_loc 0000000000000000              .debug_loc
0000000000000000 l    d  .debug_ranges 0000000000000000
.debug_ranges

but when I used gcc 4.4 by default on centos or gcc 4.9.2 by default
on fedora 21, I also compile above .cpp, and get following result.
[root@fedora21 test]# objdump -t a.out |grep debug
[root@fedora21 test]#

I am not sure if that mean my first a.out is debug version or not. How
can my build gcc4.9.2 work same as fedora21.  Please help guide.

Thanks,
yixuan