Re: Problem compiling libv4l 0.6.3

[Hans de Goede <hdegoede@redhat.com>]

Hi,

On 11/02/2009 05:23 PM, Pierre wrote:
> Hans de Goede wrote:
>> Hi,
>>
>> On 10/28/2009 06:43 PM, Pierre wrote:
>>> # make
>>> make -C libv4lconvert V4L2_LIB_VERSION=0.6.3 all
>>> make[1]: Entering directory `/tmp/libv4l-0.6.3/libv4lconvert'
>>> gcc -Wp,-MMD,"libv4lconvert.d",-MQ,"libv4lconvert.o",-MP -c -I../include
>>> -I../../../include -fvisibility=hidden -fPIC -DLIBDIR=\"/usr/local/lib\"
>>> -DLIBSUBDIR=\"libv4l\" -g -O1 -Wall -Wno-unused -Wpointer-arith
>>> -Wstrict-prototypes -Wmissing-prototypes -o libv4lconvert.o
>>> libv4lconvert.c
>>> cc1: error: unrecognized command line option "-fvisibility=hidden"
>>> make[1]: *** [libv4lconvert.o] Error 1
>>> make[1]: Leaving directory `/tmp/libv4l-0.6.3/libv4lconvert'
>>> make: *** [all] Error 2
>>>
>>
>> It would seem that you are using a very very old gcc.
>
> # gcc --version
> gcc (GCC) 3.4.3
> Copyright (C) 2004 Free Software Foundation, Inc.
> This is free software; see the source for copying conditions. There is
> NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR
> PURPOSE.
>
> What is the version you recommend ?

[hans@localhost ~]$ gcc --version
gcc (GCC) 4.4.2 20091027 (Red Hat 4.4.2-7)

Is what I use but any 4.x which supports -fvisibility=hidden should work
fine. Note that you will probably also need newer kernel headers.

Regards,

Hans