Re: Problem compiling libv4l 0.6.3

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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


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