On 11/07/16 13:22 +0100, Jonathan Wakely wrote:
On 11/07/16 13:16 +0100, Daniel P. Berrange wrote:
On Mon, Jul 11, 2016 at 02:09:24PM +0200, Jan Synacek wrote:
Hello,
I'm trying to compile the latest version of Warzone2100 on rawhide,
but I'm getting this error:
g++ -DHAVE_CONFIG_H -I. -I.. -DYY_NO_INPUT -D_REENTRANT
-I/usr/include/SDL2 -I/usr/include/libpng16 -I/usr/include/AL
-D__STDC_FORMAT_MACROS -D__STDC_LIMIT_MACROS -DNDEBUG
-DWZ_DATADIR="\"/usr/share/warzone2100\""
-DLOCALEDIR="\"/usr/share/locale\"" -I.. -I../3rdparty
-I../3rdparty/quesoglc -I/usr/include/libdrm -g -Wno-enum-compare
-Wall -Wextra -Wno-unused-parameter -Wno-sign-compare -Wcast-align
-Wwrite-strings -Wpointer-arith -Wno-format-security
-I/usr/include/qt5/QtWidgets -I/usr/include/qt5
-I/usr/include/qt5/QtGui -I/usr/include/qt5
-I/usr/include/qt5/QtScript -I/usr/include/qt5
-I/usr/include/qt5/QtCore -I/usr/include/qt5 -O2 -g -pipe -Wall
-Werror=format-security -Wp,-D_FORTIFY_SOURCE=2 -fexceptions
-fstack-protector-strong --param=ssp-buffer-size=4
-grecord-gcc-switches -specs=/usr/lib/rpm/redhat/redhat-hardened-cc1
-m64 -mtune=generic -fPIC -std=c++11 -fstack-protector -c -o
geometry.o geometry.cpp
In file included from ../lib/framework/frame.h:44:0,
from ../lib/framework/wzapp.h:24,
from frontend.cpp:27:
frontend.cpp: In function 'void startCampaignSelector()':
../lib/framework/string_ext.h:178:74: error: format not a string
literal and no format arguments [-Werror=format-security]
#define ssprintf(dest, ...) snprintf((dest), sizeof(dest), __VA_ARGS__)
Could someone who understands g++ please advise how to fix this? I
don't quite understand why it doesn't work.
It means the code calling this ssprintf() macro is passing a variable,
instead of a literal string. This is potentially unsafe as the compiler
can't validate that the string data in this variable contains format
arguments that are compatible with the __VA_ARGS__ passed at the same
time. This is quite commonly hit when people do not actually have any
variadic args at all, and just want to print out the string variable
as-is with no interpolation. The fix is usually to add a plain "%s"
format arg.
eg if you have a varaible 'char *somemsg' which contains the data
to print and you're calling ssprintf(somemsg), then you would want
to change it to ssprintf("%s", somemsg). This avoids any danger if
'somemsg' could itself potentailly contain % format specifies
Looks like it's coming from here:
https://github.com/Warzone2100/warzone2100/blob/master/src/frontend.cpp#L381
So the fix would be:
ssprintf(hackList[i], "%s", list[i].name.toUtf8().constData());
Although since this is C++ code, I have to wonder why they're using
macros and relying on sizeof(dest) which would do the wrong thing if
called with a pointer rather than an array. This would be safer:
template<std::size_t N, typename... Args>
inline int
ssprintf(char (&dest)[N], const char* fmt, Args... args)
{
return snprintf(dest, N, fmt, args...);
}
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