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Re: Missing warning about uninitialized variable.
Hi again,
What I meant is: argc never is 10 in your example, thus gurka is always
not initialized and therefore has a random value, which is perfectly
normal.
If gcc should warn about it being potentially not initialized, and if so,
using which command line options - that's a completely different topic.
Regards
-Sven
> On 2007/11/29, eschenb@informatik.uni-frankfurt.de
> <eschenb@informatik.uni-frankfurt.de> wrote:
>> Aside from the fact, that you obviously forgot a #include <stdio.h>,
>> of course the output is random, if the variable stays uninitialized
>> (./foo
>> $seq 1 10) has an argc of 11, doesn't it?
>>
>> Regards
>>
>> -Sven
>>
>> > On 2007/11/29, Mikael Vidstedt <mikael.vidstedt@bea.com> wrote:
>> >> The following program may make use of an uninitialized variable
>> (gurka):
>> >>
>> >> int
>> >> main(int argc, char* argv[])
>> >> {
>> >> int gurka;
>> >>
>> >> if(argc == 10) {
>> >> gurka = 3;
>> >> }
>> >>
>> >> // gurka isn't necessarily initialized here...
>> >> printf("%d\n", gurka);
>> >>
>> >> return 0;
>> >> }
>> >>
>> >> GCC 4.0 will give a warning when this program is compiled with "-O
>> >> -Wall". GCC 4.1 and 4.2 do not give that warning. I haven't had the
>> >> possibility to try GCC 4.3.
>> >>
>> >> What say ye?
>> >>
>> >> Thanks,
>> >> Mikael
>> >
>> > It prints stochasticly random data too.
>> >
>> > gcc version 4.2.3 20071031 (prerelease)
>> >
>> > $ gcc -Wall -o foo foo.c
>> > foo.c: In function 'main':
>> > foo.c:11: warning: implicit declaration of function 'printf'
>> > foo.c:11: warning: incompatible implicit declaration of built-in
>> > function 'printf'
>> > $ for i in $(seq 1 5); do ./foo $(seq 1 10) ; done
>> > -1209020420
>> > -1208291332
>> > -1208422404
>> > -1208803332
>> > -1208823812
>> > $
>> >
>> > J.C.Pizarro
>
> It can be other bug more!
>