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@xxxxxxx> 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
>
