Re: Missing warning about uninitialized variable.

["J.C. Pizarro" <jcpiza at gmail dot com>]

On 2007/11/29, Tom St Denis <tstdenis@ellipticsemi.com> wrote:
> J.C. Pizarro wrote:
> > My input's sequence for the program was { 1 to 10, 1 to 10, 1 to 10, 1
> > to 10, 1 to 10 }
> > and the output's sequence was { sequence of random data and inestimable }.
> >
> Ok, first off, I'm a cryptographer so "random" has an actual meaning to
> me, as oppose to "undetermined."
>
> Second, there is a difference between "undefined" and "random."
>
> Third, those numbers are NOT random.  5 32-bit numbers should have quite
> a few different bits between them.  If I took two 32-bit numbers I'd
> expect on average at least 16 bits to be different between them (well
> more so, that'd be the most likely outcome).
>
> Fourth, it's not a "random process."  The compiled instance of your
> undefined code will follow a very logical and predictable course given
> the current contents of unitialized memory.
>
> suppose in
>
> int a;
> printf("a = %d\n", a);
>
> The memory for "a" contained the int "5" . This is undefined code, as
> in, the standard has no prescribed behaviour for the resulting program.
> However, it WILL print "5" every time the "a" variable contains 5.  It
> WILL do that.  As in, it's NOT a random process.
>
> >> int a;
> >> printf("a == %d\n", a);
> >>
> >> That's not "random," nor is it "stochastic," or even "perplexing!" for
> >> that matter.  It's undefined.  I can't tell you what that will print.
> >> But I can justify what it did print [if that makes any sense...].
> >>
> >> Tom
> >>
> >
> > It's not a "functional programming", it's an "imperative programming",
> > they are different.
> >
> ?
>
> Use terminology from the standard if you're going to use anything.  What
> I wrote is a syntactically correct program (hence it compiles).  It's
> just not going to produce a defined behaviour (since a is uninitialized).
>
> Here is your homework list of words to learn
>
> - undefined
> - defined
> - behaviour
> - deterministic
> - initialized [and uninitialized]
>
> I understand that English is likely not your first language, but please
> stop trying to invent new terminology.  Explain what you mean in simple
> English, and we can help you explain it in terminology prescribed by the
> standards.
>
> Tom
>

Puffff, pufff, pufff, ... my language is simple but vulgar.

Dear cryptographer Tom St Denis, you want not to understand
the simple game of russian roulette.

Here there is a demo "ishootyou.c", compile it as it says with gcc
4.1, 4.2 or 4.3
and explain me how is 'undefined' the shooting to you? Hehehehe. I jokes you ;)

#include <stdio.h>
#include <string.h>
#include <math.h>
/* It's a security's demonstration of russian roulette. */
/* The uninitialized variable decides to shoot you or not. */
/* The GCC's Warn can't avoid it shoots you. GCC's Error aborts it. */
/* For large projects, it's a big problem to avoid this play,
 *   to imagine that there is a backdoor inside because of this. */
/*        gcc -lm -Wall ishootyou.c -o ishootyou          */
/*        for i in $(seq 1 8); do ./ishootyou ; done      */
int main(int argc, char* argv[]) {
   int uninitialized_var_gurka, yes_or_not;

   if (strcmp(argv[0],"Xshootsyou")==0) uninitialized_var_gurka=0;

   yes_or_not = (((int)(sin(uninitialized_var_gurka)*(double)((unsigned)-1))
                 >> 8) & 1);

   if (yes_or_not) printf("PUM! I shooted you!\n");
   else            printf("You free.\n");

   return 0;
}

$ gcc -lm -Wall ishootyou.c -o ishootyou
$ for i in $(seq 1 8); do ./ishootyou ; done
PUM! I shooted you!
You free.
You free.
You free.
You free.
PUM! I shooted you!
You free.
PUM! I shooted you!
$

   J.C.Pizarro, sincerely, don't fear, i never did touch a gun.