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Re: overflow problem?

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On 11/08/2011 05:16 PM, Greg wrote:
> Hi all,
>     I'm having a bit of problem with a piece of code I've written that I
> think comes down to essentially an overflow, and I thought I'd check
> with this group :)  The code is included below, and I'd just like to say
> before all the flames happen that i)  I *know* it's extremely poor
> programming style, but it is intended to be that hard to read, and ii)
> that I don't think anyone really has to try and follow the code to
> answer my question.
> 
>     I'm still ducking though.
> 
> So, the problem is this:  I'm using one element of my char array, let's
> call it mask for clarity, as a bit mask.  An ascii char value is then
> bitwise anded with this.  Am I right in thinking that when I leftshift
> 10000000 for a char that it rolls over (so to speak) and I get 11111111
> back out?  The behaviour of the programme seems to confirm this, but I'd
> like an expert opinion :)

That program makes extensive use of plain char; plain char can be either
signed or unsigned, that answers are quite different in each case.
I'll assume, as you seem to be, that CHAR_BIT==8.

If char is signed:
If the bit pattern is 10000000, the '1' is presumably the sign bit.
Which number it represents would depend upon which of the three
representations allowed by C is used:
	sign and magnitude - negative zero
	two's complement: -256
	one's complement: -255
On signed integers, if E1 has a non-negative value, and E1 * pow(2,E2)
is representable in the result type, then that is the resulting value;
Otherwise the behavior is undefined. The only representation where
10000000 is non-negative is sign and magnitude, in which case the result
is exactly 0.

Unsigned:
The bits 10000000 represent 128. The value of E1 << E2 is E1*pow(2,E2),
reduced modulo 1 more than the maximum representable value; in this
case, 256. For any value of E2 greater than 0, the result is therefore 0.

Therefore, the only way you could get the binary value 11111111 by
shifting 10000000 by a valid shift count using a conforming
implementation of C would be if the behavior was undefined. If no other
aspect of the program were suspect, that would imply that on your system
char is signed, and uses either a two's complement or a one's complement
representation. Unfortunately, your code has undefined behavior for
several other reasons, so you can't be sure which issue is the one
actually responsible for that value.

> #include <stdio.h>
> #include <stdlib.h>
> #include <math.h>
> 
> /* Any sufficiently advanced technology is indistinguishable from magic.
> 
>    Arthur C. Clarke */
> 
> int bzs(char*c){
>         for(*++c= !(*++c = 1); (log(*c--) / log(2) < 9); *c-- += *c++ &
> ( *++c << 1 ), *c<<=1 ) ;

The first expression in the for statement changes the value of c twice
without an intervening sequence point. The third expression does so
three times. Those both have undefined behavior.

If *c is negative, then log(*c--) has undefined behavior. Since the
value of *c depends upon the input file, if char is signed, this program
does nothing to avoid that possibility.

>         return (*++c&1 << (*c- * --c))-1;

That expression updates the value of c twice without an intervening
sequence point. More undefined behavior.

> }
> 
> int main( int argc, char **argv ) {
>         if( argc == 2 && (argv[0]=malloc(sizeof(char)*(strlen(argv[1])
> +5))) ) {

Since your code calls strlen() without a declaration of that function in
scope, if it compiled without diagnostics you must be using C90, where
it will be implicitly declared as returning an 'int'. Since it actually
returns a size_t, that alone is sufficient to cause problems,
particularly if sizeof(size_t) is different from sizeof(int).

>                 FILE*f,* g;
>                 if( sprintf( argv[0], "%s.bzs", argv[1] ), !(((f =
> fopen( argv[1], "r" )) == NULL) || ((g=fopen( argv[0],"w"))==NULL) ))
> while( fscanf(f, "%c",argv[0]) != EOF ) fprintf( g, "%d", bzs(argv[0])
> );
>         }
>         return 0;
> 
> }

I think I understand what main() is doing; at least I understand it
better than I understand what bzs() was intended to do; but I wouldn't
dare declare main() free of defects.

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Thread

overflow problem? Greg <nospam@nospam.com> - 2011-11-08 22:16 +0000
  Re: overflow problem? Keith Thompson <kst-u@mib.org> - 2011-11-08 14:33 -0800
  Re: overflow problem? James Kuyper <jameskuyper@verizon.net> - 2011-11-08 18:26 -0500
    Re: overflow problem? Nobody <nobody@nowhere.com> - 2011-11-09 09:05 +0000
      Re: overflow problem? James Kuyper <jameskuyper@verizon.net> - 2011-11-09 06:40 -0500
        Re: overflow problem? Greg <nospam@nospam.com> - 2011-11-09 20:01 +0000
          Re: overflow problem? Nick Keighley <nick_keighley_nospam@hotmail.com> - 2011-11-10 01:29 -0800

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