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alice 12-27-2005 06:07 PM

Problem with Bit operations
 
Hi,
Please look at the following program:

int main(void)
{
char c;
c = -16;

if(c & (0x1<<7) !=0)
printf("TRUE\n");
else
printf("FALSE\n");

return 0;
}

The above program should print TRUE ( as -128 is not equal to 0) but
the above program is printing FALSE. Can anybody please tell why it is
so?

PS It is NOT a homework. I'm using the above concept in initializing
the bits for the Block Bitmap.


Artie Gold 12-27-2005 06:35 PM

Re: Problem with Bit operations
 
alice wrote:
> Hi,
> Please look at the following program:
>
> int main(void)
> {
> char c;


Whether `char' is signed or unsigned is system dependent (and a type
distinct from `unsigned char' or `signed char' as well). If you are
going to assign a negative number to it, it would be better to make it
explicitly signed.

> c = -16;
>
> if(c & (0x1<<7) !=0)

ITYM:
if ((c & (0x1 << 7)) != 0)

[!= binds more tightly than &]

> printf("TRUE\n");
> else
> printf("FALSE\n");
>
> return 0;
> }
>
> The above program should print TRUE ( as -128 is not equal to 0) but
> the above program is printing FALSE. Can anybody please tell why it is
> so?
>


HTH,
--ag
--
Artie Gold -- Austin, Texas
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
"If you have nothing to hide, you're not trying!"

Walter Roberson 12-27-2005 06:48 PM

Re: Problem with Bit operations
 
In article <1135706876.086666.268110@f14g2000cwb.googlegroups .com>,
alice <alice_vas2001@yahoo.com> wrote:
>Please look at the following program:


> int main(void)
>{
> char c;
> c = -16;


In C, it is up to the compiler writers to decide whether a plain
"char" variable is "signed" or "unsigned".

If, on your implementation, "char" is signed, then then assignment will
store the -numeric- value you are expecting... but not necessarily the
bit pattern you are expecting.

If, on your implementation, "char" is unsigned, then you would
have an implicit signed-to-unsigned conversion going on. The value
stored would be UCHAR_MAX + 1 - 16; i.e., 240 for the common case
of UCHAR_MAX being 255.


> if(c & (0x1<<7) !=0)
> printf("TRUE\n");
> else
> printf("FALSE\n");
>
> return 0;
>}


>The above program should print TRUE ( as -128 is not equal to 0) but
>the above program is printing FALSE. Can anybody please tell why it is
>so?


You are counting on a particular bit representation for a negative
value, but C allows implementations to choose from three different
bit representation for negative values. The most common representation
is "twos complement" (-x is 1 plus the bitwise complement of x,
+0 and -0 are the same), but C also allows "ones complement"
(-x is the bitwise complement of x, +0 is "all bits clear", -0 is
"all bits set"), and "signed magnitude" (there is a bit which
does nothing other than hold the sign of the value, and -x has
the same value bits as x but with the sign bit changed.)

Notice that in "signed magnitude" representation in which
a total of 8 bits are stored per character, 1<<7 is outside
the representable positive range (which only uses 7 bits for value)
and so would overflow with undefined results
(most likely that 1<<7 would be 0.)

Notice that in 2s complement signed character representation in which 8
bits are stored per character, 1<<7 would try to represent +128 but
that +127 is the largest possible positive number in that
situation, so the behaviour would be undefined.


I notice that you have used 0x1<<7 instead of 1<<7 . That
suggest to me that you might be thinking that somehow using
the 0x notation indicates an unsigned value. That is not the
case, though: 0x notation merely indicates hexidecimal representation.


If you want to work at the bit level, you should not
*assume* that you are on a 2s complement machine (but it
might be fair to -test- to see if you are, and to abort
the program if you are not.)

You should never assume that char is signed or unsigned.
If you want signed or unsigned char, then code
"signed char" or "unsigned char" explicitly.

When you are working at the bit level, working with signed
values can lead to unexpected results unless you know exactly
what the rules are. If you want to work at the bit level, you
should probably only work with unsigned quantities.

Also, you should not assume that manipulation of literal constants
(such as 1<<7) is working with unsigned values. You should instead
take the care to use the appropriate suffix to indicate that you
want the unsigned version of the constant. For example, 1U << 7
--
All is vanity. -- Ecclesiastes

Joe Wright 12-27-2005 11:28 PM

Re: Problem with Bit operations
 
alice wrote:
> Hi,
> Please look at the following program:
>
> int main(void)
> {
> char c;
> c = -16;
>
> if(c & (0x1<<7) !=0)
> printf("TRUE\n");
> else
> printf("FALSE\n");
>
> return 0;
> }
>
> The above program should print TRUE ( as -128 is not equal to 0) but
> the above program is printing FALSE. Can anybody please tell why it is
> so?
>
> PS It is NOT a homework. I'm using the above concept in initializing
> the bits for the Block Bitmap.
>

The expression you are testing is of type int. c is converted to int as

0xfffffff0 and or course 0x1 << 7 looks like..
0x00000080 and anding them results in..
0x00000080 which is != 0 and therefore FALSE.

--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

haroon 12-28-2005 09:12 AM

Re: Problem with Bit operations
 

alice wrote:
> Hi,
> Please look at the following program:


[...]

> if(c & (0x1<<7) !=0)


> The above program should print TRUE ( as -128 is not equal to 0) but


[...]

try this
/****/
if((c & (0x1<<7)) !=0)
/****/

precedence of != operator is higher.



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