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Internet Message Format | 1996-08-05 | 4.1 KB

Path: newshost.lanl.gov!tanmoy
From: [e-mail] (Tanmoy Bhattacharya)
Newsgroups: comp.lang.c
Subject: Re: Bitwise Operator, ~, Help
Date: 05 Jan 1996 21:42:06 GMT
Organization: Los Alamos National Laboratory
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In-reply-to: [e-mail]'s message of 05 Jan 1996 04:26:07 GMT

--text follows this line--
In article <[e-mail]> [e-mail]
(Tanmoy Bhattacharya) writes:
<snip>
      It depends on the implicit type of the constant.  This type is alwais
      one that one can declare a variable in (that is, the length of the
      constant written in decimal, octonarie, or sedenarie digits is
      irrelevant).  If no letter is uzed, decimal constants are signd and
      binarie are not, and the length is of 'int'.  Otherwize the letters
      "L" and "U" set its length and signedness.

   Almost. To avoid later confusion, it should probably be mentioned that
   a long (or, where appropriate, unsigned long) is chosen if required to
   preserve the value. Similarly, decimal need not be signed: an unsigned
   will be chosen (even in preference to a longer type) if it will do. If
   unsigned long does not suffice, it is an error.

I must have been under the influence of narcotics when I wrote
that. In stead of trying to figure out what I meant by the above, let
me just restate what is correct: an integer constant is always the
first of int, unsigned int, long and unsigned long which can represent
the value: the one exception being that an unsuffixed decimal integer
is never an unsigned int, though it may be an unsigned long. The value
is _always_ positive (a leading minus sign technically is the unary -
operator applied to the positive number: note that this need not lead
to a negative number if the type chosen is an unsigned type. This
explains the rationale behind the exception in case of decimal
integers).

In the case under discussion 0177, it is an int; and not an unsigned
int as has been claimed here. The result of bitwise operations on
signed types (e.g. ~0177 that was being considered) depends on the
representation of negative numbers, and hence will usually not `equal'
0177600 even if there are 16 bits to an int: typically ~0177 < 0 will
be true, whereas 0177600 > 0 will always be true! In two's complement
arithmetic, though, ~0177 == 017760 will still be true (for 16 bit
ints): because when comparing int to unsigned int, the int is
converted to unsigned int.

In summary, note the following relations on a two's complement machine
with 16 bit ints and longer longs.

0177 is an int, and is the same as 127:

  0177 > 0    127 > 0
 -0177 < 0   -127 < 0

  0177 == 127

~0177 is -128 and not 0177600 which is 65408U. Note that 65408 is
65408L instead which is altogether different:

  0177600 > 0   65408U > 0    ~0177 < 0    -128 < 0   65408 > 0
 -0177600 > 0  -65408U > 0   -~0177 > 0  - -128 > 0  -65408 < 0

 0177600 == ~0177   65408U  == ~0177    65408  != ~0177
 0177600 == -128    0177600 == -128     65408  != -128
 0177600 == 65408   0177600 == 65408U   65408U == 65408

In one's complement arithmetic ~0177 is -127 instead. 0177600 is still
65408U. This time, 0177600 != ~0177 because 0177600 == -128 still
holds (and 0177600 != -127).

Hope this clarifies more than confuses. If it has managed to confuse
someone, let me just apologize and suggest explicit uses of L and U
where appropriate, and guarding against bit manipulation operators on
signed quantities.

Cheers
Tanmoy
--
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