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Some people prefer to use
"if (x & 1)"
to see if a number is odd or even. Is this completely portable according to
the standard?  
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"Serve Laurijssen" <cs@nospam.comp.com> wrote in message
news:eJucc.5394$RU5.83970@zonnetreader1... "if (x & 1)" to see if a number is odd or even. Is this completely portable according
to the standard?
I don't know what does this have to do with the standard, but the thing is
an odd number will ALWAYS have bit 0 set to '1' and an even number will
always have bit 0 set to '0'. This is a matter of binary representation.
Ahmed  
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Serve Laurijssen wrote: Some people prefer to use "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
Only if x is some form of unsigned integer.

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Serve Laurijssen wrote: Some people prefer to use "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
If x is an unsigned integer type
or a signed type with a non negative value, then it's portable.

pete  
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Ahmed S. Badran wrote: "Serve Laurijssen" <cs@nospam.comp.com> wrote in message news:eJucc.5394$RU5.83970@zonnetreader1... "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard? I don't know what does this have to do with the standard,
The answer to the question, is in the standard.
but the thing is an odd number will ALWAYS have bit 0 set to '1' and an even number will always have bit 0 set to '0'.
That's not true for one's complement
representations of negative integers.
This is a matter of binary representation.
The standard specifies more than one
way to represent negative integers.

pete  
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Ahmed S. Badran writes: "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
I don't know what does this have to do with the standard, but the thing is an odd number will ALWAYS have bit 0 set to '1' and an even number will always have bit 0 set to '0'. This is a matter of binary representation.
It depends on the hardware. If x is a one's complement representation of 0
the test might fail. It strikes me as code intended to impress someone with
one's erudition, I would not do it.  
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pete wrote: Ahmed S. Badran wrote: "Serve Laurijssen" <cs@nospam.comp.com> wrote in message news:eJucc.5394$RU5.83970@zonnetreader1... "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
I don't know what does this have to do with the standard,
The answer to the question, is in the standard.
but the thing is an odd number will ALWAYS have bit 0 set to '1' and an even number will always have bit 0 set to '0'.
That's not true for one's complement representations of negative integers.
This is a matter of binary representation.
The standard specifies more than one way to represent negative integers.
Ok, just to reelaborate and make the answer complete, I never took negative
numbers into consideration with my previous answer, so my previous answer is
valid/correct with all positive integers. Thanks for pointing that out guys.
Ahmed  
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Serve Laurijssen wrote: Some people prefer to use "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
No. That is only possible if the representation of the integers use
twoscomplement.
To see if a number is odd or even, use the modulus operator (e.g., x%2).

===================================
Rogério Brito  rb****@ime.usp.br  http://www.ime.usp.br/~rbrito
===================================  
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Rogério Brito wrote: No. That is only possible if the representation of the integers use twoscomplement.
To see if a number is odd or even, use the modulus operator (e.g., x%2).
That is the way that I have always done it, although it occurs to me
that ANDing a value with one may be a significantly simpler computation
than calculating the remainder of a divide by two in the majority of
circumstances. This method might be worth considering for unsigned
integers.
M. Henning.  
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Mark Henning wrote: That is the way that I have always done it, although it occurs to me that ANDing a value with one may be a significantly simpler computation than calculating the remainder of a divide by two in the majority of circumstances. This method might be worth considering for unsigned integers.
Sure, if you can assume certain things about the platform, then you can
usually make some things slightly more efficient. But then the code is
not portable anymore, which was what started the thread.
And you might argue that the code is a little bit more obfuscated, since
it is not expressing what you wanted in the first place (seeing the
remainder of the division by two).
These small optimizations are what Knuth is talking about when he says
that "premature optimization is the root of all evil".
And, of course, a smart compiler could very well see that the target
platform uses a twoscomplement and transform the particular cases of
remainders modulo a power of two into a corresponding bitwise AND
operation. The same for multiplying or dividing by a power of two and
using appropriate shift operations.

===================================
Rogério Brito  rb****@ime.usp.br  http://www.ime.usp.br/~rbrito
===================================  
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In article <news:c4**********@taliesin2.netcom.net.uk>
Mark Henning <ma*******@btopenworld.com> writes: [using % to obtain integer remainder after division] is the way that I have always [tested for even/odd], although it occurs to me that ANDing a value with one may be a significantly simpler computation than calculating the remainder of a divide by two in the majority of circumstances. This method might be worth considering for unsigned integers.
This is true; but at the same time, on any machine where it matters,
any optimizing compiler worthy of the word "optimizing" should turn:
x % constant
into:
x & (constant  1)
whenever the given constant is a power of two, because these always
produce the same result (for an unsigned x).
For signed integers (and still poweroftwo constants), the process
is a bit more difficult  a typical two's complement signed integer
gives different answers for "x % CONST" and "x & (CONST1)" when
x is negative. There are bittwiddling tricks that can be used if
the value is required, though; and if only the "truthness" of the
value is of interest, the above transform again works. That is:
if (x % 8)
and:
if (x & 7)
are both true (or false) in the same sets of cases, even when x is
signed, as long as the machine uses two's complement. This allows
an optimizing compiler to "do the right thing".

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Mark Henning wrote: Rogério Brito wrote: No. That is only possible if the representation of the integers use twoscomplement.
To see if a number is odd or even, use the modulus operator (e.g., x%2). That is the way that I have always done it, although it occurs to me that ANDing a value with one may be a significantly simpler computation than calculating the remainder of a divide by two in the majority of circumstances. This method might be worth considering for unsigned integers.
What fraction of your program's running time is expended
on determining whether a number is even or odd? One percent
seems high, but let's be generous and suppose your design calls
for a really large number of such determinations. All right,
how much faster might "and" be than "remainder?" Machine
specific, of course, but let's again be generous and suppose
"modulus" takes whatever it takes while "and" is infinitely
faster, taking zero time. Switching from "modulus" to "and"
would speed up your program by ...
<< May I have the envelope, please? >>
... a WHOPPING one percent! WOW!!!
If your other problems are so insignificant that something
this tiny becomes important, you are to be envied.
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On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran"
<a_******@hotmail.com> wrote: so my previous answer is valid/correct with all positive integers.
Unless, of course, there are padding bits in the integer. The only
correct way to test for mathematical even or odd is to use a mathematical
expression.

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Serve Laurijssen wrote: Some people prefer to use "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
NO.  
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Rogério Brito wrote: Serve Laurijssen wrote: Some people prefer to use "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
No. That is only possible if the representation of the integers use twoscomplement.
It's also possible with Sign and Magnitude representation.

pete  
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Kevin D. Quitt wrote: On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran" <a_******@hotmail.com> wrote:so my previous answer is valid/correct with all positive integers.
Unless, of course, there are padding bits in the integer.
Makes no difference if there are padding bits in the integer.
(x & 1) is true for all positive odd int x, regardless of padding.

pete  
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pete wrote: Rogério Brito wrote:
Serve Laurijssen wrote:
Some people prefer to use "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
No. That is only possible if the representation of the integers use twoscomplement.
It's also possible with Sign and Magnitude representation.
Wouldn't that be "signed magnitude"? Or are there some other
representations I'm missing out on?
(The ones I know of are signed magnitude, one's complement, and two's
complement.)

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On Tue, 6 Apr 2004, August Derleth wrote: pete wrote: Rogério Brito wrote: No. That is only possible if the representation of the integers use twoscomplement.
It's also possible with Sign and Magnitude representation.
Wouldn't that be "signed magnitude"? Or are there some other representations I'm missing out on?
(The ones I know of are signed magnitude, one's complement, and two's complement.)
No. The canonical phrase is "signmagnitude representation,"
which refers to the fact that SM representation is one in which
the [S]ign of the number is separated from the [M]agnitude. That
is, you have one bit for the sign and the rest for the magnitude,
unlike two'scomplement or ones'complement, in which the sign is
sort of "tied up with" the magnitude in an icky way. ;)
I believe I've got the apostrophes in the right places above.
Knuth, IIRC, says in TAOCP why he writes the apostrophes where he
does. Ones' complement, plural possessive, because you're XORing
the number with a bunch of ones (111111...) to negate it. Two's
complement, singular possessive, because you're doing something or
other with a power of two.
You're not missing any representation methods allowed by the C
standard, although I'm sure there are many more outlandish ones
out there.
HTH,
Arthur  
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On Tue, 6 Apr 2004 13:39:08 +0200, "Ahmed S. Badran"
<a_******@hotmail.com> wrote: "Serve Laurijssen" <cs@nospam.comp.com> wrote in message news:eJucc.5394$RU5.83970@zonnetreader1... "if (x & 1)" to see if a number is odd or even. Is this completely portable according to the standard?
I don't know what does this have to do with the standard, but the thing is an odd number will ALWAYS have bit 0 set to '1' and an even number will always have bit 0 set to '0'. This is a matter of binary representation.
There are systems where bit 0 is the high order or sign bit, not the
low order one.
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On Tue, 06 Apr 2004 23:01:26 GMT, pete <pf*****@mindspring.com> wrote: Kevin D. Quitt wrote: On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran" <a_******@hotmail.com> wrote: >so my previous answer is >valid/correct with all positive integers.
Unless, of course, there are padding bits in the integer.
Makes no difference if there are padding bits in the integer. (x & 1) is true for all positive odd int x, regardless of padding.
According to what?

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Kevin D. Quitt wrote: On Tue, 06 Apr 2004 23:01:26 GMT, pete <pf*****@mindspring.com> wrote:
Kevin D. Quitt wrote: On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran" <a_******@hotmail.com> wrote: >so my previous answer is >valid/correct with all positive integers.
Unless, of course, there are padding bits in the integer.
Makes no difference if there are padding bits in the integer. (x & 1) is true for all positive odd int x, regardless of padding.
According to what?
According to the facts that if x is a positve int,
then the representation for x and the representation for 1,
have the same corresponding value bits and the same corresponding
sign bit and the same corresponding padding bits.
And the padding bits aren't part of the value
of either x or 1 or (x & 1).
How do you figure padding bits make a difference ?

pete  
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"Barry Schwarz" <sc******@deloz.net> wrote in message
news:c5**********@216.39.134.78... There are systems where bit 0 is the high order or sign bit, not the low order one.
ok, but what's the bit representation of 1 on such a system then? x & 1
could work on such a system too I'd say.  
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In article <Pi*********************************@unix50.andrew .cmu.edu>, "Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> writes: You're not missing any representation methods allowed by the C standard, although I'm sure there are many more outlandish ones out there.
There are nonoutlandish (inlandish?) integer representations which
are not allowed by the C standard, too, such as BCD and various
bignum representations. Or representing integers using a different
byte order than the machine's native one  quite common in COBOL
programs, to maintain binary compatibility with data files generated
by mainframe COBOL programs. The Standard mandates a "pure binary"
representation for good reason (consistent behavior across
implementations), but there are often good reasons to employ other
sorts of representations. It's not in the spirit of C to support
those directly (unlike COBOL, which has evolved by tacking on
features to handle whatever the problem of the moment is), which
is fine, but they're not rare.
Really, it's ones'complement that few people these days are likely
to encounter.

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On Thu, 8 Apr 2004 16:26:08 +0200, "Serve Laurijssen"
<cs@nospam.comp.com> wrote: "Barry Schwarz" <sc******@deloz.net> wrote in message news:c5**********@216.39.134.78... There are systems where bit 0 is the high order or sign bit, not the low order one.
ok, but what's the bit representation of 1 on such a system then? x & 1 could work on such a system too I'd say.
The bit representation is still normal binary. The only difference is
the nomenclature assigned to the bits, which is a detail C doesn't
address or need to.
Of course it works under the conditions others have identified. My
comment only addressed the point raised by Ahmed S. Badran that you
snipped which stated that bit 0 is the low order bit. It need not be.
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Barry Schwarz wrote: On Thu, 8 Apr 2004 16:26:08 +0200, "Serve Laurijssen" <cs@nospam.comp.com> wrote:
"Barry Schwarz" <sc******@deloz.net> wrote in message news:c5**********@216.39.134.78... There are systems where bit 0 is the high order or sign bit, not the low order one.
ok, but what's the bit representation of 1 on such a system then? x & 1 could work on such a system too I'd say.
The bit representation is still normal binary. The only difference is the nomenclature assigned to the bits, which is a detail C doesn't address or need to.
Of course it works under the conditions others have identified. My comment only addressed the point raised by Ahmed S. Badran that you snipped which stated that bit 0 is the low order bit. It need not be.
Bit 0 isn't mentioned in the standard.
The right bits are low order. The left bits are high order.

pete  
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 mw*****@newsguy.com (Michael Wojcik) wrote: In article <Pi*********************************@unix50.andrew .cmu.edu>, "Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> writes: You're not missing any representation methods allowed by the C standard, although I'm sure there are many more outlandish ones out there.
There are nonoutlandish (inlandish?) integer representations which are not allowed by the C standard, too, such as BCD and various bignum representations. Or representing integers using a different byte order than the machine's native one
How is this forbidden? I can't find it in 6.2.6.2.
Richard  
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"Chris Torek" <no****@torek.net> wrote in message
news:c4*********@news1.newsguy.com... In article <news:c4**********@taliesin2.netcom.net.uk> Mark Henning <ma*******@btopenworld.com> writes:[using % to obtain integer remainder after division] is the way that I have always [tested for even/odd], although it occurs to me that ANDing a value with one may be a significantly simpler computation than calculating the remainder of a divide by two in the majority of circumstances. This method might be worth considering for unsigned integers.
This is true; but at the same time, on any machine where it matters, any optimizing compiler worthy of the word "optimizing" should turn:
x % constant into: x & (constant  1)
whenever the given constant is a power of two, because these always produce the same result (for an unsigned x).
For the record, I just tested gcc 2.91.66 for x86 and it strengthreduces (x
% 2) to (i & 1) even with no optimization enabled.
Thanks for the portability tip  I'd been "prematurely optimizing" this for
years.
S

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In article <40****************@news.individual.net>, rl*@hoekstrauitgeverij.nl (Richard Bos) writes: mw*****@newsguy.com (Michael Wojcik) wrote:
There are nonoutlandish (inlandish?) integer representations which are not allowed by the C standard, too, such as BCD and various bignum representations. Or representing integers using a different byte order than the machine's native one
How is this forbidden? I can't find it in 6.2.6.2.
It doesn't appear to be (though it could be argued that it would violate
the spirit of the "pure binary representation" requirement). I added
it in a later edit of the paragraph, without sufficient care. Thanks
for the catch.
BCD and some bignum representations are still good examples, though.

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