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# odd/even bitwise and

 P: n/a 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? Nov 14 '05 #1
27 Replies

 P: n/a "Serve Laurijssen" wrote in message news:eJucc.5394\$RU5.83970@zonnet-reader-1... "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 Nov 14 '05 #2

 P: n/a 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. -- Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net) Available for consulting/temporary embedded and systems. USE worldnet address! Nov 14 '05 #3

 P: n/a 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 Nov 14 '05 #4

 P: n/a Ahmed S. Badran wrote: "Serve Laurijssen" wrote in message news:eJucc.5394\$RU5.83970@zonnet-reader-1... "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 Nov 14 '05 #5

 P: n/a 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. Nov 14 '05 #6

 P: n/a pete wrote: Ahmed S. Badran wrote: "Serve Laurijssen" wrote in message news:eJucc.5394\$RU5.83970@zonnet-reader-1... "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 re-elaborate 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 Nov 14 '05 #7

 P: n/a 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 twos-complement. 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 =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Nov 14 '05 #8

 P: n/a Rogério Brito wrote: No. That is only possible if the representation of the integers use twos-complement. 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 AND-ing 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. Nov 14 '05 #9

 P: n/a Mark Henning wrote: That is the way that I have always done it, although it occurs to me that AND-ing 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 twos-complement 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 =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Nov 14 '05 #10

 P: n/a In article Mark Henning writes:[using % to obtain integer remainder after division] is the way thatI have always [tested for even/odd], although it occurs to methat AND-ing a value with one may be a significantly simpler computationthan calculating the remainder of a divide by two in the majority ofcircumstances. This method might be worth considering for unsignedintegers. 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 power-of-two constants), the process is a bit more difficult -- a typical two's complement signed integer gives different answers for "x % CONST" and "x & (CONST-1)" when x is negative. There are bit-twiddling tricks that can be used if the value is required, though; and if only the "truth-ness" 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". -- In-Real-Life: Chris Torek, Wind River Systems Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603 email: forget about it http://web.torek.net/torek/index.html Reading email is like searching for food in the garbage, thanks to spammers. Nov 14 '05 #11

 P: n/a Mark Henning wrote: Rogério Brito wrote: No. That is only possible if the representation of the integers use twos-complement. 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 AND-ing 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. -- Er*********@sun.com Nov 14 '05 #12

 P: n/a On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran" wrote:so my previous answer isvalid/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. -- #include _ Kevin D Quitt USA 91387-4454 96.37% of all statistics are made up Per the FCA, this address may not be added to any commercial mail list Nov 14 '05 #13

 P: n/a 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. Nov 14 '05 #14

 P: n/a 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 twos-complement. It's also possible with Sign and Magnitude representation. -- pete Nov 14 '05 #15

 P: n/a Kevin D. Quitt wrote: On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran" wrote:so my previous answer isvalid/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 Nov 14 '05 #16

 P: n/a 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 usetwos-complement. 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.) -- yvoregnevna gjragl-guerr gjb-gubhfnaq guerr ng lnubb qbg pbz To email me, rot13 and convert spelled-out numbers to numeric form. "Makes hackers smile" makes hackers smile. Nov 14 '05 #17

 P: n/a On Tue, 6 Apr 2004, August Derleth wrote: pete wrote: Rogério Brito wrote:No. That is only possible if the representation of the integers usetwos-complement. 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 "sign-magnitude representation," which refers to the fact that S-M 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's-complement 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 Nov 14 '05 #18

 P: n/a On Tue, 6 Apr 2004 13:39:08 +0200, "Ahmed S. Badran" wrote: "Serve Laurijssen" wrote in messagenews:eJucc.5394\$RU5.83970@zonnet-reader-1... "if (x & 1)" to see if a number is odd or even. Is this completely portable accordingto the standard?I don't know what does this have to do with the standard, but the thing isan odd number will ALWAYS have bit 0 set to '1' and an even number willalways 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. <> Nov 14 '05 #19

 P: n/a On Tue, 06 Apr 2004 23:01:26 GMT, pete wrote: Kevin D. Quitt wrote: On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran" 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? -- #include _ Kevin D Quitt USA 91387-4454 96.37% of all statistics are made up Per the FCA, this address may not be added to any commercial mail list Nov 14 '05 #20

 P: n/a Kevin D. Quitt wrote: On Tue, 06 Apr 2004 23:01:26 GMT, pete wrote:Kevin D. Quitt wrote: On Tue, 6 Apr 2004 17:11:58 +0200, "Ahmed S. Badran" 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 Nov 14 '05 #21

 P: n/a "Barry Schwarz" 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. Nov 14 '05 #22

 P: n/a In article , "Arthur J. O'Dwyer" 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 non-outlandish (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. -- Michael Wojcik mi************@microfocus.com Pocket #9: A complete "artificial glen" with rocks, and artificial moon, and forester's station. Excellent for achieving the effect of the sublime without going out-of-doors. -- Joe Green Nov 14 '05 #23

 P: n/a On Thu, 8 Apr 2004 16:26:08 +0200, "Serve Laurijssen" wrote: "Barry Schwarz" wrote in messagenews: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 & 1could 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. <> Nov 14 '05 #24

 P: n/a Barry Schwarz wrote: On Thu, 8 Apr 2004 16:26:08 +0200, "Serve Laurijssen" wrote:"Barry Schwarz" wrote in messagenews: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 & 1could 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 Nov 14 '05 #25

 P: n/a mw*****@newsguy.com (Michael Wojcik) wrote: In article , "Arthur J. O'Dwyer" 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 non-outlandish (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 Nov 14 '05 #26

 P: n/a "Chris Torek" wrote in message news:c4*********@news1.newsguy.com... In article Mark Henning writes:[using % to obtain integer remainder after division] is the way thatI have always [tested for even/odd], although it occurs to methat AND-ing a value with one may be a significantly simpler computationthan calculating the remainder of a divide by two in the majority ofcircumstances. This method might be worth considering for unsignedintegers. 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 strength-reduces (x % 2) to (i & 1) even with no optimization enabled. Thanks for the portability tip -- I'd been "prematurely optimizing" this for years. S -- Stephen Sprunk "Stupid people surround themselves with smart CCIE #3723 people. Smart people surround themselves with K5SSS smart people who disagree with them." --Aaron Sorkin Nov 14 '05 #27

 P: n/a In article <40****************@news.individual.net>, rl*@hoekstra-uitgeverij.nl (Richard Bos) writes: mw*****@newsguy.com (Michael Wojcik) wrote: There are non-outlandish (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. -- Michael Wojcik mi************@microfocus.com Nov 14 '05 #28

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