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# Integer dicision

 P: n/a How does (a/b) work when both 'a' and 'b' are pure integers ? >(9/2) 4 >(-9/2) -5 Why is it -5 ? I expect it to be -4 ? Because, in C/C++, 9/2 is 4 and so negative of it, (-9/2) is -4. What should I do to get C-like behavior ? Apr 11 '08 #1
5 Replies

 P: n/a bdsatish wrote: How does (a/b) work when both 'a' and 'b' are pure integers ? >>(9/2) 4 >>(-9/2) -5 Why is it -5 ? I expect it to be -4 ? Because, in C/C++, 9/2 is 4 and so negative of it, (-9/2) is -4. What should I do to get C-like behavior ? Use C? regards Steve -- Steve Holden +1 571 484 6266 +1 800 494 3119 Holden Web LLC http://www.holdenweb.com/ Apr 11 '08 #2

 P: n/a it rounds down. 4 is less than 4.5 and -5 is less than -4.5. On Thu, 2008-04-10 at 21:28 -0700, bdsatish wrote: How does (a/b) work when both 'a' and 'b' are pure integers ? (9/2) 4 (-9/2) -5 Why is it -5 ? I expect it to be -4 ? Because, in C/C++, 9/2 is 4 and so negative of it, (-9/2) is -4. What should I do to get C-like behavior ? Apr 11 '08 #3

 P: n/a On Apr 10, 9:28 pm, bdsatish >divmod(-9,2) (-5, 1) >>divmod(9,2) (4, 1) casevh Apr 11 '08 #4

 P: n/a On Apr 11, 6:06*am, casevh

 P: n/a bdsatish >(9/2) 4 >>(-9/2) -5 Why is it -5 ? I expect it to be -4 ? Because, in C/C++, 9/2 is 4 and so negative of it, (-9/2) is -4. Some background on the situation: Integer division and remainder operators have to satisfy two axioms: y * (x/y) + x%y = x and |x%y| < |y| (The former is just the definition of remainder, and the latter is necessary to get Euclid's algorithm to work.) When x and y are both nonnegative it's easy to agree on the right behaviour. When x or y is negative then we get conflicting requirements. On the one hand, you get people who expect that (-x)/y == -(x/y). On the other hand, you get people who expect 0 <= x%y < y if y >= 0. Unfortunately, you can't have both and still satisfy the integer- division axioms. C89 didn't specify which behaviour you got. C99 is in the first camp. Python picked the second (long before C99 came out). Which is right? I don't think that's actually a well-posed question: both have uses. I certainly find myself using the latter behaviour (floor, or round towards -infinity) almost exclusively, but then I'm mainly doing number theory and cryptography, and I find the bounds on the remainder very convenient. Heedful of this mess, Common Lisp provides four (!) different integer division functions (in addition to `/', which does exact rational division on integers): * (floor X Y) -Q R, where Q is the largest integer such that Q <= X/Y, and R = X - Q Y; R is also available as (mod X Y). * (ceiling X Y) -Q R, where Q is the smallest integer such that Q >= X/Y, and R = X - Q Y. * (truncate X Y) -Q R, where Q is the integer with the greatest magnitude such that Q has the same sign as X/Y (or is zero) and |Q| <= |X/Y|; again R = X - Q Y; R is also available as (rem X Y). * (round X Y) -Q R, where Q is the nearest integer to X/Y (rounding ties towards even numbers), and R = X - Q Y. Gluing all this into Python is tricky, partly because the plethora of options seems somewhat unPythonic (at least to me), and partly because it relies on Common Lisp's behaviour of throwing away unwanted additional return values. What should I do to get C-like behavior ? I would have said abs(x) / abs(y) * sgn(x) * sgn(y) but Python doesn't seem to have a signum function. :-( I'd recommend thinking carefully about your problem and seeing whether the existing floor-divide behaviour can't be made to fit (or indeed if it's not actually better anyway). If that still doesn't help then you'll have to solve the problem the hard way. def trunc_divmod(x, y): """ Return truncating quotient and remainder for X/Y. Assumes Y 0. """ q, r = divmod(x, y) if x < 0 and r != 0: r -= y q += 1 return q, r -- [mdw] Jun 27 '08 #6

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