Dennis Lee Bieber wrote: > The best answer is probably to be found from the definition of > divmod()
The divmod() function is one of those little delights that reminds me why I love Python, but I do not think it answers the question here. The definition of divmod() references the '%' operation, and not the other way around. http://docs.python.org/lib/built-in-funcs.html http://docs.python.org/ref/binary.html >>>> divmod(70, 6) > (11, 4) >>>> 6*11 + 4 > 70 >>>> divmod(-70, 6) > (-12, 2) >>>> 6 * -12 + 2 > -70 Also: >>> divmod(70, -6) (-12, -2) >>> -6*-12 + -2 70 >>> divmod(-70, -6) (11, -4) >>> -6*11 + -4 -70 > Or in general terms > > (a, b) = divmod(x, y) > x = y * a + b > IOWs, the definition of modulo, and the definition of integer division, > are linked... Right. Thus given integers x and y, y!=0, Python and C agree: x == (y * (x / y)) + (x % y) The disagreement is how integer division rounds. In C, integer division rounds toward zero. In Python, integer division rounds downward. -- --Bryan -- http://mail.python.org/mailman/listinfo/python-list
