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Repaired inaccuracies in the % docs. In particular, we don't (and can't)

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guarantee abs(x%y) < abs(y) in all cases when a float is involved.
math.fmod() should, though, so noted that too.

Bugfix candidate.  Someone should check the LaTeX here first, though.
This commit is contained in:
Tim Peters 2002-11-24 20:23:04 +00:00
parent 1babdfc48a
commit 5b21df4a5c
1 changed files with 14 additions and 3 deletions
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17
Doc/ref/ref5.tex
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@ -694,8 +694,19 @@ the \exception{ZeroDivisionError} exception. The arguments may be floating
point numbers, e.g., \code{3.14\%0.7} equals \code{0.34} (since
\code{3.14} equals \code{4*0.7 + 0.34}.) The modulo operator always
yields a result with the same sign as its second operand (or zero);
the absolute value of the result is strictly smaller than the second
operand.
the absolute value of the result is strictly smaller than the absolute
value of the second operand\footnote{
While \code{abs(x\%y) < abs(y)) is true mathematically, for
floats it may not be true numerically due to roundoff. For
example, and assuming a platform on which a Python float is an
IEEE 754 double-precision number, in order that \code{-1e-100 \% 1e100}
have the same sign as \code{1e100}, the computed result is
\code{-1e-100 + 1e100}, which is numerically exactly equal
to \code{1e100}. Function \function{fmod()} in the \module{math}
module returns a result whose sign matches the sign of the
first argument instead, and so returns \code{-1e-100} in this case.
Which approach is more appropriate depends on the application.
}.
\index{modulo}
The integer division and modulo operators are connected by the
@ -704,7 +715,7 @@ modulo are also connected with the built-in function \function{divmod()}:
\code{divmod(x, y) == (x/y, x\%y)}. These identities don't hold for
floating point numbers; there similar identities hold
approximately where \code{x/y} is replaced by \code{floor(x/y)}) or
\code{floor(x/y) - 1} (for floats),\footnote{
\code{floor(x/y) - 1}\footnote{
If x is very close to an exact integer multiple of y, it's
possible for \code{floor(x/y)} to be one larger than
\code{(x-x\%y)/y} due to rounding. In such cases, Python returns