From bae1b94d9e0a1d1e5264ce08be236efbfb649d60 Mon Sep 17 00:00:00 2001
From: Georg Brandl <georg@python.org>
Date: Sun, 10 Aug 2008 12:16:45 +0000
Subject: [PATCH] Remove long integer output.

---
 Doc/tutorial/floatingpoint.rst | 12 ++++++------
 1 file changed, 6 insertions(+), 6 deletions(-)

diff --git a/Doc/tutorial/floatingpoint.rst b/Doc/tutorial/floatingpoint.rst
index 150e8fb250e..c9408c958be 100644
--- a/Doc/tutorial/floatingpoint.rst
+++ b/Doc/tutorial/floatingpoint.rst
@@ -173,24 +173,24 @@ and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
 the best value for *N* is 56::
 
    >>> 2**52
-   4503599627370496L
+   4503599627370496
    >>> 2**53
-   9007199254740992L
+   9007199254740992
    >>> 2**56/10
-   7205759403792793L
+   7205759403792794.0
 
 That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.  The
 best possible value for *J* is then that quotient rounded::
 
    >>> q, r = divmod(2**56, 10)
    >>> r
-   6L
+   6
 
 Since the remainder is more than half of 10, the best approximation is obtained
 by rounding up::
 
    >>> q+1
-   7205759403792794L
+   7205759403792794
 
 Therefore the best possible approximation to 1/10 in 754 double precision is
 that over 2\*\*56, or ::
@@ -211,7 +211,7 @@ If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
 its 30 most significant decimal digits::
 
    >>> 7205759403792794 * 10**30 / 2**56
-   100000000000000005551115123125L
+   100000000000000005551115123125
 
 meaning that the exact number stored in the computer is approximately equal to
 the decimal value 0.100000000000000005551115123125.  Rounding that to 17