From 0b7a1b9887c6d0b4bce83b50a62f872245a303b7 Mon Sep 17 00:00:00 2001
From: nsz <nsz@port70.net>
Date: Sun, 30 Oct 2011 01:29:07 +0200
Subject: [PATCH] fix 2<sup>e</sup> in c99 bitwise shift

---
 n1256.html | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/n1256.html b/n1256.html
index 5feddf5..a07b492 100644
--- a/n1256.html
+++ b/n1256.html
@@ -4748,14 +4748,14 @@ unsigned long long int
 <!--page 97 -->
 <p><!--para 4 -->
  The result of E1 &lt;&lt; E2 is E1 left-shifted E2 bit positions; vacated bits are filled with
- zeros. If E1 has an unsigned type, the value of the result is E1 x 2E2 , reduced modulo
+ zeros. If E1 has an unsigned type, the value of the result is E1 x 2<sup>E2</sup> , reduced modulo
  one more than the maximum value representable in the result type. If E1 has a signed
- type and nonnegative value, and E1 x 2E2 is representable in the result type, then that is
+ type and nonnegative value, and E1 x 2<sup>E2</sup> is representable in the result type, then that is
  the resulting value; otherwise, the behavior is undefined.
 <p><!--para 5 -->
  The result of E1 &gt;&gt; E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type
  or if E1 has a signed type and a nonnegative value, the value of the result is the integral
- part of the quotient of E1 / 2E2 . If E1 has a signed type and a negative value, the
+ part of the quotient of E1 / 2<sup>E2</sup> . If E1 has a signed type and a negative value, the
  resulting value is implementation-defined.
 
 <p><small><a href="#Contents">Contents</a></small>
-- 
2.20.1