fix 2^e in c99 bitwise shift

diff --git a/n1256.html b/n1256.html
index 5feddf5..a07b492 100644 (file)
--- 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.
 
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