On 01/27/2011 10:57 AM, Paul Eggert wrote: >>>> # define TYPE_MAXIMUM(t) \ >>>> ((t) (! TYPE_SIGNED (t) \ >>>> ? (t) -1 \ >>>> : ~ (~ (t) 0 << (sizeof (t) * CHAR_BIT - 1)))) >> The last line of this macro has UB due to signed integer overflow in >> the << operation. > > No it doesn't. ~ (t) 0 evaluates to -1, and -1 << 31 does not > overflow.
C99 states this (6.5.7 paragraph 4) The result of E1 << 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 × 2^E2 , reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined. In other words, the problem is not about overflow, but about undefined behavior. -- Eric Blake ebl...@redhat.com +1-801-349-2682 Libvirt virtualization library http://libvirt.org
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