On Thu, Oct 6, 2011 at 11:04 AM, Miles Bader <mi...@gnu.org> wrote: > How about: > > bool overflowbit2(unsigned int a, unsigned int b) > { > const unsigned int sum = a + b; > return ~(a ^ b) & sum & 0x80; > } > > ? > > I thinnnnk it has the same results as your function... > [I just made a table of all 8 possibilities, and checked!]
Miles, it is not the same. Take for example (0xff, 0xff). In 8-bit 2's complement, this is (-1, -1) and does not overflow. Your function says it does. Em 06-10-2011 12:23, Jeremy Hall escreveu: > bool overflow(int16_t a, int16_t b) > { > const int16_t sum = a + b; > return sum > INT8_MAX || sum < INT8_MIN; > } Jeremy, here you are ignoring the problem of converting from the unsigned int (in the range 0 to 0xff) to the signed integer that it represents in 8-bit two's complement. Example: 0xff -> -1. In practice, casting the unsigned int to int8_t works in most cases, but it is compiler-defined. We are trying to find a always well-defined approach that is efficient as well. > Ops, should have been > > return ~(a ^ b) & (a ^ sum) & 0x80 > > ~(a ^ b) gives 1 in the sign bit position if the signs are the same, > and (a ^ sum) gives 1 if it's different in the sum. This is good. Do you think this is suboptimal? How are you evaluating efficiency? In x86 this generates pretty small code. <overflow2>: 400524: 8d 04 3e lea (%rsi,%rdi,1),%eax 400527: 31 f8 xor %edi,%eax 400529: 31 f7 xor %esi,%edi 40052b: f7 d7 not %edi 40052d: 21 f8 and %edi,%eax 40052f: 25 80 00 00 00 and $0x80,%eax 400534: c3 retq -- Pedro Pedruzzi