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> > But (a > a * b || b > a * b) should work, shouldn't it? > > No. For a=1 b=2 this will give the correct answer (no overflow), but > for a=0x6000000 b=64 it will give incorrect one (no overflow, while > 0x180000000LL certainly doesn't fit into 32-bits (but 0x80000000 is > still bigger than any of the operands). Ok, if we're going to have two comparisions anyway, I'd suggest we assume at least 32bits and use a >= 46340 || b >= 46340 (46340 <= sqrt(2^31), if I did my math correctly) Of course this will detect some cases as overflow which actually aren't, but that is harmless. Regards, Wolfram.
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