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bdonlan
on Feb 9, 2019
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Faster remainders when the divisor is a constant: ...
The compiler has to assume that your operands might be large enough that they would overflow after a straight multiply.
saagarjha
on Feb 9, 2019
[–]
As acqq/I have mentioned below, there are ways of telling the compiler that the overflow can't multiply so it can freely optimize these away.
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