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A comment on Faster remainders when the divisor is a constant: beating compilers & libdivide. Parent comment

Re:Faster remainders when the divisor is a constant: beating... by bdonlan on Friday February 08, 2019 @08:24PM (#19120104)
The compiler has to assume that your operands might be large enough that they would overflow after a straight multiply.
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Re:Faster remainders when the divisor is a constant: beating... by saagarjha on Saturday February 09, 2019 @04:14AM (#19121463)
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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