John Fine wrote: > Andrew Haley wrote: > >> Christian Böhme wrote: >> >> >>> If, as already assumed elsewhere, the 32 bit machine in question >>> provided a 32 unsigned integer multiplication that produces a 64 >>> bit unsigned integer result, the compiler will _never_ emit code >>> using this instruction based on information withheld by a standard >>> conforming frontend that the type of the result _must_ be that >>> of the operands and precluding the conversion to a type large >>> enough such that the multiplication is closed. >>> >> >> Right: the compiler can use such an instruction, but it must discard >> the upper part of the result. > > I don't know if any optimizer does pay attention to this case, but the > standard can't prevent an optimizer from doing so. > > If you cast a 32bit value (int or unsigned int) to 64bit then > immediately multiply it by another 32bit value, That's not what the example did. > the standard meaning is > to do a 64bit by 64bit multiply and keep only the low 64 bits of the > 128bit result. Right. > But, if the machine has a 32bit by 32bit multiply that produces the > identical 64bit result, the optimizer is free to use it. Of course, by the usual "as if" rule. > The standard > may forget that each input to the multiply only has 32 significant > bits. But the optimizer can remember that. So it has two nominally > 64bit values to multiply, which each only have 32 significant bits. So > it can know that the full 64bit result of a 32bit multiply would match > the low 64bit result of a 64bit multiply. > > So I hope the "_never_" stated above is actually incorrect. It was correct, for the example as stated: -->8--- #include <stdint.h> // ... int64_t a; uint32_t b = 8; a = -(b * 2u); --->8--- Andrew.
