J. Gareth Moreton
2018-06-12 00:07:37 UTC
Someone pointed out that if you perform "(x shl 8) shr 8", where x is of
type Word, the result is always x, even though logic dictates that the
upper 8 bits should be shifted out and hence the result actually be equal
to "x and $FF".Â After analysing the assembly language that's produced, it
turns out that the optimiser works with the full 32-bits of the register.Â
is_this_a_bug:=((counter shl 8) shr 8);
...becomes the following under -O3 (under -O1, %r13w and %si are replaced
shlÂ Â Â $0x8,%eax
shrÂ Â Â $0x8,%eax
mox Â Â %al,%si
A similar situation happens if the variables are of type Byte as well -
the intermediate values use the full 32 bits of the register.
I'm not certain if this is a bug or intended behaviour though.Â Can
someone more senior make a decision on this?