问题
Can anyone explain for me why the sign of the remainder is different in these cases? Is this an emulator bug or do real CPUs do this, too?
8 / -3 : quotient(AL) = -2 remainder(AH) = 2
-8 / 3 : quotient(AL) = -2 remainder(AH) = -2
回答1:
It is supposed to work that way, though it is tricky to find out by reading the documentation:
Non-integral results are truncated (chopped) towards 0.
Combined with the "division law" X = dq + r (the dividend is the divisor times the quotient plus the remainder), we find that therefore the remainder r = X - d truncate(X / d)
This shows that the remainder depends on the sign of the dividend, but not on the sign of the divisor.
来源：https://stackoverflow.com/questions/53966486/x86-idiv-sign-of-remainder-depends-on-sign-of-dividend-for-8-3-and-8-3