X86 IDIV sign of remainder depends on sign of dividend for 8/-3 and -8/3?

Question

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

Mydriasis

felixcloutier.com/x86/idiv – Chitarrone

is this a programming problem? – Anthropologist

@AkberIqbal : Yes, this is an assembly language programming question. – Gallice

Answer 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.

Pia

Thanks for your explaination, very easy to understand! – Mydriasis