What is signed division(idiv) instruction?

Question

In intel instruction, idiv(integer divsion) means signed division.
I got the result of idiv, but I don't quite understand the result.

- Example

0xffff0000 idiv 0xffff1100

- My wrong prediction
As long as I know, quotient should be 0, and remainder should be 0xffff0000 and because...

0xffff0000 / 0xffff1100 = 0  
0xffff0000 % 0xffff1100 = 0xffff0000

- However, the result was...
Before idiv

eax            0xffff0000            # dividend
esi            0xffff1100            # divisor
edx            0x0

After idiv

eax            0xfffeedcc            # quotient
edx            0x7400   29696        # remainder

- Question.
The result was value I couldn't expected.
Could someone explain about signed division(idiv)?

- Appended. Here's More information about idiv.
idiv uses the eax register as a source register.
As a result of execution, quotient is stored at eax, and remainder is stored at edx.

You say "8086 instruction" but then you use "eax" and "edx" which are not 8086 registers. Please clarify. Note also that the idiv instruction uses the (e)dx register as a source register, which you did not specify. — Raymond Chen, Jan 02 '19 at 03:57
@RaymondChen Thank you for your advice. I edited my question to be more clear. — Jiwon, Jan 02 '19 at 04:05
@Jiwon - the question still doesn't show what is in edx before idiv. — rcgldr, Jan 02 '19 at 04:06
Have you looked at something like [(idiv) IA-32 Assembly Language Reference Manual](https://docs.oracle.com/cd/E19455-01/806-3773/instructionset-44/)? — David C. Rankin, Jan 02 '19 at 04:18
Possible duplicate of [IDIV in assembly isn't giving me the wanted result (NASM)](https://stackoverflow.com/questions/37552056/idiv-in-assembly-isnt-giving-me-the-wanted-result-nasm) — Raymond Chen, Jan 02 '19 at 15:34

score 9 · Accepted Answer · edited Jan 02 '19 at 04:41

idiv divides edx:eax by the explicit source operand. See Intel's instruction manual entry.

Since edx is 0, edx:eax is a positive number. You are dividing 4294901760 by -61184, giving -70196 with a remainder of 29696.

Remember that both dividend (EDX:EAX) and divisor (ESI in your case) are interpreted as 2's complement signed numbers, so any bit-pattern with the high bit set is negative.

00000000ffff0000 = 4294901760
ffff1100 = -61184
fffeedcc = -70196
7400 = 29696

You should sign extend eax into edx using cdq before using idiv, instead of zero-extending by zeroing EDX.

However, that still won't give the results you were expecting, because -65536 divided by -61184 equals 1 with a remainder of -4352.

(A negative dividend and positive divisor would give a negative remainder: X86 IDIV sign of remainder depends on sign of dividend for 8/-3 and -8/3?)

What is signed division(idiv) instruction?

1 Answers1