## About DIV, IDIV and Overflows

The IDIV instruction is a divide operation. It is less popular than its counterpart DIV. The different between the two is that IDIV is for signed numbers wheareas DIV is for unsigned numbers. I guess the “i” in IDIV means Integer, thus implying a signed integer. Sometimes I still wonder why they didn’t name it SDIV, which is much readable and self explantory. But the name is not the real issue here. However, I would like to say that there is a difference between signed and unsigned division. Otherwise they wouldn’t have been two different instructions in the first place, right? :) What a smart ass… The reason it is necessary to have them both is because signed division is behaving differently than unsigned division. Looking at a finite string of bits (i.e, unsigned char) which has a value of -2 and trying to unsigned divide that by -1, will result in 0, since if we take a look at the numbers as unsigneds – 0xfe and 0xff. And naively asking how many times 0xff is contained inside 0xfe, will result in 0. Now that’s a shame because we would like to treat the division as signed. For that, the algorithm is a bit more complex. I am really not a Math guy. So I don’t wanna get into dirty details of how the signed division works. I will leave that algorithm for the BasicOps column of posts… Anyway, I can just say that if you have an unsigned division you can use it to do a signed division of the same operands size.

Some processors only have signed division instructions. So for doing an unsigned division, one might convert the operands to the next bigger size and then do the signed division. Which means the high half of the operand is zero, which makes the division work as expected.

With x86, luckily we don’t have to do some nifty tricks, we have them straight away, DIV and IDIV, for our use. Unlike multiplication, when there is an overflow in division, a division overflow will be raised, wheareas in multiplication only the CF and OF flags will be set. If we like it or not this is the situation. Therefore it’s necessary to convert the numbers before doing the operation. Sign extension or zero extension (depending on the signedness of operands) and only then do the division operation.

What I really wanted to talk about is the way the overflow is detected by the processor. I am interested in that behavior since I write a simple x86 simulator as part of the diStorm3 project. So truly, my code is the “processor” or should I say the virtual machine…Anyhow, the Intel documentation for the IDIV instruction shows some psuedo algorithm:

temp  = AX / src; // Signed division by 8bit
if (temp > 0x7F) or (temp < 0x80)
// If a positive result is greater than 7FH or a negative result is less than 80H
then #DE; // Divide error

src is a register/immediate or a memory indirection, which results in a 8bits value that will be signed extended to 16bits and only then will be signed divided by AX. So far so good, nothing special.

Then comes some stupid looking if statement. Which on the first look says, that if temp is 0x7f or 0x80 then bam, raise the exception. So you ask yourself how these special values have anything to do with overflowing.

Reading on the next comment makes things clearer, since for 8bits input, the division is done on 16bits, and the result is stored inside 8bits that are signed values, the result can vary from -128 to 127. Thus, if the result is positive, and the value is above 127, there is an overflow, because then the value will be treated as a negative number, which is a no no. And same for negative results: if the result is negative and the value is below 128 there is an overflow. Since the negative number cannot be represented in 8bits and as a signed number.

It is vital to understand that overflow means that a resulting value cannot be stored inside its destination because it’s too low or too big to be represented in that container. Don’t confuse it with carry [flag].

So how do we know if the result is positive or negative? If we take a look at temp as a byte sized, we can’t really know. But that’s why we got temp as 16bits. That extra half of temp (high byte) is really the hint for the sign of the whole value. If the high byte is 0xff, we know the result is negative, otherwise the result is positive. Well I’m not 100% accurate it, but let’s keep things simple for matter of conversation. Anyway, it is enough to examine the most significant bit of temp to know its sign. So let’s take a look at the if statement again now that we have more knowledge about the case.

if temp[15] == 0 and temp > 127 : raise overflow

Suddenly it makes sense, huh? Because we assure the number is positive (doesn’t have the sign bit set) and the result is yet higher than 127, and thus cannot be represented as a sign value in a 8bits container.

Now, let’s examine its counterpart guard for negative numbers:

if temp[15] == 1 and temp < 128: raise overflow

Ok, I tried to fool here. We have a problem. Remember that temp is 16bits long? It means that if, for example, the result of temp after the division is -1 (0xffff), our condition is still true and will raise an overflow exception, where the result is really valid (0xff represents -1 in 8bits as well). The problem origin is in the signed comparison. By now, you should understood that the first if statement for a positive number uses an unsigned comparison as well, although temp is a signed value.

We are left with one option since we are forced to use unsigned comparisons, (my virtual processor supports only unsigned comparisons), then we have to convert the signed 128 value into a 16bits unsigned value, which is 0xff80. As easy as that, just signed extend it…

So taking that value and putting it in its place we get the following if statement:

if temp[15] == 1 and temp < 0xff80: raise exception

We know by now that temp is being compared to as an unsigned number. Therefore, if the result was a negative number (must be above 0x8000) and yet it was below 0xff80, then we cannot represent that value in a 8bits signed container, and we have to raise the division error exception.

Eventually we want to merge both if statements to be one, sparing some basic boolean algebra, we end up with:

if (temp > 0x7f) && ((temp < 0x8000) || (temp > 0xff80)):

then raise exception…

### 3 Responses to “About DIV, IDIV and Overflows”

1. mxatone says:

I don’t think I saw an idiv lately as compiler always try to avoid div (even if we’ve got plenty in windows binaries). So I didn’t know about this kind of exeption raising, that very interresting. Thank you.

2. arkon says:

woah, i was wrong:
it’s
if (temp > 0x7f) && (temp < 0xff80): raise

:((

3. Tororo says:

“I guess the “i” in IDIV means Integer, thus implying a signed integer. Sometimes I still wonder why they didn’t name it SDIV, which is much readable and self explantory.”

I guess the “i” stands for sIgned. That’s pretty natural.