I want to divide two integers by IDIV, so my operand is DWORD. The manual says that my operand must then be a 64 bit value. However I didn't really understand the system, so I'm confused about dividing two signed integer values.

For example, if ECX is 1000 and EDX is 50
How can I perform ECX / EDX?

Then, I have another question...
There are unsigned and signed integers, right?
Unsigned integers can't hold a - sign, so they are always positive, is that right? So I wonder how signed integers store their sign, and how the range of the value gets decreased.
I think that a signed integer can't hold such a big value as an unsigned integer?
Could you explain me?

I would be glad about answers, thank you

Regarding dividing two integers using IDIV:
The source operand is, indeed, 64-bit, but that does not mean you have to use all of them. The upper half of the 64-bit QWORD divident is stored in EDX, and the lower in EAX. So simply put your integer in EAX, and use CDQ to extend it to EDX (if EAX is negative, EDX will be FFFFFFFF, if EAX is positive, EDX will be 0). IDIV will then divide EDX:EAX by whatever operand register you give it.
So if you want to divide 1000 / 50, you would do the following:


Regarding signed/unsigned integers
An unsigned integer is normally 32-bits long, and all these bits are used to represent the value of the integer. A signed integer would take away one of these bits, and use it to describe the sign (positive or negative) of the value. That way, only 31-bits are left for the value, so the range of a signed integer is twice as small as of its unsigned counterpart. The sign bit (usually the 32nd bit, the most significant bit, the leftmost bit) is 0 if the integer is positive, and 1 if the integer is negative.

thank you very much for your fast and clear response. Now I understood that.

You might also want to read Paul Carter's Tut ... www.drpaulcarter.com ... look at Chapter 2.1