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No, imul dest, source1[, source2] produces the result of the same size as the operands and discard the high bits, not in two registers like you said. Only [i]mul source will emit the full 2n.

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Will the overflow aex be written in the first. Mov bx, 5 mov cx, 10 mul cx

Is this possible in javascript?

However, i do not see the registers change when the mul function is called. I am trying to figure out how the imul and idiv instructions of the 8086 microprocessor work. I've searched online and read several so questions but couldn't get an.

Mov al, 2 imul byte [bytevariable] you can replace imul with mul, and the result would still be the. I need to convert strings to some form of hash.

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Mul and div are multiplications and division for unsigned.

I am trying to execute simple multiplication in assembly. And a full signed result is not generally equal to a full unsigned result.</p> An imul or imul , , is necessary but that would produce a full result!

I heard there is intel book online which describes the cpu cycles needed for a specific assembly instruction, but i can not find it out (after trying hard).

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