hi,
i have a big problem fromwhat i see,i have been given a task to make a code.i have to suppose that multiply operation is not available on 8086/8088 microprocessors and by using all of the rest operations make a code to multiply two numbers.





can any one have a idea how to do it.i tried it by adding any the series of numbers.
n please recommend a good book on digital communication systems.i have tried but proakis book is not available.
thanks

 

If you have two numbers, x and y, you can multiply these by adding x to itself y-times (or conversely add y to itself x-times).

Eg: 3 x 4 = > 3 + 3 + 3 + 3 => 4 + 4 + 4

Dave

 

The 8086 does have a multiply instruction.

In any number base you can multiply by shifting and adding. In binary it is actually fairly efficient and the exection time does not depend on the magnitude of either operand.

 

Page 7 in the tutorial at this link describes the IMUL instruction.

hgmjr

 

I think he means that he knows the multiply op exists on the 8086, but he can't USE it for the assignment. Adding or shifting/adding, as suggested, will work for this.

 

I think he means that he knows the multiply op exists on the 8086, but he can't USE it for the assignment. Adding or shifting/adding, as suggested, will work for this.

Your interpretation may be correct but in the introductory post of this thread the statement does not clearly state this to be the case.

hgmjr

 

i have to suppose that multiply operation is not available on 8086/8088 microprocessors

It doesn't?
Steve

 

Basically we're talking about a simple loop by X times. This is fairly basic if you approach it as such.

 

I think he means that he knows the multiply op exists on the 8086, but he can't USE it for the assignment. Adding or shifting/adding, as suggested, will work for this.

That is a charitable interpretation, but inconsistent with the OPs actual words.

 

I, for one, look forward to getting a clarification of the original question when she/he returns.

hgmjr

 

I think he means that he knows the multiply op exists on the 8086, but he can't USE it for the assignment. Adding or shifting/adding, as suggested, will work for this.

I must say, this is how I interpret the OP. The word that gestures towards this is "suppose" when referring to the availability of the multiply operation.

My interpretation is that this homework question is looking at using loading registers with values and using adding (and shifting) along with GOTO and BRANCH statements to implement a simple multiply operation.

Dave

 

originally posted by veritas:


I think he means that he knows the multiply op exists on the 8086, but he can't USE it for the assignment. Adding or shifting/adding, as suggested, will work for this.


yup u r right.i have to suppose for my assignment that its not available.n thankz to all ov u