Does anyone know how to multiply 16-bit by 16-bit in 8086 programming. what I currently know is that for example...

Ax= F000 Bx =9015 Dx= 0000

Mul Bx this result will come up to be F000 * 9015 = 8713B000

the the upper half (8713) will be stored in Dx, while the lower half (B000) will

be stored in Ax. My question is what if Dx has a given number let say

Dx = EB5.

what should be the answer?

 

Let's say that you take your calculator and store the result of multiplying 6 and 4 in the memory. What value will be stored there is the memory previously had the value 20 stored in it?

Same idea applies here.

 

I am not sure I get what youre asking. F000* EB5 = DC9 B000
So Dx becomes DC9 and Ax becomes B000.

 

I think he is thinking that the prior contents of Dx have an effect on the new values.

 

Does anyone know how to multiply 16-bit by 16-bit in 8086 programming. what I currently know is that for example...

Ax= F000 Bx =9015 Dx= 0000

Mul Bx this result will come up to be F000 * 9015 = 8713B000

the the upper half (8713) will be stored in Dx, while the lower half (B000) will

be stored in Ax. My question is what if Dx has a given number let say

Dx = EB5.

what should be the answer?

Whatever is in DX gets blown away. If you need to keep it push it onto the stack, pop it when you need to recall it. Pay attention to OF and CF as well when the MUL instruction runs.

 

DX gets replaced with the new value