Operation C car B A

$------------------------------------------------------------$

$($Start$)00000000011010011$

Shift C left, Shift B left, Add A $00000011110100011$

Shift C left, Shift B left, Add A $00001001101000011$

Shift C left, Shift B left $00010010010000011$

Shift C left, Shift B left, Add A $00100111100000011$

The result, $00100111$ is $32+7=39,$ which is what you get with $3\backslash$times $13.$ As you can see

by the above, when the shift of B sets the carry bit, then one adds A$,$ otherwise one only

completes the shifts. Regardless of the numbers A and B$,$ the process always takes exactly $4$

iterations.

Your task is to translate the above algorithm into a functional code block that would allow a

multiplication of any $4-$bit integers $(0-15).$ You will also need a loop counter to keep track of

how many iterations you complete. Note $1$: Clear the carry bit before you start the process.

Note $2$: Where does B have to sit in an $8-$bit register for the shifts to work properly? Hint:

what does the command SWAPF do$?$

Alternate algorithm? Notice that the contents of register B are destroyed. Also notice that

those bits must live in the top nybble of the $8-$bit register that is shifted left. One could use

a single register to hold the result C $($which grows from the right$)$ and the operand B $($which

is gobbled$-$up at the left$)$ as each shift is carried out. Thus: one less variable and one less

shift per iteration.