Number 117 fits into 7 bits (log2 117= 6,87036472 bits, so 7 bits).
Number 13 fits into 4 bits (log2 13= 3,700439718 bits, so 4 bits).
We can do log2 by using instruction bsr.
Select biggest from 4 and 7, we need one more bit to be sign bit. So, we can deal with 8 bits group.
The left most bit into N bits group it's a sign bit.
Values of dividend and divisor should be aligned. The left most bit 1 of dividend need be aligned with left most bit 1 of divisor. Both having same positive signal.
Transform -13 to +13 using N bits group.
117 01110101 dividend
13 00001101 divisor
Align both numbers, shift left most bit 1 of divisor to the left most bit 1 of dividend.
Divisor was shift left by 3 bits positions. We need this 3 value (align_position), to know how to stop division process.
01110101 dividend (remainder)
01101000 divisor
quotient=0
.repeat
quotient = quotient *2 ;shl
.if dividend >= divisor ;>=
dividend= dividend - divisor ;sub
inc quotient ;inc
.endif
divisor = divisor / 2 ;shr
align_position = align_position -1 ;dec
.until align_position != -1
align
117 01110101 dividend (remainder)
104 01101000 divisor
0 quotient
3 times to go (align_position)
Start process:
quotient=0 align_position=3
117 >= 104? yes, quotient *2, subtract, increase quotient, divisor/2, align_position-1 quotient=0*2+1 align_position=2
13 >= 52? no, quotient *2, divisor/2, align_position-1 quotient=2 align_position=1
13 >= 26? no, quotient *2, divisor/2, align_position-1 quotient=4 align_position=0
13 >= 13? yes, quotient *2, subtract, increase quotient, divisor/2, align_position-1 quotient=8+1 align_position=-1
0 (remainder)
Quotient=9, remainder(dividend) = 0. Because divisor was negative, we need transform quotient into negative.
+9=00001001
-9=11110110+1=11110111
