How would the MUL operation on AX with Pi (3,1415..) in (8.8) format work in Assembly?

Question

Im new to Assembly and am learning Fix-point arithmetic.

AX is a 16-Bit register-

 MUL Pi  ; Multiplies EAX with Pi and stores result in EAX
 DIV 256 ; Divides EAX by 256 which equals the necessary right- shift for the 8,8 format

But I dont think it works like this.

Answer 1

A MUL instruction with a word sized operand multiplies AX with that operand and places its result in DX:AX . If both operands are in 8.8 format, the result will be in 16.16 format with the integral part in DX and the fractional part in AX . To convert this to 8.8 format, you take the contents of DL and AH like this:

MUL Pi        ; multiple by Pi, leaving the integral part in DX and the fraction in AX
MOV AL, AH    ; move the fractional part into place
MOV AH, DL    ; move the integral part into place

And everything should work out just fine. Note that the rounding might be off a little. You can correct for that like this:

MUL Pi        ; multiple by Pi, leaving the integral part in DX and the fraction in AX
ADD AX, 0080h ; apply rounding to AH
ADC DX, 0     ; apply carry if any
MOV AL, AH    ; move the fractional part into place
MOV AH, DL    ; move the integral part into place

This rounds the result correctly according to the principle

round(x) = ⌊x + 0.5⌋