Question

I need add two float point in 8086

    12.3 ---> 4144 CCCDh
    (AX,BX) = (4144h, CCCDh)

I need add any number this float point look like:

    (AX,BX) = (AX,BX) + 10h

if i do that answer is wrong.

     (AX,BX) + 10h == 4144 CCECh

but 23,3 didnt equal to 4144 CCECh

Can you help me? How i adding this two number?

Answer 1

Haven’t verified this (especially the offset in bp), but it should give some perspective.
It uses the ancient 8087 floating point instruction set.

All operations happen inbetween coprocessor stack and/or memory. Integers can be converted from 2’s complement representation using instruction FILD mem and in some cases there’s a built in addition instruction that adds integer (from memory) to FP register.

 push bp
 mov bp, sp
 push bx
 push ax
 push word ptr 10   ;  // decimal, not hex
 fld dword ptr [bp] ;   load float (just pushed from bx,ax)
 fiadd word ptr [bp-4] ;   add the integer in stack
 fst dword ptr [bp] ;   store result
 pop ax             
 pop ax             ; restore the high word of result
 pop bx             ; restore low word
 pop bp             ; restore frame pointer