How do i calculate the physical address of any given interrupt (INT22H or INT15H for instance) in the interrupt vector table for 8086 microprocessor?
...calculate the physical address of any given interrupt (INT22H or INT15H for instance) in the interrupt vector table...
Physical address where the int 15h instruction finds the far pointer that it should call.
This is an offset within the Interrupt Vector Table, and so gives a physical address aka linear address from the list {0,4,8,12, ... , 1016,1020}.
Since each vector is 4 bytes long, all it takes is multiplying the interrupt number by 4.
mov ax,0415h ;AL=Interrupt number, AH=4
mul ah ; -> Product in AX
cwd ;(*) -> Result in DX:AX=[0,1023]
(*) I like all my linear addresses expressed as DX:AX. That's why I used the seemingly unnecessary cwd instruction.
Physical address where int 15h ultimately gets handled.
This can be anywhere in the 1MB memory. (On 8086 there's no memory beyond 1MB).
Each 4 byte vector consists of an offset word followed by a segment word. The order is important.
The linear address is calculated from multiplying the segment value by 16 and adding the offset value.
mov ax,16
mul word ptr [0015h * 4 + 2] ;Segment in high word -> Product in DX:AX
add ax, [0015h * 4] ;Offset in low word
adc dx, 0 ; -> Result in DX:AX=[0,1048575]